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natulia [17]
2 years ago
12

Can you factor the common factor out of this problem? 3p+12q-15q^2r^2

Mathematics
2 answers:
motikmotik2 years ago
6 0
Greatest common factor is 3
ioda2 years ago
5 0
The answer after being factored is, -3( 5q^{2}  r^{2} -p-4q)

When we are factoring, all we are doing is finding a GCF and taking it from the expression and dividing it into every term. Let's break our expression into parts first

3p

12q

-15q^{2}  r^{2}

Now, each number is divisible by 3, but since our biggest number is a negative, we will make our GCF -3 instead of 3. So let's divide each part by -3.

3p / -3 =-p

12q/-3=-4q

-15q ^{2}  r^{2} /-3=5q^{2}  r^{2}

Now let's put it together in an expression:

5q^{2}  r^{2} -p-4q

When we factor we always put our factored expression in parenthesis with the term the expression was factored by on the outside like so:


-3(5q^{2} r^{2} -p-4q)

Thank you for the question I hope this helped! Have an amazing day! :D
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In your own words explain how you would solve this two step equation for the variable x. 9x-7=-7
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Add 7 to both sides

Step-by-step explanation:

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What is factor ? someone explain pls !!
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Step-by-step explanation:

Factor, in mathmatechis, a number or algebric expression that divides another number or expression evenly.

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Solve the system by using a matrix equation (Picture provided)
Katyanochek1 [597]

Answer:

Option b is correct (8,13).

Step-by-step explanation:

7x - 4y = 4

10x - 6y =2

it can be represented in matrix form as\left[\begin{array}{cc}7&-4\\10&-6\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}4\\2\end{array}\right]

A= \left[\begin{array}{cc}7&-4\\10&-6\end{array}\right]

X= \left[\begin{array}{c}x\\y\end{array}\right]

B= \left[\begin{array}{c}4\\2\end{array}\right]

i.e, AX=B

or X= A⁻¹ B

A⁻¹ = 1/|A| * Adj A

determinant of A = |A|= (7*-6) - (-4*10)

                                    = (-42)-(-40)

                                    = (-42) + 40 = -2

so, |A| = -2

Adj A=  \left[\begin{array}{cc}-6&4\\-10&7\end{array}\right]

A⁻¹ =  \left[\begin{array}{cc}-6&4\\-10&7\end{array}\right]/ -2

A⁻¹ =  \left[\begin{array}{cc}3&-2\\5&-7/2\end{array}\right]

X= A⁻¹ B

X=  \left[\begin{array}{cc}3&-2\\5&-7/2\end{array}\right] *\left[\begin{array}{c}4\\2\end{array}\right]

X= \left[\begin{array}{c}(3*4) + (-2*2)\\(5*4) + (-7/2*2)\end{array}\right]

X= \left[\begin{array}{c}12-4\\20-7\end{array}\right]

X= \left[\begin{array}{c}8\\13\end{array}\right]

x= 8, y= 13

solution set= (8,13).

Option b is correct.

3 0
3 years ago
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