Simplify square root of (1-cos)(1+cos)/cos^2

Answer:

6 to the tenth power over 7 to the sixth power

Step-by-step explanation:

Given phrase,

6 to the fifth power over 7 cubed all raised to the second power,

[tex]\implies (\frac{6^5}{7^3})^2[/tex]

By using [tex](a^m)^n=a^{mn}[/tex]

[tex]=\frac{6^{5\times 2}}{7^{3\times 2}}[/tex]

[tex]=\frac{6^{10}}{7^6}[/tex]

= 6 to the tenth power over 7 to the sixth power

Simplify the expressions

(6⁵/7³)² = 2143588816/117649

(6⁷/7¹⁰) = 279936/282475249

(6¹⁰/7⁶) = 60466176/117649

(6³/7) = 216/7

(12⁵/14³) = 90855/1001

To simplify the given expressions, we can calculate the numerical values and perform the necessary operations. Let's evaluate each expression:

(6⁵/7³)²:

First, calculate the numerator and denominator:

Numerator: 6⁵ = 6 × 6 × 6 × 6 × 6 = 7776

Denominator: 7³ = 7 × 7 × 7 = 343

Now, substitute the values into the expression and square the result:

(7776/343)² = (7776/343) × (7776/343) = 2143588816/117649

The simplified form is 2143588816/117649.

(6⁷/7¹⁰):

Calculate the numerator and denominator:

Numerator: 6⁷ = 6 × 6 × 6 × 6 × 6 × 6 × 6 = 279936

Denominator: 7¹⁰ = 7 × 7 × 7 × 7 × 7 × 7 × 7 × 7 × 7 × 7 = 282475249

Substitute the values into the expression:

279936/282475249

This expression cannot be simplified further.

(6¹⁰/7⁶):

Calculate the numerator and denominator:

Numerator: 6¹⁰ = 6 × 6 × 6 × 6 × 6 × 6 × 6 × 6 × 6 × 6 = 60466176

Denominator: 7⁶ = 7 × 7 × 7 × 7 × 7 × 7 = 117649

Substitute the values into the expression:

60466176/117649

This expression cannot be simplified further.

(6³/7):

Calculate the numerator and denominator:

Numerator: 6³ = 6 × 6 × 6 = 216

Denominator: 7

Substitute the values into the expression:

216/7

This expression cannot be simplified further.

(12⁵/14³):

Calculate the numerator and denominator:

Numerator: 12⁵ = 12 × 12 × 12 × 12 × 12 = 248832

Denominator: 14³ = 14 × 14 × 14 = 2744

Substitute the values into the expression:

248832/2744 = 90855/1001

The simplified form is 90855/1001.

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Simplify 6 to the fifth power over 7 cubed all raised to the second power. 6 to the seventh power over 7 to the tenth power 6 to the tenth power over 7 to the sixth power 6 cubed over 7 12 to the fifth power over 14 cubed

(6⁵/7³)²

(6⁷/7¹⁰)

(6¹⁰/7⁶)

(6³/7)

(12⁵/14³)

The vertical asymptotes of the function y = 2cot(3x) + 4 are  A.x = π/3  C. x = 2π D.x = 0

The function is given as:

y = 2cot(3x) + 4

The above function is a cotangent function, represented as:

y = Acot(Bx +C) + D

By comparison, we have:

B = 3

The vertical asymptotes are then calculated using:

[tex]x = \frac{\pi}{B}n[/tex], where n are integers

Substitute 3 for B

[tex]x = \frac{\pi}{3}n[/tex]

Using the above format, the vertical asymptotes in the options are  A.x = π/3  C. x = 2π D.x = 0

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Answer:

AB overbar congruent to CD overbar

Explanation:

The question is asking whether Line segment AB is CONGRUENT to line segment CD.

The meaning of congruent is having the same shape and size.

Congruent ≅

Element ∈

Equal =

Similar ~

In conclusion, you could say:

AB ≅ CD

I cannot type the lines over the top AB and CD but they are there.

