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3 years ago

6

Tell whether the angles are adjacent or vertical. Then find the value of x.

2 answers:

goldenfox [79]3 years ago

8 0

The angles are verticals and the value of x is 128°.

SVEN [57.7K]3 years ago

8 0

Answer:

The angles are vertical
X= 128

Hope that helps :)

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Solve each equation.<br> 1/3a = -5 a = ?<br> 12 - b = 12.5 b = ? <br> -0.1 = -10c c = ?

Rina8888 [55]

Answer:

a = -15, b = -.5, c = -1

Step-by-step explanation:

1/3a = -5 --> multiply each side with 3 --> a = -15

12 - b = 12.5 --> subtract 12 from both sides --> -b = .5 --> multiply each side with -1 to get b positive --> b = -.5

.1 = -10c --> multiply each side with -10 --> c = -1

8 0

3 years ago

Find the sum. 2/5 + 1/2

dybincka [34]

2/5 + 1/2

4+5 / 10 • look for the LCD of 5 and 2
which is 10 then 5 goes 2
times into 10 the multiply by 2
= 4. Then 2 goes 5 times into
10 then multiply by 1 = 5.

Therefore the sum of 2/5 + 1/2 =9/10

5 0

3 years ago

Read 2 more answers

When a sunfloweris 2 weeks old it is 38 centimeters tall when it is 6 weeks old it is 114 cm tall how tall is it when it is 3 ½

Anna11 [10]

2 wk --- 38 cm
3.5 wk --- x cm
x = 3.5 wk x 38 cm
---------------------
2 wk
x= 66.5 cm

5 0

3 years ago

Find the angle between the given vectors. Round your answer, in degrees, to two decimal places. u=⟨2,−6⟩u=⟨2,−6⟩, v=⟨4,−7⟩

NISA [10]

Answer:

$\theta = 108.29$

Step-by-step explanation:

Given

$u =$

$v =$

Required:

Calculate the angle between u and v

The angle $\theta$ is calculated as thus:

$cos\theta = \frac{u.v}{|u|.|v|}$

For a vector

$A =$

$A = a * b$

$cos\theta = \frac{u.v}{|u|.|v|}$ becomes

$cos\theta = \frac{.}{|u|.|v|}$

$cos\theta = \frac{2*6+4*-7}{|u|.|v|}$

$cos\theta = \frac{12-28}{|u|.|v|}$

$cos\theta = \frac{-16}{|u|.|v|}$

For a vector

$A =$

$|A| = \sqrt{a^2 + b^2}$

So;

$|u| = \sqrt{2^2 + 6^2}$

$|u| = \sqrt{4 + 36}$

$|u| = \sqrt{40}$

$|v| = \sqrt{4^2+(-7)^2}$

$|v| = \sqrt{16+49}$

$|v| = \sqrt{65}$

So:

$cos\theta = \frac{-16}{|u|.|v|}$

$cos\theta = \frac{-16}{\sqrt{40}*\sqrt{65}}$

$cos\theta = \frac{-16}{\sqrt{2600}}$

$cos\theta = \frac{-16}{\sqrt{100*26}}$

$cos\theta = \frac{-16}{10\sqrt{26}}$

$cos\theta = \frac{-8}{5\sqrt{26}}$

Take arccos of both sides

$\theta = cos^{-1}(\frac{-8}{5\sqrt{26}})$

$\theta = cos^{-1}(\frac{-8}{5 * 5.0990})$

$\theta = cos^{-1}(\frac{-8}{25.495})$

$\theta = cos^{-1}(-0.31378701706)$

$\theta = 108.288386087$

<em></em> $\theta = 108.29$ <em> (approximated)</em>

4 0

2 years ago

A2b -b2 a=4 and b= -7

cestrela7 [59]

Step-by-step explanation:

hope this helps you dear friend.

8 0

3 years ago