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ser-zykov [4K]
3 years ago
13

Angelica is using synthetic division to divide 2x^3-9x^2-21x+10 by x+2 Which answer is correct?

Mathematics
2 answers:
NeTakaya3 years ago
7 0

Answer:

Option 1 is correct

Step-by-step explanation:

Given:

Polynomial: 2x^3-9x^2-21x+10

Divide by x+2

Synthetic division: Divide a polynomial by a polynomial.

Write the coefficient of polynomial in highest degree to lowest degree.

Factor x+2 , divide by -2

Set of synthetic division:

-2  | 2     -9     -21       10 |

              -4       26      -10

    | 2     -13      5        0 |

First put down 2 at bottom row and then multiply by -2 and write down -9. Add -9 and 4 .

Repeat steps till end.

statuscvo [17]3 years ago
3 0

Answer:

A

Step-by-step explanation:

Argument

C and D are wrong right off the top. When you have x + 2 in synthetic division the divisor (2) must change signs.

B is incorrect because you don't change the signs of the polynomial that is being divided into. So that only leaves A and it is correct.

Summary

  • Change the sign of the binomial doing the dividing (2 goes to - 2)
  • Leave the polynomial's coefficients alone.
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Marty made a $220 bank deposit using $10 bills and $5 bills. She gave the teller a total of 38 bills, how many $5 bills were in
Luda [366]

ANSWER: 32 five-dollar bills

======

EXPLANATION:

Let x be number of $5 bills

Let y be number of $10 bills

Since we have total of 38 bills, we must have the sum of x and y be 38

x + y = 38 (I)

Since the total amount deposited is $220, we must have the sum of 5x and 10y be 220 (x and y are just the "number of" their respective bills, so we multiply them by their value to get the total value):

5x + 10y = 220 (II)

System of equations:

\left\{ \begin{aligned} x + y &= 38 && \text{(I)} \\ 5x + 10y &= 220 && \text{(II)} \end{aligned} \right.

Divide both sides of equation (II) by 5 so our numbers become smaller

\left\{ \begin{aligned} x + y &= 38 && \text{(I)} \\ x + 2y &= 44 && \text{(II)} \end{aligned} \right.

Rearrange (I) to solve for y so that we can substitute into (II)

\begin{aligned} x + y &= 38 && \text{(I)} \\ y &= 38 - x \end{aligned}

Substituting this into equation (II) for the y:

\begin{aligned} x + 2y &= 44 && \text{(II)} \\ x + 2(38 - x) &= 44\\ x + 76 - 2x &= 44 \\ -x &= -32 \\ x &= 32 \end{aligned}

We have 32 five-dollar bills

======

If we want to finish off the question, use y = 38 - x to figure out number of $10 bills

y = 38 - 32 = 6

32 five-dollar bills and 6 ten-dollar bills

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3 years ago
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