(I know this question is probably old, and i am also tying this so I remember as well, but the other answer didn't have a bit bigger explanation so if anyone comes across this i hope this helped. :)

If two or more objects are the same copy in length and shape then that will be said to be congruent thus AB overbar is congruent to CD overbar and AB overbar is equal to CD overbar are the equivalent thus options (B) and (C) is correct.

If two figures are exactly the same in sense of their length side all things then they will be congruent.

In other meaning, if you can copy a figure then that copy and the original figure will be congruent.

All line segments are in the same shape and have degrees as one in the equation therefore only one criterion which is length is needed to prove congruency.

So, congruent lines are lines whose lengths are the same.

The sign of congruency is ≅ so AB ≅ CD.

Hence " AB overbar is congruent to CD overbar and AB overbar is equal to CD overbar are the equivalent to AB ≅ CD".

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To find the interest Candis would make from a 15,400% interest rate on a $220 investment for one year, we calculate using the simple interest formula, resulting in $33,880.

The question asks us to determine how much interest Candis would make from a payday loan with an effective interest rate of 15,400% if she invested $220 for a year. To calculate the interest earned, we can use the formula for simple interest which is I = Prt, where I is interest, P is principal amount (the initial amount of money), r is the annual interest rate (in decimal form), and t is the time in years.

Converting 15,400% to a decimal, we get 154. Then, apply the formula:

I = $220 × 154 × 1

This gives us:

I = $33,880

Therefore, Candis would make $33,880 in interest after one year, which corresponds to option D.

The third derivative of the function is [tex]y^{'''}=\dfrac{\left(3y^2+12xy-x\right)y''+y'\left(6yy'+12xy'+12y-1\right)-\left(6y-1\right)y'-6yy'}{\left(3y^2+x\right)^2}-\dfrac{2\left(6yy'+1\right)\left(\left(3y^2+12xy-x\right)y'-y\cdot\left(6y-1\right)\right)}{\left(3y^2+x\right)^3}[/tex]and the value at the point x = 1 is 42

How to determine the third derivative at the point x = 1

From the question, we have the following parameters that can be used in our computation:

[tex]x^2 + xy + y^3 = 1[/tex]

Differentiate implicitly

So, we have

[tex]3y^2y' + xy'+y+2x=0[/tex]

Make y' the subject of formula

So, we get

[tex]y'=-\dfrac{y+2x}{3y^2+x}[/tex]

Differentiate the second time

Using a graphing tool, we have

[tex]y''=\dfrac{\left(3y^2+12xy-x\right)y'-6y^2+y}{\left(3y^2+x\right)^2}[/tex]

Differentiate the third time to get the third derivative

Using a graphing tool, we have

[tex]y^{'''}=\dfrac{\left(3y^2+12xy-x\right)y''+y'\left(6yy'+12xy'+12y-1\right)-\left(6y-1\right)y'-6yy'}{\left(3y^2+x\right)^2}-\dfrac{2\left(6yy'+1\right)\left(\left(3y^2+12xy-x\right)y'-y\cdot\left(6y-1\right)\right)}{\left(3y^2+x\right)^3}[/tex]

Recall that

x = 1

Calculating y, we have

[tex]1^2 + (1)y + y^3 = 1[/tex]

[tex]1 + y + y^3 = 1[/tex]

[tex]y^3 + y = 0[/tex]

Factorize

[tex]y(y^2 + 1) = 0[/tex]

So, we have

y = 0 or [tex]y^2 + 1 = 0[/tex]

The equation [tex]y^2 + 1 = 0[/tex] will give a complex solution

So, we have

x = 1 and y = 0

Calculating y', we have

[tex]y'=-\dfrac{0+2(1)}{3 * 0^2+1}[/tex]

[tex]y'=-\dfrac{2}{1}[/tex]

y' = -2

Calculating y", we have

[tex]y''=\dfrac{\left(3y^2+12y-1\right)y'-6y^2+y}{\left(3y^2+1\right)^2}[/tex]

[tex]y''=\dfrac{\left(3(0)^2+12(1)(0)-1\right)(-2)-6(0)^2+0}{\left(3(0)^2+1\right)^2}[/tex]

