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3 years ago

How to find x in a triangle with only two known lengths

Mathematics

2 answers:

kow [346]3 years ago

8 0

A triangle is 180 degrees. X+first length+second length=180. Then solve

Sidana [21]3 years ago

7 0

The whole triangle equals 180 so take the other two lengths and subtract it from 180 and get the last length

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What is the coefficient of q in the sum of (2/3q-3/4) and(-1/6q-r)?

Harlamova29_29 [7]

Answer:

The coefficient is $\frac{1}{2}$

Step-by-step explanation:

The given sum is

$(\frac{2}{3}q -\frac{3}{4})+(-\frac{1}{6}q-r)$

We can choose to add only the coeffients of $q$ or simplify the whole sum.

$=\frac{2}{3}q -\frac{1}{6}q-\frac{3}{4})-r)$

We collect the LCM for the $q$ terms;

$=\frac{4-1}{6}q-\frac{3}{4})-r)$

This will give us;

$=\frac{3}{6}q-\frac{3}{4})-r)$

Which is the same as

$=\frac{1}{2}q-r-\frac{3}{4}$

The coefficient is $\frac{1}{2}$

6 0

3 years ago

Read 2 more answers

Consider the polynomial, 5z - 9x' + 2z11 +6.

Yuri [45]

Answer:

11 ; 2 ; 4 ; 6

Step-by-step explanation:

1) look at the constant with the highest degree (11)

2) look at the coefficent that mutiplies the constant with the highest degree (2)

3) Count the terms that are separated by minus or plus (4)

4) look at the therm without variable (6)

5 0

3 years ago

A conservative estimate of the number of stars in the universe is 6 x 10^22. The average human can see about 3,000 stars at nigh

Ilia_Sergeevich [38]

$2 \times 10^{19}$ times more stars are there in universe compared to human eye can see

<h3><u>Solution:</u></h3>

Given that, conservative estimate of the number of stars in the universe is $6 \times 10^{22}$

The average human can see about 3,000 stars at night with only their eyes

To find: Number of times more stars are there in the universe, compared to the stars a human can see

Let "x" be the number of times more stars are there in the universe, compared to the stars a human can see

Then from given statement,

$\text{Stars in universe} = x \times \text{ number of stars human can see}$

<em><u>Substituting given values we get,</u></em>

$6 \times 10^{22} = x \times 3000\\\\x = \frac{6 \times 10^{22}}{3000}\\\\x = \frac{6 \times 10^{22}}{3 \times 10^3}\\\\x = 2 \times 10^{22-3}\\\\x = 2 \times 10^{19}$

Thus $2 \times 10^{19}$ times more stars are there in universe compared to human eye can see

5 0

3 years ago

Cold antoje help me answering this please :).<br>Fake answer will be reported.

Harrizon [31]

Answer:

158.1

Step-by-step explanation:

Distance is the square root of (x1-x2)^2+(y1-y2)^2

square root of (-4-146)^2+(2-52)^2

square root of 150^2+50^2

square root of 22500+2500

square root of 25000

50sqrt10

Rounding is 158.1

5 0

3 years ago

Find the slope of the line passing through the point (-6,8) and (-9,1)

Ipatiy [6.2K]

Hello!

Y2-y1/x2-x1

1-9/8-6
-8/2

The slope is -4

Hope this helps!

-bambi

5 0

2 years ago