Place parentheses in the following equation to make it true.
6+6/6x6-6= 0

Answer:

37

Step-by-step explanation:

tan(theta)=perpendicular/base

tan(theta)=3/4

theta=arctan(3/4)=37

Answer:

37°

Step-by-step explanation:

for this question you have to use the tan ratio since the opposite of that angle has been given as 3 and the adjacent has also been given which is 4..I will represent the unknown angle using x so,

tan x=opposite/adjacent

tanx=3/4

tanx=0.75

x=tan inverse of 0.75

x=36.9 or 37°

I hope this helps

Answer:

33/14

Step-by-step explanation:

[tex] - 10 \frac{2}{7} + ( - 4 \frac{4}{11} )[/tex]

[tex] = - \frac{72}{7} \div - ( \frac{48}{11} )[/tex]

[tex] = \frac{72}{7} \times \frac{11}{48} [/tex]

[tex] = \frac{3}{7} \times \frac{11}{2} [/tex]

[tex] = \frac{33}{14} [/tex]

stan dreamcatcher

Answer:

45

Step-by-step explanation:

CHECK ANSWER:

45  *5 = 225

45+5 = 50

50 < 225

sum < times

I hope this helped !

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:

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:

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]

Step-by-step explanationIf you are reading this say

thank u