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3 years ago

What is the quotient in the polynomial form?

Mathematics

1 answer:

Oxana [17]3 years ago

8 0

First step of a synthetic divison is that we need to carry down the leading coefficient. Here the leading coefficient is 2. So, carry down 2 at the bottom.

Next step is to multiply the divisor -3 with this carry down number 2. So, we have got 3*(-2)= -6 which will place atthe bottom of the next coefficient 4.

Next step is to add this column.

Now repeat the same method again till the last colum.

At the end we have got 0 after the addition. Which means the remainder is 0.

So, the quotient is 2x^2-2x+2.

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How do you reduce 36/100 using prime factorization?

Nutka1998 [239]

36 and 100 are both divisible by 4 which gives 9 over 25. That's all you can do to simplify it.
Hope this helps!

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3 years ago

A box of chocolates contains five milk chocolates, five dark chocolates, and five white chocolates. You randomly select and eat

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2 years ago

For a craft each student will need five rubber bands.There.There are eight students.Rubber bands come in bags of nine. How many

matrenka [14]

Answer:

40 rubber bands are needed

5 rubber bands will be left from 5 bags of of rubber band

Step-by-step explanation:

Each student will need five rubber bands.

There are eight students

Total rubber bands needed = rubber bands per student × number of students

= 5 × 8

= 40

Total rubber bands needed = 40

Rubber bands come in bags of nine.

Number of bags needed = Total rubber bands needed / number of rubber bands in each bag

= 40 / 9

= 4.44 bags

It is impossible to get decimal number of rubber bands bag, so, the next whole number after 4 is 5

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= 45 rubber bands

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Left over rubber bands = 45 - 40

= 5 rubber bands

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3 years ago

Vocabulary Is every relation also a function? Explain.

mash [69]

Answer:

It is B

Step-by-step explanation:

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3 years ago

Read 2 more answers

Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview

svetlana [45]

Answer: The required solution is

$y=(-2+t)e^{-5t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.$

So, the general solution of the given equation is

$y(t)=(A+Bt)e^{-5t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.$

According to the given conditions, we have

$y(0)=-2\\\\\Rightarrow A=-2$

and

$y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.$

Thus, the required solution is

$y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.$

6 0

3 years ago