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sweet [91]
3 years ago
13

What is the value of x? Enter your answer in the box. x =

Mathematics
2 answers:
enyata [817]3 years ago
8 0

Answer:

x = 27

Step-by-step explanation:

Since they are corresponding they have to be equal.

4x+7 = 5(x-4)

4x+7 = 5x-20 distribute the 5

7 = 5x-4x-20 bring the 4x over

7 = x-20 simplify the x

7+20= x bring the twenty over

27=x

iris [78.8K]3 years ago
4 0

Answer:

27 =x

Step-by-step explanation:

The two angles are vertical angles and vertical angles are equal

4x+7 = 5(x-4)

Distribute

4x+7 = 5x-20

Subtract 4x from each side

4x+7-4x = 5x-4x-20

7 = x-20

Add 20 to each side

7+20 = x-20+20

27 =x

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Can you help me to solve this
stealth61 [152]

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\text{Triangle 1:}\\\\x=30\sqrt{2},\\\\\text{Triangle 2:}\\x\approx 36.18,\\?\approx12.37

Step-by-step explanation:

*Notes (clarified by the person who asked this question):

-The triangle on the right has a right angle (angle that appears to be a right angle is a right angle)

-The bottom side of the right triangle is marked with a question mark (?)

<u>Triangle 1 (triangle on left):</u>

Special triangles:

In all 45-45-90 triangles, the ratio of the sides is x:x:x\sqrt{2}, where x\sqrt{2} is the hypotenuse of the triangle. Since one of the legs is marked as 30, the hypotenuse must be \boxed{30\sqrt{2}}

It's also possible to use a variety of trigonometry to solve this problem. Basic trig for right triangles is applicable and may be the simplest:

\cos 45^{\circ}=\frac{30}{x},\\x=\frac{30}{\cos 45^{\circ}}=\frac{30}{\frac{\sqrt{2}}{2}}=30\cdot \frac{2}{\sqrt{2}}=\frac{60}{\sqrt{2}}=\frac{60\cdot\sqrt{2}}{\sqrt{2}\cdot \sqrt{2}}=\frac{60\sqrt{2}}{2}=\boxed{30\sqrt{2}}\\

<u>Triangle 2 (triangle on right):</u>

We can use basic trig for right triangles to set up the following equations:

\cos 20^{\circ}=\frac{34}{x},\\x=\frac{34}{\cos 20^{\circ}}\approx \boxed{36.18},

\tan 20^{\circ}=\frac{?}{34},\\?=34\tan 20^{\circ}\approx \boxed{12.37}

We can verify these answers using the Pythagorean theorem. The Pythagorean theorem states that in all right triangles, the following must be true:

a^2+b^2=c^2, where c is the hypotenuse of the triangle and a and b are two legs of the triangle.

Verify 34^2+(34\tan20^{\circ})^2=\left(\frac{34}{\cos20^{\circ}}\right)^2\:\checkmark

3 0
3 years ago
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