I do not understand how to do this problem please help!

6. Below is a graph that represents the total profits for a third month. Write the equation of the line that represents this graph. Show your work or explain how you determined the equations.

The first month they made 1200 and the second month they made 1500...

graph a line using the slope-intercept form:
Manipulate the equation into slope-intercept form, y = mx + b.
Identify and plot the y-intercept of the line. Remember, the y-intercept is b. Don't forget: The sign goes with the number.
Identify and use the slope of the line to find a second point. Remember, the slope is m. Don't forget: Starting at the y-intercept, the numerator tells you the rise (count up if positive and down if negative), the denominator tells you the run (count right.)
Draw a straight line through the two points to complete the graph.
Linear Functions:
A linear function is a function that is defined by a linear equation.
To write a linear function with function notation, first write it in slope-intercept form, and then replace y with f(x).
The graph of a linear function is all points (x, f(x)), where x is in the domain of the function.

The height of the telephone pole is approximately 20.79 feet, calculated using tangent function with a 30° angle.

To find the height of the telephone pole, we can use trigonometry. We'll use the tangent function since we have the opposite side and the adjacent side of the right triangle formed by the person, the pole, and the ground.

Let [tex]\( h \)[/tex] be the height of the pole.

We have the tangent of the angle:

[tex]\[ \tan(30^\circ) = \frac{h}{36} \][/tex]

Now, we can solve for [tex]\( h \):[/tex]

[tex]\[ h = 36 \times \tan(30^\circ) \][/tex]

Let's calculate:

[tex]\[ h = 36 \times \tan(30^\circ) \]\[ h = 36 \times 0.5774 \] (rounded value of tan(30°))\[ h \approx 20.7864 \][/tex]

So, the height of the telephone pole is approximately 20.79 feet.

Final answer:

The simplified form of the expression (n⁵)(n²) is found by adding the exponents since the base is the same, resulting in n⁷.

Explanation:

The simplified form of the expression (n⁵)(n²) is achieved by applying the laws of exponents. Since both terms have a common base of n, when multiplying two exponentials with the same base, we simply add their exponents.

The addition of 5 and 2 (the exponents) provides n⁵⁺² which simplifies to n⁷.

Hence, the simplified form of (n⁵)(n²) is n⁷.

To write log8 3 as a logarithm of base 5, we can use the change of base formula: loga b = logc b / logc a. Applying this formula, we have log5 3 = log8 3 / log8 5. Therefore, the correct option is (log8)5/(log3)5.

The expression log83 represents the logarithm of 3 with base 8. To write it as a logarithm with base 5, we can use the change of base formula:

logab = logcb / logca

Applying this formula, we have:

log53 = log83 / log85

Therefore, the correct option is (log8)5/(log3)5.

Answer:

Part 4) Right triangle

Part 5) Kite

Step-by-step explanation:

Part 4) What kind of triangle is made by connecting the points A(0, –6), B(3, –6), and C(3, –2)?

Using a graphing tool    

see the attached figure N [tex]1[/tex]

The triangle of the figure is not equilateral------> The triangle does not have three equal sides

The triangle of the figure is a right triangle------>The triangle  has an angle of [tex]90\°[/tex]

The triangle of the figure is not isosceles------> The triangle does not have two equal sides

The triangle of the figure is not a right and isosceles

Part 5) What type of quadrilateral is formed by connecting the points [tex](0, 9), (3, 6), (0, 1), and (-3, 6)[/tex]?                

Using a graphing tool    

see the attached figure N [tex]2[/tex]        

The figure is not a rhombus------> All sides are not congruent  

The figure is not a trapezoid-----> has not parallel sides

The figure is a kite------> Two disjoint pairs of consecutive sides are congruent and the diagonals meet at a right angle

Answer:      U7L7

Polygons in the coordinate plane

1.D

2.A

3.D

4.B

5.C

Step-by-step explanation:

The final equation is vf = [tex]\frac{(m1v1 + m2v2)}{(m1 + m2)}[/tex]

The given equation represents the conservation of momentum, which can be written as :

m1v1 + m2v2 = (m1 + m2) vf

To solve for vf, follow these steps :

Both students are correct because polynomials can be grouped in different ways to factor. Both ways result in a common binomial factor between the groups. Using the distributive property , this common binomial term can be factored out. Each grouping results in the same two binomial factors.

Answer:

1

Step-by-step explanation:

6 Is rounded to 10 like 0.6 is rounded to 1

Answer:

Its D) 8t + 16 = 200

Answer:  The correct option is (d) [tex]8t+16=200.[/tex]

Step-by-step explanation:  Given that Benito is selling T-shirts for $8 each for his school fund-raiser and he has sold 16 T-shirts till now.

We are to find the number of T-shirts that he still need to sell to reach his goal of $200 in sales.

The number of T-shirts still he needs to sell is represented by t.

