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torisob [31]
3 years ago
10

Divide -2x^3-5x^2+4x+2 by x+2

Mathematics
1 answer:
erik [133]3 years ago
4 0

<u>Answer</u><u>: </u>

The correct option is C.  -2x^2-x+6, R(-10)

<u> Explanation: </u>

<u>Given:</u>

Dividend = -2x^3-5x^2+4x+2

Divider =(x+2)

<u>Solution</u>:

Now by using the division method , we will find out the quotient and remainder of the given equation  

And the equation will become  

-2x^3-5x^2+4x+2=(x+2)(-2x^2-x+6)-10

So , We will get the quotient =-2x^2-x+6

And Remainder = -10

Hence the Correct option which perfectly satisfy the given equation is C.

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Oduvanchick [21]
Infinite amount of solutions

See attached image

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Hope this helps :)

6 0
3 years ago
Please help me I’m crying right now
Semenov [28]

Answer:

24ft²

Step-by-step explanation:

Dude. Go back and relearn the concept man. This is the third time you've been asking the same subject.

Area of triangle = ½BH = ½(6)(8) = 24

5 0
4 years ago
Read 2 more answers
5x-3=21-x please someone answer ​
wolverine [178]

Answer:

X=4 Hope it helps Lol

Step-by-step explanation:

  5*x-3-(21-x)=0

Step by step solution :

Step  1  :

Pulling out like terms :

1.1     Pull out like factors :

  6x - 24  =   6 • (x - 4)

Equation at the end of step  1  :

Step  2  :

Equations which are never true :

2.1      Solve :    6   =  0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation :

2.2      Solve  :    x-4 = 0

Add  4  to both sides of the equation :

                     x = 4

One solution was found :

                  x = 4

5 0
3 years ago
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In a certain​ chemical, the ratio of zinc to copper is 4 to 19. A jar of the chemical contains 836 grains of copper. How many gr
VLD [36.1K]
136 grams of zinc

836/19=44
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3 0
3 years ago
Luis has 3 1/3 yards of ribbon to use for two separate presents, one of which is very large. If Luis uses 2 1/2 yards to decorat
never [62]

Answer:

\frac{5}{6} yards

Step-by-step explanation:

Luis had 3\frac{1}{3} yards of ribbon in total to use for two presents. For the first present Luis used 2\frac{1}{2} yards of the ribbon.

Since she had 3\frac{1}{3} yards in total and she used  2\frac{1}{2} yards from it, the amount of ribbon left for second present can be obtained by taking out(subtracting)  2\frac{1}{2} yards from 3\frac{1}{3}.

So, the number of yards of ribbon left for second present will be:

3\frac{1}{3}-2\frac{1}{2}\\\\ =\frac{10}{3}-\frac{5}{2}\\\\ =\frac{10(2)-5(3)}{2 \times 3}\\\\ =\frac{20-15}{6}\\\\ =\frac{5}{6}

This means Luis had \frac{5}{6} yards of ribbon for the second present.

5 0
4 years ago
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