Which of the
x
xx-values are solutions to the following inequality?
x
>
111
x>111x, \gt, 111
Choose all answers that apply:
Choose all answers that apply:

(Choice A)
A
x
=
97
x=97x, equals, 97

(Choice B)
B
x
=
101
x=101x, equals, 101

(Choice C)
C
x
=
119
x=119x, equals, 119

Answer:

I htink x ≈ 8

Step-by-step explanation:

Answer:

X is approximately 7.8.

Step-by-step explanation:

You can use SOH-CAH-TOA to help figure out what function (sin, cos, tan) you need to use in order to figure out the missing side.

For this one, we can see the angle is pointing to the opposite side (x length), and we have been given the hypotenuse (18). So we want to use the sin function.

[tex]sin\ (angle)=\frac{opposite}{hypotenouse}[/tex]

[tex]sin (26)=\frac{x}{18}[/tex]

[tex]0.438=\frac{x}{18}[/tex]

[tex]7.890... = x[/tex]

Using Pythagorean theorm, you can figure out the other side if need be :)

For reference:

[tex]sin (angle)=\frac{opposite}{hypotenuse}[/tex]

[tex]cos(angle)=\frac{adjacent}{hypotenuse}[/tex]

[tex]tan(angle)=\frac{opposite}{adjacent}[/tex]

Answer:

A " (1,-2)

B " (4,0)

C " (6,-3)

Step-by-step explanation:

Hope it helped.

° ° °

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:

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:

4

Step-by-step explanation:

Let's simplify step-by-step.

−2z2+4z+2z2

Combine Like Terms:

=−2z2+4z+2z2

=(−2z2+2z2)+(4z)

=4z

Answer:

4z

Hope this helps <3 Need thanks?

Comment /hearthelp

[tex]\\ \sf\longmapsto -2z^2+4z+2z^2[/tex]

[tex]\\ \sf\longmapsto -2z^2+2z^2+4z[/tex]

[tex]\\ \sf\longmapsto (-2+2)z^2+4z[/tex]

[tex]\\ \sf\longmapsto 0z^2+4z[/tex]

[tex]\\ \sf\longmapsto 4z[/tex]

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.

the answer is

A. 28, 25, 22