what is -2(3x+12y-5-17x-16y+4) simplified

Answer:

a) the output of mine #1 in 5 days

b) x1·v1 +x2·v2 = (240, 2824)

c) x1 = 544/315 ≈ 1.727; x2 = 1482/315 ≈ 4.705

Step-by-step explanation:

a) If v1 represents the production of mine #1 for 1 day, then 5v1 represents that mine's production for 5 days.

__

b) The production of each mine, multiplied by the number of days of production, adds together to give the total desired production:

x1·v1 + x2·v2 = (240, 2824)

__

c) Treating the vector components separately, the vector equation gives rise to two linear equations:

30x1 +40x2 = 240

600x1 + 380x2 = 2824

These can be solved by any of the usual methods. My favorite for numbers that are large or relatively prime is Cramer's rule and/or a graphing calculator. The above equations can be reduced to standard form to make the numbers slightly more manageable:

3x1 +4x2 = 24

150x1 +95x2 = 706

By Cramer's rule, ...

x1 = (4·706 -95·24)/(4·150 -95·3) = 544/315

x2 = (24·150 -706·3)/315 = 1482/315

The vector 5v1 represents the output of 5 days of operation at mine #1. The vector equation x1v1 + x2v2 = (240, 2824) gives the number of days each mine should operate to hit certain production targets. The specific solution is not given.

a) 5v1 would represent the output of 5 days of operation at mine #1. Specifically, it would mean that in 5 days, mine #1 produces 150 metric tons of copper and 3000 kilograms of silver.

b) The vector equation we need to represent the situation could look something like this: x1v1 + x2v2 = (240, 2824). Here, x1 and x2 represent the number of days each mine should operate and v1 and v2 are the vectors that represent the daily output of each mine. The solution to this vector equation would give the number of days each mine needs to operate in order to fit these production targets.

c) Since we are not supposed to solve the equation, we will simply write it again here for reference: x1v1 + x2v2 = (240, 2824).

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Answer:

  (c) 16/9·π²·r⁶

Step-by-step explanation:

The water displaced is equivalent to the volume of the sphere, given as

  V = (4/3)π·r³

The product of two identical displaced volumes will be ...

  V² = ((4/3)π·r³)² = (4/3)²·π²·r⁶

  = 16/9·π²·r⁶ . . . . . . matches choice (c)

[tex]v[/tex] starts at the point (-5, 0) and ends at the point (7, 2). It points in the same direction as the vector [tex]w[/tex] where

[tex]w=(7,2)-(-5,0)=(12,2)[/tex]

which starts at the origin and ends at (12, 2). Its direction [tex]\theta[/tex] is such that

[tex]\tan\theta=\dfrac2{12}=\dfrac16[/tex]

[tex]w[/tex] terminates in the first quadrant, so both [tex]w[/tex] and [tex]v[/tex] have direction

[tex]\theta=\tan^{-1}\dfrac16\approx9.5^\circ[/tex]

Answer:

You can view a kite as 4 triangles

Step-by-step explanation:

A geometric kite can easily be viewed as 4 triangles.  The formula to calculate the area of a kite (width x height)/2 is very similar to the one of a triangle (base x height)/2.

According to the formula to calculate the area of a kite, we would get:

(36 x 30)/2 = 540.

If we take the approach of using 4 triangles, we could imagine a shape formed by 4 triangles measuring 18 inches wide with a height of 15.

The area of each triangle would then be: (18 x 15)/2 = 135

If we multiply this 135 by 4... we get 540.

Answer:

Draw a vertical line to break the kite into two equal triangles with a base of 36 and a height of 15. Use the formula A = 1/2bh to find the area of each. The sum of the areas is the area of the kite.

Step-by-step explanation:

Answer:

The measure of DE is 12

Step-by-step explanation:

