Evaluate the functions in the given points.
a. f(x) = x

2 − 6 for f(0), f(−6), f(√6), f(
1
2
)

b. f(x) = x

3 + 2x for f(0), f(−2), f(−1), f(
1
2
)

c. f(x) =
1−2x
3
for f(0), f(−2), f(2), f (
1
2
) , f(m), f(m − 1), f(−m)

d.f(x) = |x| −
1
x
for f(−1), f(2), f(h), f (
1
2
) , f(−h + 2) where h ≠ 0

Answer:

A " (1,-2)

B " (4,0)

C " (6,-3)

Step-by-step explanation:

Hope it helped.

° ° °

Answer:

Theta = 37.9 degrees

AC = 11.4

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

tan theta = opp/ adj

tan theta = 7/9

tan ^ -1 tan theta = tan ^-1 (7/9)

theta =37.87498

To 1 decimal place

Theta = 37.9 degrees

We use the Pythagorean theorem to find AC

a^2 + b^2 = c^2

7^2 + 9^2 = AC^2

49+81 = AC^2

130 = AC^2

Taking the square root of each side

sqrt(130) = sqrt(AC^2)

AC = 11.40175

Rounding to 1 decimal place

AC = 11.4

Answer:

Step-by-step explanation:

Tan 22 = [tex]\frac{opposite \ side}{adjacent \ side}[/tex]

[tex]tan \ 22 = \frac{30}{x}\\\\0.4040 = \frac{30}{x}\\\\0.4040*x = 30\\\\x = \frac{30}{0.4040}\\\\x = 74.25\\\\x = 74.3[/tex]

Answer:

Sin= opp/hyp

Sin22= 30/x

0.3746=30/x

cross multiply

0.3746x=30

make x the subject of the formula

x= 30/0.3746

x=80

Answer:

-117 is irrational number

Answer:

Irrational

Step-by-step explanation:

Irrational number can't be written as a faction, -11pie can't be written as a fraction. Therefore it is a irrational number.

Answer:

$687.5

Step-by-step explanation:

Cost of the rug per square foot = $22

Length of the rug = 31.25 feet

Total amount Mrs brown will pay = Cost of the rug per square foot × Length of the rug

= 31.25 × 22

= $687.5

Total amount Mrs brown will pay = $687.5

Answer:

AB

Step-by-step explanation:

The segments that are parallel need to be in the same direction ( up and down)

The segments that are parallel are FH, AB, GC

Answer:

AB

Step-by-step explanation:

since is a cube all of the angles are 90 degees and this only possibel whn the line That a vertical a parrelllt to each other

Answer:

k = 14

Step-by-step explanation:

Prime factorize 32

32 = 2 * 2 * 2 * 2 * 2  = 2⁵

[tex]\frac{2^{12}}{2^{\frac{k}{2}}}= 32\\\\\frac{2^{12}}{2^{\frac{k}{2}}}=2^{5}\\\\2^{12-\frac{k}{2}}=2^{5}[/tex]

Both sides base are same.So, compare exponents

[tex]12-\frac{k}{2}=5\\[/tex]

Subtract 12 from both side

[tex]-\frac{k}{2}=5-12\\\\-\frac{k}{2}=-7\\[/tex]

Multiply both sides by (-2),

[tex](-2)*(-\frac{k}{2})=-7*(-2)\\\\k = 14[/tex]

Answer:

11 4-passenger cars

Step-by-step explanation:

let 'x' represent the number of 4-passenger cars,then 'x-3' represent the number of 6-passenger cars.

92=4x+6(x-3)

92=4x+6x-18

92=10x-18

92+18=10x

110=10x

x=110/10

x=11

Therefore there are 11 4-passenger cars

Answer:

Infinite solutions

Step-by-step explanation:

x + 4y = 7 --------------(I)

8y = 14 - 2x

⇒ 2x + 8y = 14 -----------(II)

[tex]\frac{a_{1}}{a_{2}}=\frac{1}{2}\\\\\frac{b_{1}}{b_{2}}=\frac{4}{8}=\frac{1}{2}\\\\\frac{c_{1}}{c_{2}}=\frac{7}{14}=\frac{1}{2}\\\\\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=\frac{1}{2}[/tex]

So, infinite solutions

Answer:

G

Step-by-step explanation:

Given the 2 equations

x + 4y = 7 → (1)

8y = 14 - 2x → (2)

make x the subject of (1) by subtracting 4y from both sides

x = 7 - 4y

Substitute x = 7 - 4y into (2)

8y = 14 - 2(7 - 4y) ← distribute

8y = 14 - 14 + 8y

8y = 8y

Since both sides are equal this indicates the system has infinite solutions

Answer:

6.8 gallions i believe. im not quite sure

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.

Answer:

I found x for you, right there.

Answer:  x is 12 and y is 5

Step-by-step explanation:

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.

[tex]\\ \sf\longmapsto 2x^2+3x^2+3(-1)=5x.x+5x.1[/tex]

[tex]\\ \sf\longmapsto 5x^2-3=5x^2+5x[/tex]

[tex]\\ \sf\longmapsto 5x^2-5x^2-3=5x[/tex]

[tex]\\ \sf\longmapsto 5x=-3[/tex]

[tex]\\ \sf\longmapsto x=\dfrac{-3}{5}[/tex]