3х + 5
Invers of function f

Answer:

O D. 3x2 + 2x-3

Answer:

A.7x2-5x questions 2of 20

Answer:

a=13

Step-by-step explanation:

-90=-6a-12

-6a=-78

a=13

Answer:

a=13 well as u can see the other comment already gave an explanation of y it is a=13 and i completely agree with it have a nice afternoon,night,or day to u

Step-by-step explanation:

Answer:

1441.08in^3

Step-by-step explanation:

9514 1404 393

Answer:

  2. (only)

Step-by-step explanation:

The Pythagorean theorem tells you the sum of the squares of the two legs of a right triangle is equal to the square of the hypotenuse. To determine if these are right triangles, determine if that condition is met.

  1. 3^2 +5^2 = 9 + 25 = 34 ≠ (√35)^2 . . . . not a right triangle

  2. 5^2 +4^2 = 25 +16 = 41 = (√41)^2 . . . a right triangle

  3. 6^2 +8^2 = 36 +64 = 100 ≠ (√10)^2 . . . . not a right triangle

  4. 3^2 +3^2 = 9 +9 = 18 ≠ (3√3)^2 = 27 . . . . not a right triangle

Answer:

Step-by-step explanation:

∠OST = 90° as ST ⊥ OS (tangent is perpendicular to radius at same point)

m∠OSP = 1/2(180° - m∠SOP) = 90° - 96°/2 = 42° (sum of interior angles of the triangle SOP)

m∠PST = 90° - m∠OSP = 90° - 42° = 48° (angle addition postulate)

Correct choice is B

Answer:

The area of the circle is 100 square units.

Step-by-step explanation:

We are given that the circumference of a circle is 20π, and we want to determine its area.

Recall that the circumference of a circle is given by the formula:

[tex]\displaystyle C = 2\pi r[/tex]

Substitute:

[tex]20 \pi = 2 \pi r[/tex]

Solve for the radius:

[tex]\displaystyle r = \frac{20\pi}{2\pi} = 10[/tex]

The area of a circle is given by:

[tex]\displaystyle A = \pi r^2[/tex]

Since the radius is 10 units:

[tex]\displaystyle A = \pi (10)^2[/tex]

Evaluate:

[tex]\displaystyle A = 100\pi\text{ units}^2[/tex]

In conclusion, the area of the circle is 100 square units.

Step-by-step explanation:

Hi there!

Please see the answer in the picture.

Hope it helps!

1. Approach

One is given a trigonometric equation with and one is asked to prove that it is true. Using the attached image, combined with the knowledge of trigonometry, one can evaluate each trigonometric function. Then one can simplify each ratio to solve. To yield the most accurate result, one has to each of the ratios in a fractional form, rather than simplifying it into a decimal form. Remember the right angle trigonometric ratios, these ratios describe the relationship between the sides and angles in a right triangle. Such ratios are as follows,

[tex]sin(\theta)=\frac{opposite}{hypotenuse}\\\\cos(\theta)=\frac{adjacent}{hypotenuse}\\\\tan(\theta)=\frac{opposite}{adjacent}\\\\csc(\theta)=\frac{hypotenuse}{opposite}\\\\sec(\theta)=\frac{hypotenuse}{adjacent}\\\\cot(\theta)=\frac{adjacent}{opposite}[/tex]

Please note that the terms (opposite) and (adjacent) are relative to the angle uses in the ratio, however the term (hypotenuse) refers to the side opposite the right angle, this side never changes its name. Use these ratios to evaluate the trigonometric functions. Then simplify to prove the identity.

2. Problem (9)

[tex]\frac{sin(60)+cos(30)}{1+sin(30)+cos(60)}=sin(60)[/tex]

As per the attached image, the following statements regarding the value of each ratio can be made:

[tex]sin(60)=\frac{\sqrt{3}}{2}\\\\cos(30)=\frac{\sqrt{3}}{2}\\\\sin(30)=\frac{1}{2}\\\\cos(60)=\frac{1}{2}[/tex]

Substitute,

[tex]\frac{sin(60)+cos(30)}{1+sin(30)+cos(60)}=sin(60)[/tex]

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

Simplify,

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

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

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

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

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

Thus, this equation is true.

2. Problem (10)

Use a similar strategy to evaluate this equation,

[tex]\frac{1-cos(30)}{sin(30)}=\frac{1-cot(60)}{1+cot(60)}[/tex]

Use the attached image to evaluate the ratios.

[tex]cos(30)=\frac{\sqrt{3}}{2}\\\\sin(30)=\frac{1}{2}\\\\cot(60)=\frac{1}{\sqrt{3}}[/tex]

Substitute,

[tex]\frac{1-cos(30)}{sin(30)}=\frac{1-cot(60)}{1+cot(60)}[/tex]

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

Simplify,

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

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

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

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

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

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}+1}}[/tex]

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

Rationalize the denominator,

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

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}+1}*\frac{\sqrt{3}-1}{\sqrt{3}-1}[/tex]

[tex]2-\sqrt{3}=\frac{(\sqrt{3}-1)^2}{3-1}[/tex]

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

[tex]2-\sqrt{3}=\frac{4-2\sqrt{3}}{2}[/tex]

[tex]2-\sqrt{3}=2-\sqrt{3}[/tex]

Therefore, this equation is also true.

Answer: {1,2} is the answer.

Step-by-step explanation:

A-B

{1,2,3,4}-{3,4,5,6)

= {1,2}

Answer:

A " (1,-2)

B " (4,0)

C " (6,-3)

Step-by-step explanation:

Hope it helped.

° ° °

Answer:

3rd graph

Step-by-step explanation:

Answer: Hundreds and tens place values (the two copies of '6')

Explanation:

We're looking for where the digits are the same, which would be those two copies of '6'

The first 6 on the left is in the hundreds place. It represents 600

The other 6 is in the tens place, and it represents 60

The jump from 60 to 600 is "times 10".

Answer:

7/0

Step-by-step explanation:

7/0 = 7

A rational number is written in the form of p/q, where q is not equal to zero. Therefore it is not a rational number.

Answer:

7/0

Step-by-step explanation:

a irrational can not be written as a fraction

Answer:

Step-by-step explanation:

This is the problem we need to solve:

[tex]z=\frac{x_i-\bar{x}}{\sigma}[/tex] and we have everything but the z-score (which we find from a table) with our main unknown being the standard deviation.

If the probability that a random variable that is less than 11 is .67, we first have to find the z-score from the table that is closest to .67, and there are 2:

P(z < .43) = .66640 and P(z < .44) = .67003

We'll use z = .44

[tex].44=\frac{11-10}{\sigma}[/tex] and

[tex].44=\frac{1}{\sigma}[/tex] and

[tex]\sigma=\frac{1}{.44}[/tex] so

σ = 2.27 (check it; it works!)

Step-by-step explanationIf you are reading this say

thank u