* Lets study the figure to solve the problem

- There are two intersected circles B and C

- BA is a radius of circle B and CE is a radius of circle C

- AD is a tangent to the two circles touch circle B in A and touch

 circle C in E

- The line center BC intersects the tangent AD at D

- There are two triangles in the figure Δ BAD and Δ CED

* Now lets solve the problem

∵ AD is a tangent to circles B and C

∵ BA and CE are radii

∴ BA ⊥ AD at A

∴ CE ⊥ AD at E

- Two lines perpendicular to the same line, then the two lines are

 parallel to each other

∴ BA // CE

- From the parallelism

∴ m∠ABD = m∠ECD ⇒ corresponding angles

∴ m∠BAD = m∠CED ⇒ corresponding angles

- In any two triangles if their angles are equal then the two triangles

 are similar

- In the two triangles BAD and CED

∴ m∠ABD = m∠ECD ⇒ proved

∴ m∠BAD = m∠CED ⇒ proved

∵ ∠D is a common angle of the two triangle

∴ The two triangle are similar

- There are equal ratios between their sides

∴ BA/CE =AD/ED = BD/CD

∵ BD = 50 , AD = 40 , CD = 15

∴ 40/ED = 50/15 ⇒ using cross multiplication

∴ ED(50) = 15(40)

∴ 50 ED = 600 ⇒ divide both sides by 50

∴ ED = 12

For this case we must simplify the following expression:

[tex]3-2y-1 + 5x ^ 2-7y + 7 + 4x ^ 2[/tex]

We combine similar terms, taking into account that equal signs are added and the same sign is placed, while different signs are subtracted and the sign of the major is placed.

[tex]3-1 + 7-2y-7y + 5x ^ 2 + 4x ^ 2 =\\9-9y + 9x ^ 2[/tex]

Answer:

[tex]9-9y + 9x ^ 2[/tex]

Answer:

Step-by-step explanation:

The quadratic formula of a quadratic equation

[tex]ax^2+bx+c=0\\\\\text{If}\ b^2-4ac>0\ \text{then the equation has two solutions}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

[tex]\text{If}\ b^2-4ac=0\ \text{then the equation has one solution}\ x=\dfrac{-b}{2a}[/tex]

[tex]\text{If}\ b^2-4ac<0\ \text{then the equation has no solution}[/tex]

We have:

[tex]x^2-3x-10=0\to a=1,\ b=-3,\ c=-10\\\\b^2-4ac=(-3)^2-4(1)(-10)=9+40=49>0\\\\x=\dfrac{-(-3)\pm\sqrt{49}}{2(1)}=\dfrac{3\pm7}{2}\\\\x=\dfrac{3-7}{2}=\dfrac{-4}{2}=-2\\or\\x=\dfrac{3+7}{2}=\dfrac{10}{2}=5[/tex]

It is a rational expression.

Answer: 2048

Step-by-step explanation:

Tn = arⁿ⁻¹

T6 = ar⁶⁻¹

T6 = ar⁵

T6 = 2*4⁵

T6= 2048

ANSWER

[tex]a_{6} = 2048[/tex]

EXPLANATION

The general term of a geometric sequence is given by,

[tex] a_{n} = a_{1} ( {r})^{n - 1} [/tex]

The first term of the geometric sequence is 2

[tex]a_{1} =2[/tex]

The common ratio is 4. This means r=5.

The nth term of the sequence is

[tex]a_{n} = 2 ( {4})^{n - 1} [/tex]

The 6th term is

[tex]a_{6} = 2 ( {4})^{6 - 1} [/tex]

[tex]a_{6} = 2 ( {4})^{5} [/tex]

[tex]a_{6} = 2048[/tex]

Answer:

807.8 in^2

Step-by-step explanation:

The total area of the box is the sum of the areas of all faces of the box. The top, bottom, front, and back faces are rectangles 18 in long. The end faces each consist of a rectangle and a triangle. We can compute the sum of these like this:

The areas of top, bottom, front, and back add up to be 18 inches wide by the length that is the perimeter of the end: 2·5in +2·8 in + 9.6 in = 35.8 in. That lateral area is ...

(18 in)(35.6 in) = 640.8 in^2

The area of the triangle on each end is equivalent to the area of a rectangle half as high, so we can compute the area of each end as ...

(9.6 in)(8.7 in) = 83.52 in^2

Then the total area is the lateral area plus the area of the two ends:

640.8 in^2 + 2·83.52 in^2 = 807.84 in^2 ≈ 807.8 in^2

Check the picture below.

Answer:

  200 ft²

Step-by-step explanation:

Each face is an isosceles right triangle with a hypotenuse of length 10 ft. The area of each of those triangles is

  A = 1/4·h² . . . . where the h in this formula is the hypotenuse length

So, the area of the four faces (the lateral area of the pyramid is 4 times this, or ...

