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pantera1 [17]
3 years ago
8

Pls help! Due soon. I will give brainiest *no links*

Mathematics
1 answer:
Artyom0805 [142]3 years ago
5 0

Answer:

#2 a.

#3 c.

#4 a.

#5 c.

#6 b.

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X−3(5x−6)=3−10x What is x, and how do you find x?
Fudgin [204]

Answer:

x=3or-1

Step-by-step explanation:

x-3 (5x-6)=3-10x

5x^2-15x-6x+18=3-10x

collect like terms

5x^2+18-3=15x+6x-10x

5x^2+15=11x

5x^2-11x+15=0

factorise

(x-3) (5x+5) =0

...... x=3 or -1

pls like if helpful nd rank me as most helpful ans

7 0
4 years ago
Measurements of angles
Likurg_2 [28]

Answer:

x = 60

Step-by-step explanation:

Since this triangle is an isosceles triangle, the angle next to the 60 degrees is 60 as well. The interior angles of a triangle adds up to 180 degrees, so:

60 + 60 + x = 180
120 + x = 180
x = 60
Hope this helps :)

8 0
3 years ago
What is the recursive rule for the following sequence: -9, -2, 5, 12, ….
tino4ka555 [31]

Answer:

The answer is option (C)=an-1+7

Step-by-step explanation:

A recursive rule is a formula that in which each term is expressed as a function of its preceding term(s), meaning in order to get to the nth term you have to express it in a form of the term that comes before it. In our case the a(n-1) term

So for the sequence -9, -2, 5, 12

The nth term is any number on the sequence and

  • -2 is the a(n-1) term for -9
  • 5 is the a(n-1) term for -2
  • 12 is the a(n-1) term for 5

So we need to find out what we have to do to the preceding term to get the next.

To get -2 from -9 we have to add 7 to -9; -9+7=-2

To get 5 from -2 we have to add 7 to -2; -2+7=5

To get 12 from 5 we add 7 to 5; 7+5=12

So the recursive rule would be= a n-1+7

5 0
3 years ago
Find the slope and y-intercept of the line that is perpendicular to y=3x-3 and passes through the point (-8,-2)
kiruha [24]
I think this is right

3 0
3 years ago
If a, b, c are in A.P. show that<br>a (b + c)/bc,b(c + a) /ca, c(a-b )/bc<br>are in A.P.<br>​
vfiekz [6]

Answer:

Step-by-step explanation:

\frac{a(b+c)}{bc} ,\frac{b(c+a)}{ca} ,\frac{c(a+b)}{ab} ~are~in~A.P.\\if~\frac{ab+ca}{bc} ,\frac{bc+ab}{ca} ,\frac{ca+bc}{ab} ~are~in~A.P.\\add~1~to~each~term\\if~\frac{ab+ca}{bc} +1,\frac{bc+ab}{ca} +1,\frac{ca+bc}{ab} +1~are~in~A.P.\\if~\frac{ab+ca+bc}{bc} ,\frac{bc+ab+ca}{ca} ,\frac{ca+bc+ab\\}{ab} ~are~in~A.P.\\\\divide~each~by~ab+bc+ca\\if~\frac{1}{bc} ,\frac{1}{ca} ,\frac{1}{ab} ~are ~in~A.P.\\if~\frac{a}{abc} ,\frac{b}{abc} ,\frac{c}{abc} ~are~in~A.P.\\if~a,b,c~are~in~A.P.\\which~is~true.

3 0
3 years ago
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