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pentagon [3]
3 years ago
9

Please help 15 points and brainliest!

Mathematics
1 answer:
Umnica [9.8K]3 years ago
3 0
C because the proportions are equal and multiplied by 4/4
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Find the value of 6x - 5 given that −17x -9 = 8
Nikitich [7]
Okay, So We Need To Solve For X, So We Can Solve The Expression:
6x - 5.

Lets Begin.

-17x - 9 = 8.

So, We First Need To Add 9 To Both Sides. 

-17x = 8 + 9

8+9 = 17.

-17x = 17

Divide.

-17x/-17 = 1x = x.

So:

X = 17/-17

17/-17 = -1

So:

X = -1

Now, Plug In The Value For The Expression.

6 * (-1) - 5

6 * (-1) = -6

(-6) - 5 = -11.

So, Our Value For 6x - 5 Is -11.

5 0
3 years ago
What is the value of x in the figure below? I need help with learning the math also
Archy [21]
E) sqrt(30)

This is a nice problem. It was helpful to read the topic at the top.

Notice that the two right triangles are similar. Their sides are in proportion.

That is the side '3' is what 'x' is in the other triangle (you need to rotate it!)

Then 'x' is similar to '10'

3/x=x/10 ---> x^2=30, x=sqrt(30)

If you are patient, you can check that for sqrt(30) pythagoras theorem works for all the triangles.

Best

Then
4 0
3 years ago
What is the decimal multiplier to decrease by 2.7%?
MatroZZZ [7]

Multiplying by 1 keeps it the same. A number greater than 1 would be an increase and a number below 1 would be a decrease.

The decrease is 2.7% which is written as 0.027 as a decimal.

Subtract that from 1:

1 - 0.027 = 0.973

The multiplier would be 0.973

6 0
3 years ago
WILL GIVE BRAINLIEST
cricket20 [7]

Answer:(–4.5, –2.5)

(4.5, 6)

(1.3, 3.5)

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
The angle θ lies in Quadrant II .
Andreyy89

let's keep in mind that, in the II Quadrant, cosine is negative and sine, is positive.

cosine is adjacent/hypotenuse, however the hypotenuse is simply a radius unit, and thus is never negative, so in the -(2/3) the negative must be the numerator, -2.


\bf cos(\theta )=\cfrac{\stackrel{adjacent}{-2}}{\stackrel{hypotenuse}{3}}\impliedby \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{3^2-(-2)^2}=b\implies \pm\sqrt{5}=b\implies \stackrel{\textit{II Quadrant}}{+\sqrt{5}=b}~\hfill tan(\theta )=\cfrac{\stackrel{opposite}{\sqrt{5}}}{\stackrel{adjacent}{-2}} \\\\\\ ~\hspace{34em}

6 0
4 years ago
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