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mixer [17]
2 years ago
6

PLZZZZZZzzzzzZzZzZzZzZZz BRAINLIESTTTTTTTT

Mathematics
2 answers:
Cloud [144]2 years ago
5 0

Answer:

A) 3, B) 8 and 4, C) 12

Step-by-step explanation:

There are 3 terms: 8x^2, 4x, and 12.

The coefficiemts are numbers that come before a variable. That's why the coefficients are 8 and 4.

The constant is the number that doesn't come with a variable, thus 12.

olasank [31]2 years ago
3 0

Answer:

Part A: 3 terms

Part B: 8 and 4

Part C: 12

Step-by-step explanation:

There are 3 terms, each separated by a + sign.

8 and 4 both have variables attached to them, so they are the coefficients.

12 is the constant because it doesn't have any variables attached to it.

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Find the sum: (3x2 + 5x − 8) + (5x2 − 13x − 5)
musickatia [10]

Answer:

(3x^2+5x-8)+(5x^2-13x-5)=8x^2-8x-13

Step-by-step explanation:

The given polynomials are,

3x^2+5x-8,5x^2-13x-5

In polynomial addition and subtraction only like terms can be added or subtracted.

So,

=(3x^2+5x-8)+(5x^2-13x-5)

=3x^2+5x-8+5x^2-13x-5

=3x^2+5x^2+5x-13x-8-5

=8x^2-8x-13

7 0
3 years ago
Read 2 more answers
What polynomial has roots of -4, 1, and 6?
Delicious77 [7]
X^3-3x^2-22x+24
the first option is the answer

solution: (x + 4)(x - 1)(x -6) = (x^2 - x + 4x -4)(x - 6)
(x^2 + 3x -4)(x - 6) = x^3 -6x^2 + 3x^2 - 18x - 4x + 24 = x^3 - 3x^2 - 22x + 24
3 0
3 years ago
What is the area of this figure?<br> 6 mi<br> 5 mi<br> 2 mi<br> 2 mi<br> 11 mi<br> 9 mi<br> ורן 7
Alborosie
I can’t see the image maybe if you redo it
5 0
3 years ago
The bookstore at SunTan U had a sale on school t-shirts. Thursday, they sold 52 of the marked down t-shirts for a total of $482.
inysia [295]

Answer:

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Step-by-step explanation:.........................................................................................................................

7 0
3 years ago
Prove that an = 4^n + 2(-1)^nis the solution to
olga nikolaevna [1]

Answer:

See proof below

Step-by-step explanation:

We have to verify that if we substitute a_n=4^n+2(-1)^n in the equation a_n=3a_{n-1}+4a_{n-2} the equality is true.

Let's substitute first in the right hand side:

3a_{n-1}+4a_{n-2}=3(4^{n-1}+2(-1)^{n-1})+4(4^{n-2}+2(-1)^{n-2})

Now we use the distributive laws. Also, note that (-1)^{n-1}=\frac{1}{-1}(-1)^n=(-1)(-1)^{n} (this also works when the power is n-2).

=3(4^{n-1})+6(-1)^{n-1}+4(4^{n-2})+8(-1)^{n-2}

=3(4^{n-1})+(-1)(6)(-1)^{n}+4^{n-1}+(-1)^2(8)(-1)^{n}

=4(4^{n-1})-6(-1)^{n}+8(-1)^{n}=4^n+2(-1)^n=a_n

then the sequence solves the recurrence relation.

4 0
3 years ago
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