The, according to the given information, we have

[tex]8(t+16)=200\\\\\Rightarrow 8t+128=200\\\\\Rightarrow 8t=200-128\\\\\Rightarrow 8t=72\\\\\Rightarrow t=\dfrac{72}{8}\\\\\Rightarrow t=9.[/tex]

So, the number of T-shirts that he still needs to sell is 9.

Now, we can see from the above steps of solution for t that options (a), (b) and (c) gives the correct value of t, but from option (d), we get

[tex]8t+16=200\\\\\Rightarrow 8t=200-16\\\\\Rightarrow 8t=184\\\\\Rightarrow t=23\neq 9[/tex]

So, option (d) does NOT solve for the correct value of t.

Thus, (d) is the correct option.

we know that

The standard form of the equation of the line is

[tex] Ax + By = C [/tex]

we have

[tex] y = \frac{1}{6}x + 4 [/tex]

Multiply by [tex] 6 [/tex] both sides

[tex] 6y = x+24 [/tex]

Subtract x both sides

[tex] -x+6y = 24 [/tex]

therefore

the answer is

the equation of the line in standard form is equal to

[tex] -x+6y = 24 [/tex]

You made a typo in your question.

The equation should be y = 2x - 3.

y = 2(-1) - 3

y = -2 - 3

y = - 5

The trains will meet at 6:40 P.M. after traveling at different speeds towards each other.

The trains will meet at 6:40 P.M.

To calculate the time when the trains will meet, we need to determine the time it takes for the second train to cover the distance that the first train has already covered.

The first train covers the 200-mile distance in 200 / 75 = 2.67 hours, which is 2 hours and 40 minutes.

The second train leaves at 5:00 P.M., so adding 2 hours and 40 minutes gives us 7:40 P.M. as the time the second train reaches the meeting point.

By then, the first train has been traveling for 3 hours and 40 minutes, making the meeting time 6:40 P.M.

Answer:

Question 1). Option C.

Question 2) Option C.

Question 3) Option D.

Step-by-step explanation:

Question 1), A. 7 - (9x + 3) = -9x -4

7 - 9x - 3 = -9x - 4

-9x + 4 = -9x - 4

Left hand side(L.H.S.)≠ Right hand side(R.H.S.)

Therefore, it's not an identity

B). 6m - 7 = 7m + 5 -m

6m -7 = 6m + 5

Again L.H.S.≠R.H.S.

So, it's not an identity.

C). 10p + 6 - p = 12p - 3(p - 2)

9p + 6 = 12p - 3p + 6

9p + 6 = 9p + 6

L.H.S.=R.H.S.

Therefore, it's an identity.

D). 3y + 2 = 3y - 2

L.H.S. ≠ R.H.S.

Therefore, it's not an identity.

Question 2. Part A. 7v + 2 = 8v - 3

7v - 8v = -2 - 3

- v = - 5

v = 5

Part B. 3x - 5 = 3x + 8 - x

3x - 5 = 2x + 8

3x - 2x = 8 + 5

x = 13

Part C. 4y + 5 = 4y - 6

This equation has no solution.

Part D. 7z + 6 = -7z - 5

7z + 7z = -6 - 5

14z = -11

z = [tex]-\frac{11}{14}[/tex]

Question 3). 5 + 7x = 11 + 7x

This equation has same coefficient of variable x on both the sides of the equation.

Therefore, equation has no solution.

Option D. no solution is the correct option.

The correct option for different parts are as follows:

Part (1): [tex]\boxed{\bf option (c)}[/tex]

Part (2): [tex]\boxed{\bf option (c)}[/tex]

Part (3): [tex]\boxed{\bf option (d)}[/tex]

Further explanation:

Part (1):

Option (a)

Here, the equation is [tex]7-(9x+3)=-9x-4[/tex].

Now, solve the above equation as follows:

[tex]\begin{aligned}7-(9x+3)&\ _{=}^{?}-9x-4\\7-9x-3&\ _{=}^{?}-9x-4\\4-9x&\neq-9x-4\end{aligned}[/tex]  

Here, left hand side (LHS) is not equal to right hand side (RHS).

Therefore, the given equation is not an identity.

This implies that option (a) is incorrect.

Option (b)

Here, the equation is [tex]6m-5=7m+5-m[/tex].

Now, solve the above equation as follows:

[tex]\begin{aligned}6m-5\ &_{=}^{?}\ 7m+5-m\\6m-5 &\neq6m+5\end{aligned}[/tex]  

Here, left hand side (LHS) is not equal to right hand side (RHS).

Therefore, the given equation is not the identity.

This implies that option (b) is incorrect.

Option (c)

Here, the equation is [tex]10p+6-p=12p-3(p-2)[/tex].

Now, solve the above equation as follows:

[tex]\begin{aligned}10p+6-p\ &_{=}^{?}\ 12p-3(p-2)\\9p+6\ &_{=}^{?}\ 12p-3p+6\\9p+6&\neq9p+6\end{aligned}[/tex]  

Here, left hand side (LHS) is equal to right hand side (RHS).

