What is the range of this function?

Which is true about the degree of the sum and difference of the polynomials 3x5y – 2x3y4 – 7xy3 and –8x5y + 2x3y4 + xy3?

Alekssandra [29.7K]

<span>First we have to find the sum and the difference of those polynomials- The sum is: ( 3 x^5y - 2 x^3y^4 - 7 xy^3 ) + ( - 8 x^5y + 2 x^3y^4 + xy^3 ) = 3 x^5 - 2 x^3y^4 - 7xy^3 - 8 x^5y + 2 x^3y^4 + xy^3 = - 5 x^5y - 6 xy^3. And the difference: ( 3 x^5y - 2 x^3y^4 - 7 xy^3 ) - ( - 8 x^5y + 2 x^3y^4 + xy^3 ) = 3 x^5y - 2 x^3y^4 - 7 xy^3 + 8 xy^5 - 2 x^3y^4 - xy^3 = 11 xy^5 - 4 x^3y^4 - 8xy^3. The highest exponent in both polynomials is 5. Answer: The degree of the polynomials is 5.</span>

7 0

3 years ago

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Ms. Steven creates a drawing in her kitchen. Her drawing is 7cm in length and 4cm in width. The scale on the drawing is 1cm = 3f

marysya [2.9K]

Answer:

7cm=7×3ft=21ft

4cm=4×3ft=12ft

here

length=21ft

breadth=12ft

now

area= l×b

=21ft×12ft

=252ft^2

6 0

3 years ago

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Help with this<br> twentycharacters

olya-2409 [2.1K]

Answer:

f(x) = 150-25x

Step-by-step explanation:

200-50=150.

Take the 150 and insert it there.

Each lesson is $25, so you multiply 25 by 2 every lesson. Example= 150-25=125-25=100-75 and so on.

6 0

3 years ago

Suppose yoko borrows $3500 at an interest rate of 4% compounded each year Find the amount owed at the end of 1 year

olchik [2.2K]

Amount owed at the end of 1 year is 3640

<h3><u>Solution:</u></h3>

Given that yoko borrows $3500.

Rate of interest charged is 4% compounded each year

Need to determine amount owed at the end of 1 year.

In our case :

Borrowed Amount that is principal P = $3500

Rate of interest r = 4%

Duration = 1 year and as it is compounded yearly, number of times interest calculated in 1 year n = 1

<em><u>Formula for Amount of compounded yearly is as follows:</u></em>

$A=p\left(1+\frac{r}{100}\right)^{n}$

Where "p" is the principal

"r" is the rate of interest

"n" is the number of years

Substituting the values in above formula we get

$\mathrm{A}=3500\left(1+\frac{4}{100}\right)^{1}$

$\begin{array}{l}{A=\frac{3500 \times 104}{100}} \\\\ {A=35 \times 104} \\\\ {A=3640}\end{array}$

Hence amount owed at the end of 1 year is 3640

5 0

3 years ago

The height of Mountain P is 1,086 feet.The height of Mountain Q is 4 times the height of Mountain P.The area model shown below r

earnstyle [38]

Answer:

A=4000, B=80, C=24

Step-by-step explanation:

You forgot to put the correct area model, I attached it to the answer.

We have the fact that Mountain Q is 4 times the height of Mountain P. That's the "4" we have in the left side of our model. It's like having a multiplication table, next to the "4" we have "A" and upper this we have "1000", the only thing we have to do is multiplify 4*1000=4000. The next letter we have is B and below it we have "320", we divided it by 4, 320/4=80. The last letter we have is C, and is below a "6", we only have to multiplify it by 4, 6*4=24.

At the end we only sum our

A + 320 + c = 4344 (4 times the height of Mountain P).
1000 + B + 6 = 1086(the height of the Mountain P).

8 0

3 years ago