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shtirl [24]
3 years ago
10

After you have the system all lined up, what do you want to look for or create?

Mathematics
1 answer:
vfiekz [6]3 years ago
8 0
The answer is b. Hope this helps
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Factorise fully 2ab^3+ 8a^2b
fredd [130]

That is the answer :)

8 0
3 years ago
What polynomial has roots of −6, 1, and 4?
tester [92]

Answer:

The correct option is C.

Step-by-step explanation:

polynomial has roots of −6, 1, and 4.

We can write it as:

(x+6)(x-1)(x-4)

Now multiply the terms:

First multiply first and second bracket:

{x(x+6)-1(x+6)}(x-4)

{(x^2+6x-x-6)} (x-4)

Solve the like terms:

{(x^2+5x-6)}(x-4)

x(x^2+5x-6) -4(x^2+5x-6)

x^3+5x^2-6x-4x^2-20x+24

Solve the like terms:

x^3+x^2-26x+24

Hence it is proved that the correct option is C....

6 0
3 years ago
Alisha has a $15,000 car loan with a 6 percent interest rate that is compounded annually. How much will she have paid at the end
Harlamova29_29 [7]
6 percent intrest per year for 5 years I think is the same as 30 percent for 5 years
so $15,000 x .3 is $4500
$15,000 + $4500 = $19,500
not 100 percent sure if this is correct but I think its right. 
Sorry if I am wrong not 100% sure.

8 0
3 years ago
The rectangle below has an area of 8x^5+12x^3+20x^28x 5 +12x 3 +20x 2 8, x, start superscript, 5, end superscript, plus, 12, x,
neonofarm [45]

Answer:

The width is: w(x)=4x^2

The length is l(x)=2x^2+3x+5

Step-by-step explanation:

The given rectangle has area given algebraically by the function:

a(x)=8x^5+12x^3+20x^2

The width of the rectangle is the greatest common factor of 8x^5,  12x^3 and 20x^2

That is the width is: w(x)=4x^2

We now divide the area by the width to obtain the length of the rectangle:

l(x)=\frac{8x^5+12x^3+20x^2}{4x^2}

This simplifies to:

l(x)=\frac{8x^5}{4x^2}+\frac{12x^3}{4x^2}+\frac{20x^2}{4x^2}

l(x)=2x^2+3x+5

8 0
3 years ago
Read 2 more answers
A day on saturn is 42.6% of a day on earth. how many hours are there in a day on saturn?
raketka [301]

Answer:

10 full hours (10.224 is exact)

Step-by-step explanation:

4 0
3 years ago
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