3х + 5
Invers of function f

Answer:

1441.08in^3

Step-by-step explanation:

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

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:

6.8 gallions i believe. im not quite sure

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:

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:

A " (1,-2)

B " (4,0)

C " (6,-3)

Step-by-step explanation:

Hope it helped.

° ° °

Answer:

The equation is true.

Step-by-step explanation:

In order to solve this problem, one must envision a right triangle. A diagram used to represent the imagined right triangle is included at the bottom of this explanation.  Please note that each side is named with respect to the angle is it across from.

Right angle trigonometry is composed of a sequence of ratios that relate the sides and angles of a right triangle. These ratios are as follows,

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

One is given the following equation,

[tex]\frac{sin(A)+sin(B)}{cos(A) +cos(B)}+\frac{cos(A)-cos(B)}{sin(A)-sin(B)}=0[/tex]

As per the attached reference image, one can state the following, using the right angle trigonometric ratios,

[tex]sin(A)=\frac{a}{c}\\\\sin(B)=\frac{b}{c}\\\\cos(A)=\frac{b}{c}\\\\cos(B)=\frac{a}{c}[/tex]

Substitute these values into the given equation. Then simplify the equation to prove the idenity,

[tex]\frac{sin(A)+sin(B)}{cos(A) +cos(B)}+\frac{cos(A)-cos(B)}{sin(A)-sin(B)}=0[/tex]

[tex]\frac{\frac{a}{c}+\frac{b}{c}}{\frac{b}{c}+\frac{a}{c}}+\frac{\frac{b}{c}-\frac{a}{c}}{\frac{a}{c}-\frac{b}{c}}=0[/tex]

[tex]\frac{\frac{a+b}{c}}{\frac{a+b}{c}}+\frac{\frac{b-a}{c}}{\frac{a-b}{c}}[/tex]

Remember, any number over itself equals one, this holds true even for fractions with fractions in the numerator (value on top of the fraction bar) and denominator (value under the fraction bar).

[tex]\frac{\frac{a+b}{c}}{\frac{a+b}{c}}+\frac{\frac{b-a}{c}}{\frac{a-b}{c}}[/tex]

[tex]\frac{\frac{a+b}{c}}{\frac{a+b}{c}}+\frac{\frac{-(a-b)}{c}}{\frac{a-b}{c}}[/tex]

[tex]1+(-1)=0[/tex]

[tex]1-1=0[/tex]

[tex]0=0[/tex]

Answer: {1,2} is the answer.

Step-by-step explanation:

A-B

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

= {1,2}

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:

Let ABC = 73.6

Complement = ABD = 16.4

ABx = unknown angle

ABx + (ABx + 73.6) = 90

ABx = 16.4 / 2 = 8.2

The angles are 8.2 and (8.2 + 73.6) = 90

Answer: [tex]w=\frac{33}{28}[/tex]

Step-by-step explanation:

To solve for w, we want to isolate w.

[tex]-\frac{9}{7}=-\frac{2}{3}w-\frac{1}{2}[/tex]            [add both sides by 1/2]

[tex]-\frac{11}{14}=-\frac{2}{3}w[/tex]                 [multiply both sides by -3/2]

[tex]w=\frac{33}{28}[/tex]

Now we know that [tex]w=\frac{33}{28}[/tex].

Answer:

[tex]\sf w=\dfrac{33}{28} \\[/tex]

Step-by-step explanation:

[tex]\sf -\dfrac{9}{7} =-\dfrac{2w}{3} -\dfrac{1}{2}[/tex]

First, take -2w/3 to the left side.

[tex]\sf -\dfrac{9}{7}+\dfrac{2w}{3} = -\dfrac{1}{2}[/tex]

Then, add 9/7 to both sides.

[tex]\sf \dfrac{2w}{3} = -\dfrac{1}{2}+\dfrac{9}{7}[/tex]

Make the denominators the same and add the fractions.

[tex]\sf \dfrac{2w}{3} = -\dfrac{1*7}{2*7}+\dfrac{9*2}{7*2}\\\\\sf \dfrac{2w}{3} = -\dfrac{7}{14}+\dfrac{18}{14}\\\\\sf \dfrac{2w}{3} = \dfrac{-7+18}{14}\\\\\sf \dfrac{2w}{3} = \dfrac{11}{14}[/tex]

Use cross multiplication.

[tex]\sf 2w*14=11*3\\\\28w=33[/tex]

Divide both sides by 28.

[tex]\sf w=\dfrac{33}{28} \\[/tex]

Answer:

3rd graph

Step-by-step explanation:

Step-by-step explanationIf you are reading this say

thank u