[tex]y''=\dfrac{\left2}{1}[/tex]

y" = 2

Calculating y", we have

[tex]y^{'''}=\dfrac{\left(3y^2+12xy-x\right)y''+y'\left(6yy'+12xy'+12y-1\right)-\left(6y-1\right)y'-6yy'}{\left(3y^2+x\right)^2}-\dfrac{2\left(6yy'+1\right)\left(\left(3y^2+12xy-x\right)y'-y\cdot\left(6y-1\right)\right)}{\left(3y^2+x\right)^3}[/tex]

Simplifying the denominators, we have

[tex](3y^2 + x)^2 = (3(0)^2 + 1)^2 = 1[/tex]

[tex](3y^2 + x)^3 = (3(0)^2 + 1)^3 = 1[/tex]

So, we have

[tex]y^{'''}=\dfrac{\left(3y^2+12xy-x\right)y''+y'\left(6yy'+12xy'+12y-1\right)-\left(6y-1\right)y'-6yy'}{1}-\dfrac{2\left(6yy'+1\right)\left(\left(3y^2+12xy-x\right)y'-y\cdot\left(6y-1\right)\right)}{1}[/tex]

Divide

[tex]y^{'''}=[\left(3y^2+12xy-x\right)y''+y'\left(6yy'+12xy'+12y-1\right)-\left(6y-1\right)y'-6yy']-[2\left(6yy'+1\right)\left(\left(3y^2+12xy-x\right)y'-y\cdot\left(6y-1\right)\right)][/tex]

Simplifying each term:

[tex](3y^2+12xy-x)y''+y'(6yy'+12xy'+12y-1)-(6y-1)y'-6yy' = (3(0)^2+12(1)(0)-(1))(2) + (-2)(6(0)(-2) +12(1)(-2) + 12(0) - 1) - (6(0) - 1)(-2) - 6(0)(-2)[/tex]

[tex](3y^2+12xy-x)y''+y'(6yy'+12xy'+12y-1)-(6y-1)y'-6yy' = 46[/tex]

Also, we have

[tex]2(6yy'+1)((3y^2+12xy-x)y'-y(6y-1)) = 2(6(0)(-2) + 1)((3(0)^2 + 12(1)(0) - 1)(-2) - 0(6(0)-1))[/tex]

[tex]2(6yy'+1)((3y^2+12xy-x)y'-y(6y-1)) = 4[/tex]

So, the expression becomes (by substitution)

[tex]y^{'''}= 46 -4[/tex]

This gives

[tex]y^{'''}= 42[/tex]

Hence, the third derivative at the point x = 1 is 42

Answer:

Option B is correct.

The approximate area of regular octagon is, 101 square ft.

Step-by-step explanation:

Given: A regular octagon has a radius of 6 ft and a side length of 4.6 ft.

To find the area of a regular octagon(A) of side length a is given by :

[tex]A=2\cdot(1+\sqrt{2})a^2[/tex]

Given the length of side, a= 4.6 ft

Substitute the value of a=4.6 ft in the given formula of area:

[tex]A=2\cdot(1+\sqrt{2})\cdot(4.6)^2[/tex] or

[tex]A=(2+2\sqrt{2})\cdot (21.16)[/tex] or

[tex]A=(2+2.828)\cdot(21.16)[/tex]

Simplify:

[tex]A=4.828\cdot 21.16 =102.16048[/tex] square ft.

therefore, the approximate area of regular octagon is, 101 square ft

The shaded region is by assumption the region which is not covered by the semicircle in in given rectangles.

The area of the shaded region is given by 15.48 sq. ft.

A semicircle is a circle cut in half. Thus, one circle produces two semicircle.

Firstly we will find the area of the rectangle and then subtract the area of the semicircle to find the are of the shaded region.

Since the radius of the semicircle is equal to width of the rectangle(6 ft), thus the length of the diameter of the circle( twice the radius which is 12 ft) serves as length of the considered rectangle.