  A = 4·1/4·(10 ft)² = 100 ft²

Of course, the base area is simply the area of the square base, the square of its side length:

  A = (10 ft)² = 100 ft²

So, the total area is the sum of the lateral area and the base area:

  total area = 100 ft² +100 ft² = 200 ft²

_____

If you think about this for a little bit, you will realize the pyramid must have zero height. That is, the slant height of a face is exactly the same as the distance from the center of an edge to the center of the base. "Not drawn to scale" is a good description.

Answer:

200 [tex]ft^{2}[/tex]

Step-by-step explanation:

Answer:

  about 10.24 ft

Step-by-step explanation:

The formula for the volume of a cylinder is ...

  V = πr²h . . . . where h is the height and r is the radius

The formula for the volume of a sphere is ...

  V = (4/3)πr³ = πr²·(4/3r) . . . . equivalent to a cylinder of height 4/3r

__

We have a cylinder of height 3d = 3(2r) = 6r. It has half a sphere on top, so the equivalent height of that is (1/2)·(4/3r) = 2/3r.

Then our total volume is equivalent to a cylinder with radius r and height (6 2/3)r = (20/3)r. That is, ...

  22,500 ft³ = πr²·(20/3)r = (20π/3)r³

Multiplying by the inverse of the coefficient of r³, then taking the cube root, we have ...

  r = ∛(22,500·3/(20π)) ft ≈ 10.24 ft

The radius of the silo should be about 10.24 feet.

Answer:

10.24  ft

Step-by-step explanation:

Standard form for linear equations is in the form ax + by = c. Thus, 2x + 4y = -3 is already in standard form.

Answer:

It is already in standard form

Step-by-step explanation:

Standard form is ax+by=c

2x=ax

4y=by

-3=c

Nothing needs to be changed

Answer:

$77.18

Step-by-step explanation:

Fill in your equation like this:

[tex]B=70(1+.05)^2[/tex] and

[tex]B=70(1.05)^2[/tex] and

[tex]B=70(1.1025)[/tex] so

B = $77.18

Answer:

  7500 meters

Step-by-step explanation:

Isabella walked 2 × 2.5 km = 5 km. Together, they walked ...

  2.5 km + 5 km = 7.5 km = 7.5×1000 m = 7500 m

Cole and Isabella walked 7500 meters altogether.

_____

"kilo-" is a prefix meaning "one thousand". So one kilometer is 1000 meters. Then 7.5 kilometers is 7.5 times 1000 meters, or 7500 meters.

is noteworthy that the leading term has a negative coefficient, meaning this parabola is opening downwards like a "camel hump", so it reaches a maximum point and then goes back down, and of course the maximum point is at its vertex.

[tex]\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ H(x)=\stackrel{\stackrel{a}{\downarrow }}{-\frac{1}{2}}x^2\stackrel{\stackrel{b}{\downarrow }}{+4}x\stackrel{\stackrel{c}{\downarrow }}{-5} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{4}{2\left( -\frac{1}{2} \right)}~~,~~-5-\cfrac{4^2}{4\left( -\frac{1}{2} \right)} \right)\implies \left( 4~,~-5+\cfrac{16}{2} \right)\implies (4~~,~~3)[/tex]

Answer:

(4, -3)

Step-by-step explanation:

I'm assuming that you meant  H(x) = -(1/2)x^2 + 4x - 5.  Here, the coefficients of this quadratic are a = -1/2, b = 4 and c = -5.

The axis of symmetry is x = -b/(2a).  This axis goes through the vertex.  Here the axis of symmetry is  x = -(4) / [ 2(-1/2) ], or x = 4.

Evaluating H(x) at x = 4 gives us the y value of the vertex.  It is:

H(4) = (-1/2)(4)^2 + 4(4) - 5, or H(4) = -8 + 16 - 5, or 3.

We know that this function must have a max because a is - and therefore the graph opens down.

The vertex and the maximum is (4, 3).

Answer:

That is a VERY famous math problem first published in 1637 by Pierre Fermat.  He said that in an equation of the type

x^n + y^n = z^n

you will ONLY find solutions when "n" is no greater than 2.  He said he had a proof but the margin in his note book was too small to fit it in.

(It is now believed he never had any such proof.)

