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3 years ago

Find the value of x. A.20 B.25 C.18.75 D.22.5

Mathematics

2 answers:

miss Akunina [59]3 years ago

7 0

The two equations are vertical angle, which would be equal to each other.

Set the equations to equal and solve for x:

5x -10 = 3x +40

Subtract 3x from both sides:

2x -10 = 40

Add 10 to both sides:

2x = 50

Divide both sides by 2:

x = 50 / 2

x = 25

The answer is B. 25

Natalka [10]3 years ago

3 0

Answer:

B. 25

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i hate myself :/

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Step-by-step explanation:

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5 0

3 years ago

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Find the value of x of this question

mel-nik [20]

<h3>Let's understand the concept:-</h3>

Here angle B is 90°
So $\triangle ABC$ and $\triangle ABD$ Are right angled triangle
So we use Pythagoras thereon for solution

<h3>Required Answer:-</h3>

First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

We need to find base=b

According to Pythagoras thereon

${\boxed{\sf b^2=h^2-p^2}}$

Substitutethe values

$\longrightarrow$ $\sf b^2=10^2-p^2$

$\longrightarrow$ $\sf b={\sqrt {10^2-8^2}}$

$\longrightarrow$ $\sf b={\sqrt{100-64}}$

$\longrightarrow$ $\bf b={\sqrt {36}}$

$\longrightarrow$ $\sf b=6$

$\therefore$ $\overline{BC}=6cm$

BD=BC+CD

$\longrightarrow$ $BD=9+6$

$\longrightarrow$ $BD=15cm$

Now in $\triangle ABD$

Perpendicular=p=8cm

Base =b=15cm

We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

${\boxed {\sf h^2=p^2+b^2}}$

Substitute the values

$\longrightarrow$ $\sf h^2=8^2+15^2$

$\longrightarrow$ $\sf h={\sqrt {8^2+15^2}}$

$\longrightarrow$ $\sf h={\sqrt {64+225}}$

$\longrightarrow$ $\sf h={\sqrt {289}}$

$\longrightarrow$ $\sf h=17cm$

$\therefore$ ${\underline{\boxed{\bf x=17cm}}}$

8 0

3 years ago

What is the measure? Would appreciate any help!

Luba_88 [7]

Answer:

If I'm not wrong, angle O should be 62°!

Step-by-step explanation:

since triangle MON and triangle DAY are equal, if we add 75° and 43° it would be 118°. We then subtract 118° from 180° and it should be 62°. :)

6 0

3 years ago

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