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3 years ago

Find the sum and classify the polynomial based on degree and number of terms.

1 answer:

8 0

$3n^2(5m^2 -2n + 1) + 4m^2 - 11n^4 - 9 = 15m^2n^2 - 6n^3 + 3n^2 + 4m^2 - 11n^4 - 9 =\\15m^2n^2 -11n^4 -6n^3 + 3n^2 + 4m^2 - 9.\\$The degree with respect to the variables m, y is 2.\\With respect to m is 2, with respect to n is 4$

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Create a quadratic polynomial function f(x) and a linear binomial in the form (x − a).

Tamiku [17]

Our quadratic polynomial function is:

f(x) = 4x^2 + 2x + 5

our linear binomial form is:
x - a = -2
hence:
a = x + 2

and so we have:
f(x) = 4x^2 + 2x + 5
f(a) = 4(x + 2)^2 + 2<span>(x + 2)</span> + 5
= 4(x^2 + 4x + 4) + 2x + 4 + 5
f(a) = 4x^2 + 18x + 25

3 0

4 years ago

Ludmilka [50]

28

Explanation: 728/26=28

6 0

3 years ago

Read 2 more answers

3x3x-5y-7=0 <br> X=2k-1. <br> Y=k

pshichka [43]

Answer:

Step-by-step explanation:

We know that the system of equations

x+b

y=c

x+b

y=c

has infinitely many solutions, if

Here, a

=2,b

=3,c

=4,a

=k+2,b

=6,c

=3k+2.

Therefore, the given system of equations will have infinitely many solutions, if

k+2

3k+2

⇒

k+2

and

3k+2

⇒

k+2

and

3k+2

⇒k+2=4 and 3k+2=8

⇒k=2 and k=2

⇒k=2

Hence, the given system of equations will have infinitely many solutions, if k=2.

8 0

3 years ago

When you divide a unit fraction by a whole number number, will the quotient be greater than or less than the unit fraction?

Salsk061 [2.6K]

Answer:

it will be less than

Step-by-step explanation:

3 0

4 years ago

Read 2 more answers

A company offers a raffle whose grand prize is a $35,000 new car. Additional prizes are a $1,100 television and $600 computer. T

anastassius [24]

Answer:

-$ 10.13

Step-by-step explanation:

Given,

The cost price of each ticket = $ 22,

So, the value of each ticket other than price ticket = - $ 22, ( negative sign shows loss,i.e. if we don't get the price we will have the loss of $ 22 )

Now, there are 3 tickets in which first is of $35,000 price, second is of $1,100 and third is of $ 600,

So, the value of first ticket = 35000 - 22 = $ 34978,

Value of second ticket = 1100 - 22 = $ 1078,

Value of third ticket = 600 - 22 = $ 578,

Also,

$\text{Probability}=\frac{\text{Favourable outcome}}{\text{Total outcome}}$

Hence, by the above information we can make a table for the given situation,

Number 2997 1 1 1

Price -$ 22 $ 34978 $ 1078 $ 578

Probability 2997/3000 1/3000 1/3000 1/3000

Therefore, the expected value of a ticket

$=-22\times \frac{2997}{3000}+34978\times \frac{1}{3000}+578\times \frac{1}{3000}$

$=-$10.126$

≈ - $ 10.13

5 0

3 years ago