Anyway, we can find an infinite number of solutions when n = 2

3^2 + 4^2 = 5^5

5^5 + 12^2 = 13^2

but you cannot find any solutions when n = 3 or higher.

https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem

Finally in 1995, (358 YEARS after Fermat first published this theorem), the British mathematician Andrew Wiles published his own proof of this theorem.

https://en.wikipedia.org/wiki/Andrew_Wiles

Step-by-step explanation:

Final answer:

Fermat's Last Theorem is a famous mathematical conjecture proposed by Pierre de Fermat in the 17th century, stating that there are no three positive integers a, b, and c that satisfy the equation a^n + b^n = c^n for any integer value of n greater than 2. The theorem remained unproven for over 350 years until Andrew Wiles presented a proof in 1994.

Explanation:

Fermat's Last Theorem is a famous mathematical conjecture proposed by Pierre de Fermat in the 17th century. The theorem states that there are no three positive integers a, b, and c that satisfy the equation a^n + b^n = c^n for any integer value of n greater than 2. In other words, there are no whole number solutions to this equation when n is greater than 2.

This conjecture remained unproven for over 350 years and became one of the most elusive problems in mathematics. However, in 1994, the British mathematician Andrew Wiles presented a proof of Fermat's Last Theorem, which is considered one of the most significant achievements in the history of mathematics.

Answer:

30 mm

Step-by-step explanation:

The volume of a rectangular prism is the product of its three dimensions. To find the missing dimension, divide the volume by the product of the two that are given:

(165000 mm^3)/((100 mm)(55 mm)) = 165000/5500 mm^3/mm^2 = 30 mm

Answer:

30mm

Step-by-step explanation:

30mm

Answer:

C

Step-by-step explanation:

If the hexagon is regular, that means that all of its sides are the same length, all of its interior angles are equal, as are the central angles formed by the triangles within in it. It is one of these triangles that we are concerned about.

If the regular hexagon is inscribed in a circle with a radius of 6, that means that the radius of the hexagon is also 6. The radii of the regular hexagon start at its center and go to each one of the 6 pointy ends (vertices). There are 6 sides so that means that there are 6 radii. That also means that each pair of radii create a triangle. There are 6 triangles inside this hexagon, and all of them are congruent. Because there are 6 central angles and because the degree measure around the outside of a circle is 360 degrees, we can find the vertex angle of each one of these 6 triangles by dividing 360 by 6 to get 60 degrees. The Isosceles Triangle Theorem tells us that if two sides of a triangle are congruent, then the angles opposite those congruent sides are also congruent. So 180 - 60 (the vertex angle) = 120, and 120 divided in half is 60. So this is an equilateral/equiangular triangle. Since all the angles measure 60, that means that all the sides measure the same, as well. So they all measure 6 cm. That gives us that one side of the hexagon measures 6 cm and we will need that for the formula for the area of said hexagon. The altitude of one of those equilateral triangles serves as the apothem that we also need for the area of the hexagon. If we split one of those triangles in half at the altitude, the base will measure 3 and the vertex angle will measure 30 degrees. In the Pythagorean triple for a 30-60-90, the side across from the 30 angle measures x, the side across from the 60 angle measure x times the square root of 3, and the hypotenuse measure 2x. That means that the apothem (which is the altitude of this triangle) is the length across from the 60 angle. So if x measures 3, then the side across from the 60 measures [tex]3\sqrt{3}[/tex]

The formula for the area of a regular polygon is

[tex]A=\frac{1}{2}ap[/tex]

where a is the apothem and p is the perimeter around the hexagon. We found one side to be 6 cm, so 6 times the 6 sides of the hexagon is 36 cm. The apothem is [tex]3\sqrt{3}[/tex]

so putting it all together in our formula looks like this:

[tex]A=\frac{1}{2}(3\sqrt{3})(36)[/tex]

Do the math on that and you will get

[tex]A=54\sqrt{3} cm^2[/tex]

93.53 cm².

The area of a regular hexagon inscribed in a circle of radius 6 cm can be calculated using the formula for the area of a regular hexagon.

Calculate the area of one of the equilateral triangles formed by the hexagon inscribed in the circle.

Multiply the area of one triangle by 6 to find the total area of the regular hexagon.

In this case, the area of the regular hexagon is approximately 93.53 cm².

30.8% ≅ 31%. The probability in percent that a given graduate student is on financial aid is 31%.

The key to solve this problem is using the conditional probablity equation P(A|B) = P(A∩B)/P(B). Conditional probability is the probability of one event occurring with some relationship to one or more other events.