Thus, we have:

[tex]\text{Area of the given rectangle\:} = 6 \times 12 = 72 \: \rm ft^2[/tex]

Since the semicircle is having radius of 6 ft, thus:

[tex]\text{Area of semicircle} = \dfrac{\pi r^2}{2} = \dfrac{3.14 \times 6^2}{2} = 56.52 \: \rm ft^2[/tex]

Thus, area of the shaded region will be equal to area of rectangle - area of semicircle = 72 - 56.52 = 15.48 sq. ft.

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Look at the picture.

Answer: A. (3.5, -4)

To find the perpendicular line's equation, first find the negative reciprocal of the original line's slope. Next, use the point-slope form with the given point. Lastly, rearrange the equation into standard form, resulting in x + 4y = 22.

To find the equation of a line that is perpendicular to another line and passes through a given point, you need to perform a series of steps. The first line's equation is given as 4x - y = 2. Firstly, solve for y to put it in slope-intercept form, y = mx + b. Here, the equation becomes y = 4x - 2, so the slope (m) is 4. The slope of the perpendicular line will be the negative reciprocal of this, which is -1/4.

The next step is to use the point-slope form of the line, which is y - y1 = m(x - x1), where (x1, y1) is the point through which the line passes. For the point (2,5), the equation of the line is y - 5 = -1/4(x - 2). Multiplying both sides by 4 to clear the fraction gives 4y - 20 = -x + 2.

Finally, rearrange the equation to get it into standard form, Ax + By = C, giving us x + 4y = 22. This is the standard form of the equation we were seeking.

area = H/2*(b1+b2)

8.1 = 1.5/2*(6.7 +b2)

8.1=0.75*(6.7+b2)

10.8=6.7+b2

b2=10.8-6.7

b2=4.1m

Answer:

Step-by-step explanation:

the answer is a

p=(5.25-5.00)/5.00

p=0.25/5.00

p=0.05

p = 5% increase

Answer:37

Step-by-step explanation:

12•12=144

35•35=1225

1225+144=1369

Square root 1369=37

there is a 1/2 probability of getting heads on any one flip

 since the first one landed on heads you have a 1/2 probability f getting a 2nd one

Probability = 1/2

Answer:

A. 0.135%

Step-by-step explanation:

We have been given that adult male heights are normally distributed with a mean of 70 inches and a standard deviation of 3 inches. The average basketball player is 79 inches tall.  

We need to find the area of normal curve above the raw score 79.

First of all let us find the z-score corresponding to our given raw score.

[tex]z=\frac{x-\mu}{\sigma}[/tex], where,

[tex]z=\text{z-score}[/tex],

[tex]x=\text{Raw-score}[/tex],

[tex]\mu=\text{Mean}[/tex],

[tex]\sigma=\text{Standard deviation}[/tex].

Upon substituting our given values in z-score formula we will get,

[tex]z=\frac{79-70}{3}[/tex]

[tex]z=\frac{9}{3}[/tex]

[tex]z=3[/tex]

Now we will find the P(z>3) using formula:

[tex]P(z>a)=1-P(z<a)[/tex]

[tex]P(z>3)=1-P(z<3)[/tex]

Using normal distribution table we will get,

[tex]P(z>3)=1-0.99865 [/tex]

[tex]P(z>3)=0.00135[/tex]

Let us convert our answer into percentage by multiplying 0.00135 by 100.

[tex]0.00135\times 100=0.135%[/tex]

Therefore, approximately 0.135% of the adult male population is taller than the average basketball player and option A is the correct choice.

Answer:

Step-by-step explanation:

To solve this problem, we need to use the Intersecting Chords Theorem which states "when two chords intersect each other inside a circle, the products of their segments are equal".

Applying this theorem, we have

[tex]AE \times EB = CE \times ED[/tex]

Where [tex]AB=AE+EB[/tex] and [tex]CD=CE+ED[/tex], also [tex]AB \cong CD[/tex], which means

[tex]AE+EB=CE+ED[/tex]

However, if both chords are equal, then their arcs are also equal, that's the easiest way to deduct it, that is

[tex]arc(AB) \cong arc(CD)[/tex]

Because an arc is defined by its chord basically, and in this case they are congruent.