PLZ HELP ME ASAP PLZZ

Answer:

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Answer:

x=61.71428 or 61 5/7

Step-by-step explanation:

We can use a ratio to solve

7 pieces              12 pieces

-----------------   = ---------------

36 seconds          x seconds

Using cross products

7x = 36*12

7x = 432

Divide by 7

7x/7 = 432/7

61 5/7

x=61.71428

Answer:

66

Step-by-step explanation:

Answer:

O D. 3x2 + 2x-3

Answer:

A.7x2-5x questions 2of 20

Answer:

A^7

Step-by-step explanation:

Answer:

p = -5

every element in the sequence is created by subtracting 3 from the pervious element.

Step-by-step explanation:

I am not sure you put every necessary information here.

but I'd the visible information is truly everything, then it's is trivial.

the difference between 4 and 1 is ... well, -3. meaning we subtract 3 from 4 to get 1.

we suspect this is the rule and keep trying.

1 -3 = -2

hey it works.

and -8 -3 = -11

hey, also correct.

and the difference between -2 and -8 is -6, and when we place another item in between (p), we cut that in half again to -3. so, it is all consistent.

therefore,

p = -2 -3 = -5

the rule is

an = an-1 - 3

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:

i use logical reasoning

Answer:

1≤f(x)≤5

Step-by-step explanation:

-1≤x≤1

-2≤2x≤2 (Multiplied by 2 both side)

-2+3≤2x+3≤2+3 (Adding three both sides)

1≤f(x)≤5

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:

the vertical line is:

x = 5

The horizontal line is:

y = -9

Step-by-step explanation:

A vertical line has a fixed x-value, while a horizontal line has a fixed y-value.

Then we can write a vertical line as:

x = a

and a horizontal line as

y = b

Then, if we want a vertical line that passes through the point (5, -9), remember that the x-vale will be fixed, then we fix the x-value at the same x-value of the point, which we know that is 5, then the vertical line that passes through the point (5, -9) is:

x = 5

While the horizontal line that passes through the point (5, -9) will be a line with the y-value fixed at the y-value of the point, which we know is -9

Then the horizontal line is:

y = -9

Step-by-step explanationIf you are reading this say

thank u