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Mandarinka [93]

2 years ago

7

Which of these operations is not closed for polynomials?

1 answer:

alexira [117]2 years ago

6 0

Answer:

B. Division

Step-by-step explanation:

Subtraction is closed for polynomials subtracting two polynomials will result in another polynomial. Same goes with multiplication. However, if you divide polynomials, you are able to get a result that is not always a polynomial. Therefore, division is the answer.

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Add, using the rule for order of operations if necessary:<br> 15+ [2 + (-4)] + (−9)?

schepotkina [342]

$15 + ( - 2) + - 9 = 13 - 9 = 4$

#Goodluck!

8 0

1 year ago

What is factored form of 4x3 + 3x2 + 8x + 6?

Ira Lisetskai [31]

The answer is (4x+3) (x^2+2)

3 0

3 years ago

Cylinder and a cone have the same diameter: 8 inches. The height of the cylinder is 5 inches. The height of the cone is 15 inche

ss7ja [257]

Answer:

$V_{cylinder}=251.2\ in^3,$

$V_{cone}=251.2\ in^3,$

$V_{cylinder}=V_{cone}.$

Step-by-step explanation:

The volume of the cylinder is

$V_{cylinder}=\pi r^2 h,$

where r is the radius of a base circle and h is the cylinder's height.

If cylinder has diameter of 8 inches, then its radius is 4 inches. The height of the cylinder is 5 inches. Thus, the volume of the cylinder is

$V_{cylinder}=3.14\cdot 4^2\cdot 5=251.2\ in^3.$

The volume of the cone is

$V_{cone}=\dfrac{1}{3}\pi r^2 h,$

where r is the radius of a base circle and h is the cone's height.

If cone has diameter of 8 inches, then its radius is 4 inches. The height of the cone is 15 inches. Thus, the volume of the cone is

$V_{cylinder}=\dfrac{1}{3}\cdot 3.14\cdot 4^2\cdot 15=251.2\ in^3.$

As you can see the cylinder and the cone have the same volume.

4 0

3 years ago

An oil tanker can be emptied by the main pump in 5 hours. An auxiliary pump can empty the tanker in 7 hours. If the main pump is

r-ruslan [8.4K]

Answer:

so the answer will be 1

Step-by-step explanation:

An oil tanker can be emptied by the main pump in 5 hours.

An auxilary pump can empty the tanker in 14 hours.

If the main pump is started at 7pm, when should the auxilary pump be started

so that the tanker is emptied by 11pm?

:

Let t = no. of hrs to run the Aux pump

:

The main pump will run 4 hr (7 - 11 pm)

:

Let the completed job = 1; (an empty Tanker)

:

The shared work equation

4%2F5 + t%2F14 = 1

Multiply equation by 70 to get rid of the equation, results:

14(4) + 5t = 70

56 + 5t = 70

5t = 70 - 56

5t = 14

t = 14%2F5

t = 2.8 hrs to run the aux pump

:

2.8 hr = 2 + .8(60) = 2 hrs 48 min

:

Subtract 2:48 from 11:00 = 8:12 PM start the aux pump

;

;

Check solution

4/5 + 2.8/14 =

.8 + .2 = 1

6 0

3 years ago

Rasek [7]

Answer:

B. 1

Step-by-step explanation:

Multiply the first term by $x^a$ , the second term by $x^b$ , the third term by $x^c$ .

<u>The first term becomes:</u>

$x^a/(x^a+x^b+x^c)$

<u>The second term becomes:</u>

$x^b/(x^a+x^b+x^c)$

<u>The third term becomes:</u>

$x^c/(x^a+x^b+x^c)$

<u>Their sum is:</u>

$(x^a+x^b+x^c)/(x^a+x^b+x^c) = 1$

Correct choice is B

3 0

2 years ago