From the table we can see P(A∩B) which is the intersections of event A and event B, in this case the intersection is the amount of graduates students receiving financial aid 1879. Then P(A∩B) = 1879/10730.

From the table we can see P(B) which is the probability of the total of students  undergraduate and graduate and the total of the students of the university data. Then P(B) = 6101/10730

P(A|B) = (1879/10730)/(6101/10730) = 0.307982

Round to nearest thousandth 0.308

Multiplying by 100%, we obtain 30.8% or ≅ 31%

Answer:

• ΔCFB ~ ΔEDB by the AA similarity

• mCE = 46°

Step-by-step explanation:

No lengths are marked equal on the diagram, so we cannot assume any of the chords is the same length as any other. Then there is no evidence that the conditions for SAS congruence are met for the given triangles. Likewise, there is no evidence that arcs DE and CF are the same length, which they would have to be to have measure 108°.

The angles with vertices C and E subtend the same arc, so have equal measures. Likewise for the angles with vertices D and F. The angles CBF and EBD are vertical angles, so also congruent. Hence the two triangles are AA similar.

The angle labeled 72° is half the sum of the measures of arcs CE and DF, so we have ...

(CE + 98°)/2 = 72°

CE = 144° -98° = 46° . . . . . multiply by 2 and subtract 98°

Answer:

Answer:

• ΔCFB ~ ΔEDB by the AA similarity

• mCE = 46°

Step-by-step explanation:

Answer:

  see below

Step-by-step explanation:

Put the (x, y) values of point F into the equation for line AG and see if they work:

  y = (b/(a+c))x

For (x, y) = (a+c, b), this is ...

  b = (b/(a+c))(a+c) = b·(a+c)/(a+c) = b·1 = b . . . . . a true statement

Yes, F lies on line AG.

Answer:

D

Step-by-step explanation:

The surface area of the sphere is given by the equation

[tex]A=4\pi * r^{2}[/tex],

where A is the surface area and r is the radius.

We want to find the volume of the sphere, which is given by the equation

[tex]V = \frac{4}{3} * \pi * r^{3}[/tex],

where V is the volume and r is the radius.

Looking at these equations, we see that they both involve the sphere's radius. If we know what r is, we can calculate the volume.

We know that the sphere's surface area is [tex]36 \pi[/tex]. Plugging that in for A in the surface area equation, we get

[tex]36 \pi=4\pi * r^{2}[/tex], then divide by [tex]\pi[/tex]

[tex]36 = 4 * r^{2}[/tex], then divide by 4

[tex]r^{2} = 9[/tex], then take the square root of both sides

[tex]r = 3[/tex]

So the radius of the sphere is 3. Plugging this into the volume equation,

[tex]V = \frac{4}{3} * \pi * 3^{3}[/tex], simplify terms

[tex]V = \frac{4}{3} * \pi * 27[/tex], multiply [tex]\frac{4}{3}[/tex] by 27

[tex]V = 36 * \pi[/tex]

So the volume of the sphere is [tex]36\pi[/tex].

Answer:

36π ft^3

Step-by-step explanation:

The surface area (S) of a sphere can be defined as:

S = 4×π×r^2 = 36×π

Solve for r to get the radius of the sphere:

r = ([tex]\sqrt{36/4}[/tex] = 3

The voluem (V) of a sphere can be defined as:

V= (4/3)×π×r^3

The volume of the sphere is:

V = (4/3)×π×(3^3) = 36π ft^3

The volume of the sphere can be calculated from the surface area given.

ANSWER

C. 7

EXPLANATION

We want to evaluate

[tex]2 - ( - 8) + ( - 3) [/tex]

We use the order of operations PEDMAS.

Dealing with the parenthesis first,we have

[tex]2 - - 8+- 3[/tex]

Note that:

[tex] - - = + [/tex]

and

[tex] - + = - [/tex]

Our expression now becomes:

[tex]2 + 8 - 3[/tex]

Next, we add to get:

[tex]10 - 3[/tex]

We finally subtract to get,

[tex]7[/tex]

The correct answer is C

The answer is: C. 7

To solve the problem, first, we need to consider the signs out and inside of the parenthesis.

