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Romashka [77]
2 years ago
13

How do you factor 130x - 13. Please show work

Mathematics
2 answers:
kipiarov [429]2 years ago
8 0

Answer:

13(10x-1)

Step-by-step explanation:

130x-13

pick 13 from expression

13(10x-1)

Nana76 [90]2 years ago
5 0

Answer

this.

Step-by-step explanation:

130x-13

13(10x-1)

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15. Find the value of x.<br> (68 - x)<br> 3x
masha68 [24]

Answer:

17=x

Step-by-step explanation:

In this case, we need to use Vertical Angles Theorem to solve this problem. Basically, because of the position of the angles, they are congruent.

68-x = 3x

Add x to both sides

68=4x

Divide by 4

17=x

Hope this helps!

3 0
3 years ago
Read 2 more answers
Plssss help <br> I need AC
Zolol [24]

Answer:

AC = 18

Step-by-step explanation:

Using the cosine ratio in the right triangle and the exact value

cos30° = \frac{\sqrt{3} }{2} , then

cos30° = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{AC}{12\sqrt{3} } = \frac{\sqrt{3} }{2} ( cross- multiply )

2 AC = 12\sqrt{3} × \sqrt{3} = 12 × 3 = 36 ( divide both sides by 2 )

AC = 18

3 0
2 years ago
Read 2 more answers
What is the value of x?
rodikova [14]
1) expand the brackets 
6(-3 - x) - 2x = 14
-18 - 6x - 2x = 14
simplify
-18 - 8x = 14
then add 18 to both sides to get the x's on there own
-8x = 32
then divide by -8
x = 32/-8
x = -4

2) multiply everything by 7x
2 + (6 x 7x) + ( 17x × 7x) = 18 x 7x
2 + 42x + 119x^2 = 126x
put everything on one side
119x^2 - 84x + 2 = 0
then use quadratic equation to solve

x = 0.024671849 or 0.6812105039

 
6 0
3 years ago
Anyone in 10th grade please help me because I have to pass this later at 10 pm:(
Alenkinab [10]

18) 19m^8n^8-\left(4m^5n\right)\left(3m^3n^7\right)

=19m^8n^8-4m^5n\cdot \:3m^3n^7

=19m^8n^8-12m^8n^8

=7m^8n^8

19) \left(-5cd\right)\left(-3c^4d\right)-\left(7c^2d^2\right)\left(2c^3\right)

=5cd\cdot \:3c^4d-7c^2d^2\cdot \:2c^3

=15c^5d^2-14c^5d^2

=c^5d^2

6 0
3 years ago
Factor the following trinomial. (Hint: Solve the equations first.)<br><br> x²-2x-2
Alex_Xolod [135]

Answer:

(x-(1+\sqrt{3}))(x-(1-\sqrt{3}))  

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

x^{2} -2x-2=0  

so

a=1\\b=-2\\c=-2

substitute in the formula

x=\frac{-(-2)(+/-)\sqrt{-2^{2}-4(1)(-2)}} {2(1)}

x=\frac{2(+/-)\sqrt{12}} {2}

x=\frac{2(+/-)2\sqrt{3}} {2}

x_1=\frac{2(+)2\sqrt{3}} {2}=1+\sqrt{3}

x_2=\frac{2(-)2\sqrt{3}} {2}=1-\sqrt{3}

therefore

x^{2} -2x-2=(x-(1+\sqrt{3}))(x-(1-\sqrt{3}))  

5 0
3 years ago
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