Therefore, the given equation is an identity.

This implies that option (c) is correct.

Option (d)

Here, the equation is [tex]3y+2=3y-2[/tex].

Now, the above equation is as follows:

[tex]3y+2\neq3y-2[/tex]  

Here, left hand side (LHS) is not equal to right hand side (RHS).

Therefore, the given equation is not an identity.

This implies that option (d) is incorrect.

Therefore, equation in option (c) is an identity.  

Part (2):

Option (a)

Here, the equation is [tex]7v+2=8v-3[/tex].

Now, solve the above equation as follows:

[tex]\begin{aligned}7v+2&=8v-3\\7v-8v&=-2-3\\-v&=-5\\v&=5\end{aligned}[/tex]  

Thus, the value of [tex]v[/tex]is [tex]5[/tex].

Therefore, the given equation has a solution.

This implies that option (a) is incorrect.

Option (b)

Here, the equation is [tex]3x-5=3x+8-x[/tex].

Now, solve the above equation as follows:

[tex]\begin{aligned}3x-5&=3x+8-x\\3x-3x+x&=8+5\\x&=13\end{aligned}[/tex]  

Thus, the value of [tex]x[/tex] is [tex]5[/tex].

Therefore, the given equation has a solution.

This implies that option (b) is incorrect.

Option (c)

Here, the equation is [tex]4y+5=4y-6[/tex].

Now, solve the above equation as follows:

[tex]\begin{aligned}4y+5&=4y-6\\4y-4y&=-6-5\\0&\neq-11\end{aligned}[/tex]

Thus, the given equation has no solution.

This implies that option (c) is correct.

Option (d)

Here, the equation is [tex]7z+6=-7z-5[/tex].

Now, solve the above equation as follows:

[tex]\begin{aligned}7z+6&=-7z-5\\7z+7z&=-5-6\\14z&=-11\\z&=-\dfrac{11}{14}\end{aligned}[/tex]

Thus, the value of [tex]z[/tex] is [tex]-\frac{11}{14}[/tex].

Therefore, the given equation has a solution.

This implies that option (d) is incorrect.

Therefore, equation in option (c) does not have solution.

Part (3):

The equation is [tex]5+7x=11+7x[/tex].

Solve the above equation as follows:

[tex]\begin{aligned}5+7x&=11+7x\\7x-7x&=11-5\\0&\neq6\end{aligned}[/tex]

Therefore, the given equation has no solution.

Option (a)

Here, the value of [tex]x[/tex] is [tex]0[/tex].

But the equation [tex]5+7x=11+7x[/tex] has no solution.

So, option (a) is incorrect.

Option (b)

Here, the value of [tex]x[/tex] is [tex]14[/tex].

But the equation [tex]5+7x=11+7x[/tex] has no solution.

So, option (b) is incorrect.

Option (c)

In option (c) it is given that there are infinite number of solutions.

But the equation [tex]5+7x=11+7x[/tex] has no solution.

So, option (c) is incorrect.

Option (d)

In option (d) it is given that the solution does not exist.

As per our calculation the equation [tex]5+7x=11+7x[/tex] does not have any solution.

So, option (d) is correct.

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Answer details:

Grade: Middle school

Subject: Mathematics

Chapter: Linear equations

Keywords: Linear equations, linear equation in one variable, linear equation in two variable, slope of a line, equation of the line, function, real numbers, ordinates, abscissa, interval, open interval, closed intervals, semi-closed intervals, semi-open intervals, sets, range domain, codomain.

The range of the reflected function consists of these reflected y-values. The range of the reflected function is (-4, -2, 1).

The range of a function reflected over the x-axis can be determined by considering the y-values of the original domain. In this case, the given domain points are (3,4), (2,-1), and (-1,2). The reflected points will have the same x-coordinates but opposite y-coordinates. Therefore, the reflected range will be the set of y-values obtained by changing the signs of the y-values in the original domain.

The original domain points have y-values of 4, -1, and 2. When reflected over the x-axis, these y-values become -4, 1, and -2, respectively.

Thus, the range of the reflected function consists of these reflected y-values. The range of the reflected function is (-4, -2, 1).

Learn more about range and reflection in functions here:

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To find the arithmetic means in the given sequence, we fill in the missing numbers using the average of the numbers before and after them. The arithmetic means in the given sequence are 160, 175, and 190.

To find the arithmetic means in the given sequence, we need to fill in the missing numbers.

Given sequence: 145, __, __, __, 205.

The arithmetic mean is the average of two numbers. We can find the missing numbers by taking the average of the numbers before and after them.

First, let's find the missing number between 145 and 205. The average of 145 and 205 is 175. So, the missing numbers are:

145, 175, 175, 175, 205.

To find the last missing number, we can take the average of the two numbers after it: (175 + 205) / 2 = 190.

So, the arithmetic means in the given sequence are 160, 175, and 190.

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