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olga2289 [7]
3 years ago
13

How to Factor 3x^2-x-10

Mathematics
1 answer:
adoni [48]3 years ago
5 0

Solution

( − 2)(3 + 5)

Step-by-step solution:

Use the sum-product pattern

3x^2−−10

3^2+5−6x−10

Common factor from the two pairs

3^2+5−6−10

(3+5)−2(3+5)

Rewrite in factored form

(3+5)−2(3+5)

(−2)(3+5

hope this was good! <3

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Which equation represents the line that passes through the point (4, -5) and is perpendicular to the line x + 2y = 5?
Lerok [7]

Answer:

y = 2x - 13

Step-by-step explanation:

Equation of a line is y = mx + c, m is the gradient and c is the intercept

The line passes through points 4 and -5, x is 4 and y is -5

-5 = 4m + c

When two lines are perpendicular, the products of their gradients are equal to -1, m1 * m2 = -1

x + 2y = 5

2y = -x + 5

y = (-1/2 * x) + 5

therefore m = -1/2

m1 * m2 = -1

m * -1/2 = -1

-m = -2 , therefore m = 2

-5 = 4 * 2 + c

c = -5 - 8, which is -13

Therefore the equation for the line is

y = 2x - 13

7 0
3 years ago
Thirty times the square of a non-zero number is equal to 8 times the number. What is the number?
astra-53 [7]
Hello here is a solution :

4 0
3 years ago
Find the number between 30 and 70 that is divisible by 8, and when divided by 5 has a remainder of 1
monitta
30+70÷8÷5-1=
=30.75
hope this helps
7 0
3 years ago
6y-5=11 what the answer I’m having a hard time finding it
QveST [7]

6y-5=11

Move -5 to the other side. Sign changes from -5 to +5.

6y-5+5=11+5

6y=11+5

6y=16

Divide by 6 for both sides to get y by itself.

6y/6=16/6

Cross out 6 and 6, divide by 6 and then becomes 1*1*y=y

y=16/6

Reduce 16/6 by dividing by 2

16/2=8

6/2=3

Answer: y=8/3 or y=2 2/3

6 0
3 years ago
Read 2 more answers
Write (x+7)(x−2) as a sum
Dennis_Churaev [7]

Answer:

x^2+5x-14

Step-by-step explanation:

you multiply both equations so that you get:

x^2-2x+7x-14

and then you simplify it to:

x^2+5x-14

<em><u>If you found this helpful please give it a thanks </u></em>

<em><u>I would also appreciate it if you give me a brainliest</u></em>

8 0
3 years ago
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