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asambeis [7]
3 years ago
15

3(4x-5)+4(2x+6) simplify

Mathematics
2 answers:
astra-53 [7]3 years ago
5 0

Answer:

20+9

Step-by-step explanation:

Brums [2.3K]3 years ago
3 0

Answer: 20x+9 (I THINK that’s the answer but I’m not completely sure so...)

Step-by-step explanation: (12x-15) + (8x+24)

12x+8x= 20x

-15+24 (or 24-15)= 9

20x+9

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Two factors of -48 have a difference of 19 the factor with the greater absolute value is positive what is the sum of the factors
MA_775_DIABLO [31]
1). Two factors of -48 => x * y = -48

=> the absolute values of the two numbers may be:

48 and 1

24 and 2

16 and 3

12 and 4

6 and 8 

2). Have a differencie of 19 = > x - y = 19


That makes that the numbers  be + 16 and - 3


Because (+16) * (-3) = - 48

And +16 - (-3) = +16 + 3 = 19.


3) The sum of the factors is +16 + (-3) = 16 - 3 = 13.


So the answer is the option C. 13.
4 0
3 years ago
-x + 3y = -23<br> -8x - 3y = -22<br> using elimination
Nikolay [14]
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4 0
2 years ago
Read 2 more answers
How do I find the Q1 and Q3?<br><br> 0,0,1,2,2,3,4,4,4,4,5,6,6,7,7
Angelina_Jolie [31]

Answer:

Q1 = 2

Q3 = 6

Step-by-step explanation:

Mathematically, we have

Q1 = (n + 1)/4 th term

where n is the number of terms

By the count, we have n as 15

Q1 = (15 + 1)/4

Q1 = 4th term

Looking at the arrangement, the 4th term is 2

For Q3

Q3 = 3(n + 1)/4 th term

n = 15

Q3 = 3 * 4 = 12th term

The 12th term is 6

So that is the 3rd quartile

5 0
2 years ago
OF bisects EOF. EOF=y+30 and FOG=3y-50. solve for y.
Alexeev081 [22]

It always helps to draw a picture. Given the information, Segment OF is the center line that is bisecting this angle.

Since it's bisecting (cutting in half)... we can simply set the two angles equal to each other.

y+30=3y-50
-2y+30=-50
-2y=-80
y = 40

C)40.
3 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}\\\\&#10;\begin{array}{ccccccccc}&#10;&&x_1&&y_1&&x_2&&y_2\\&#10;%  (a,b)&#10;&A&(~ 2 &,& 8~) &#10;%  (c,d)&#10;&C&(~ 6 &,& 2~)&#10;\end{array}~~~ &#10;%  distance value&#10;d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}&#10;\\\\\\&#10;AC=\sqrt{(6-2)^2+(2-8)^2}\implies AC=\sqrt{4^2+(-6)^2}&#10;\\\\\\&#10;AC=\sqrt{16+36}\implies AC=\sqrt{52}\implies AC=\sqrt{4\cdot 13}&#10;\\\\\\&#10;AC=\sqrt{2^2\cdot 13}\implies AC=2\sqrt{13}

\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\&#10;\begin{array}{ccccccccc}&#10;&&x_1&&y_1&&x_2&&y_2\\&#10;%  (a,b)&#10;&A&(~ 2 &,& 8~) &#10;%  (c,d)&#10;&B&(~ 16 &,& 2~)&#10;\end{array}\\\\\\&#10;AB=\sqrt{(16-2)^2+(2-8)^2}\implies AB=\sqrt{14^2+(-6)^2}&#10;\\\\\\&#10;AB=\sqrt{196+36}\implies AB=\sqrt{232}\implies AB=\sqrt{4\cdot 58}&#10;\\\\\\&#10;AB=\sqrt{2^2\cdot 58}\implies AB=2\sqrt{58}

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

8 0
3 years ago
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