We must remember the following rule:

[tex]--=+\\+-=-[/tex]

We are given the following expression:

[tex]2-(-8)+(-3)[/tex]

Then, we can rewrite it using the rule of the signs, we have:

[tex]2--8+-3=2+8-3=2+8-3[/tex]

[tex]2+8-3=10-3=7[/tex]

Hence, the correct option is C. 7.

Have a nice day!

ANSWER

[tex]9(\pi - \frac{ \sqrt{3} }{2} )[/tex]

Approximately, A=20

EXPLANATION

The circle has radius r=3 units.

The height of the triangle is ,

[tex]h = 6 \cos(60 \degree) = 3[/tex]

The base of the triangle is

[tex]b = 6 \sin(60 \degree) = 3 \sqrt{3} [/tex]

The area of the triangle is

[tex] \frac{1}{2} bh[/tex]

[tex] = \frac{1}{2} \times 3 \sqrt{3} \times 3[/tex]

[tex] = \frac{9}{2} \sqrt{3} [/tex]

The area of the circle is

[tex]\pi {r}^{2} [/tex]

[tex] = {3}^{2} \pi[/tex]

[tex] = 9\pi[/tex]

The difference between the area of the circle and the triangle is

[tex]9\pi - \frac{9}{2} \sqrt{3} = 9(\pi - \frac{ \sqrt{3} }{2} )[/tex]

Answer:

Difference = 20.47 square  units

Step-by-step explanation:

Points to remember

Area of circle = πr²

Where r - Radius of circle

Area of triangle = bh/2

Where b - Base and h- Height

It is given a circle with radius 3 units

And a right angled triangle with angles 30, 60 and 90 and hypotenuse = 6 units

To find the area of circle

Here r = 3 units

Area =  πr²

 = 3.14 * 3 * 3

 = 28.26 square units

To find the area of triangle

Here sides are in the ratio Base : Height : hypotenuse = 1 : √3 : 2

 = Base :  Height : 6

 = 3 : 3√3 : 6

Base b = 3 and height h  = 3√3

Area = bh/2

 = (3 * 3√3)/2

 = 7.79 square units

To find the difference

Difference = 28.26 - 7.79

 = 20.47 square  units

Answer:

x=0, x=2

Step-by-step explanation:

Here is the graph: desmos.com/calculator/thpubranfo

As we can see, the two functions have two points of intersection.  (0, 3) and (2, -1).

If the value of the functions is the same. Then the value of x will be 0 and 2.

Functions are found all across mathematics and are required for the creation of complex relationships.

Graph the functions on the same coordinate plane.

f(x) = x² - 4x + 3

g(x) = -x² + 3

If the value of the functions is the same. Then the value of x will be

                       f(x) = g(x)

           x² - 4x + 3 = -x² + 3

x² - 4x + 3 + x² - 3 = 0

               2x² - 4x = 0

                x (x - 2) = 0

                          x = 0, 2

The graphs are given below.

Then the correct options are B and D.

More about the function link is given below.

https://brainly.com/question/5245372

#SPJ2

Answer:

Step-by-step explanation:

Let "a" represent the number of adult (highest price) tickets sold. Then 400-a is the number of student tickets, and the revenue is ...

  16a +10(400 -a) = 4600

  6a = 600 . . . . . . . . . . . . . . . simplify, subtract 4000

  a = 100 . . . . . . . . . . . . . . . . . divide by the coefficient of a

100 adult and 300 student tickets were sold.

_____

Note that the above can be described by the verbal reasoning: If all the tickets sold were the (lower price) student tickets, revenue would be $4000. It was actually $600 more than that. Each adult ticket sells for $6 more than a student ticket, so there must have been $600/$6 = 100 adult tickets sold.

_____

Another way to work this problem is as a "mixture" problem. The average selling price per ticket is $4600/400 = $11.50. The differences between this price and the adult and student ticket prices are 4.50 and 1.50, so the ratio of student tickets to adult tickets is 4.50:1.50 = 3:1. That is, there were 300 student tickets sold and 100 adult tickets sold.

Answer:

(f+g)(x) = x^2 +10x +7

Step-by-step explanation:

(f+g)(x) = f(x) +g(x) = (15x +7) +(x^2 -5x)

= x^2 +x(15 -5) +7

= x^2 +10x +7

Answer: 4) 0.8/(1/6)=4.8 miles per hour

6) 100%-16%=84%

500*0.84=420 mL

Step-by-step explanation:

4) rate=miles/hour