X-14+2x+20=90 Solve for X

Mathematics

2 answers:

Salsk061 [2.6K]3 years ago

6 0

Answer:

Step-by-step explanation:

X-14+2x+20=90

×+2x=90+14-20

X+2x=84

3x=84

X=84÷3

X=28

sleet_krkn [62]3 years ago

3 0

x - 14 + 2x + 20 = 90

First, combine like terms: x and 2x, -14 and 20.

x + 2x = 3x

-14 + 20 = 6

3x + 6 = 90

Subtract 6 from both sides.

3x = 84

Divide both sides by 3

x = 28

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Simplify 24÷(-2)(3)+7<br><br> a. -29<br> b. 3<br> c. 11

bekas [8.4K]

Answer:

a. -29

Step-by-step explanation:

The expression is:

24÷(-2)(3)+7

We can see that there is division, multiplication and addition involved. We will use the mathematical DMAS rule to solve the problem. So, division will be done first,

Division will give us:

24÷(-2)*(3)+7

= -12 * 3 + 7

After division, multiplication will be done:

Multiplication will give us:

= -36 +7

which equals to -29

So, option a is the correct answer ..

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4 years ago

Name the quadrilateral whose diagonals are perpendicular bisectors of each other

dexar [7]

Rhombus or square.

Hope it helps!

5 0

3 years ago

The vertices of ∆ABC are A(2, 8), B(16, 2), and C(6, 2). what is the perimeter and area in square units

ad-work [718]

Check the picture below.

the triangle has that base and that height, recall that A = 1/2 bh.

now as for the perimeter, you can pretty much count the units off the grid for the segment CB, so let's just find the lengths of AC and AB,

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&C&(~ 6 &,& 2~)
\end{array}~~~ 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AC=\sqrt{(6-2)^2+(2-8)^2}\implies AC=\sqrt{4^2+(-6)^2}
\\\\\\
AC=\sqrt{16+36}\implies AC=\sqrt{52}\implies AC=\sqrt{4\cdot 13}
\\\\\\
AC=\sqrt{2^2\cdot 13}\implies AC=2\sqrt{13}$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&B&(~ 16 &,& 2~)
\end{array}\\\\\\
AB=\sqrt{(16-2)^2+(2-8)^2}\implies AB=\sqrt{14^2+(-6)^2}
\\\\\\
AB=\sqrt{196+36}\implies AB=\sqrt{232}\implies AB=\sqrt{4\cdot 58}
\\\\\\
AB=\sqrt{2^2\cdot 58}\implies AB=2\sqrt{58}$

so, add AC + AB + CB, and that's the perimeter of the triangle.