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olya-2409 [2.1K]

3 years ago

7

write a sample problem in which you factor a monomial from a polynomial. Then write and factor another problem that is a quadrat

ic expression.

1 answer:

k0ka [10]3 years ago

4 0

First answer:
-1 + x^4
result:
(x-1)(x+1)(x^2+1)

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Bob can paint a fence in 12 hours, and working with Jen can paint a fence in 4 hours. How long would it have taken Jen to paint

butalik [34]

Answer:

6 hours

Step-by-step explanation:

For Bob

12 hours = 1 fence

1 hour = 1/12 fence

Together Bob and Jen

4 hours = 1 fence

1 hour = 1/4 fence

Now, for Jen:

1 hour = 1/4 - 1/12= 3/12 - 1/12 = 2/12=1/6

Find the amount of time Jen will need:

1/6 fence = 1 hour

1 fence = 6 hour

8 0

3 years ago

what is an equation in the point-slope form of the line that passes through (-3,-1) and has a slope of 2?

LuckyWell [14K]

Answer:

well it would have to be y=2x-3

Step-by-step explanation:

yup

4 0

3 years ago

Find the mean and mode for this set of data.<br>5, 6, 8, 3, 5, 7, 4, 11, 5, 7

dangina [55]

Answer:the mean is 6.1 the mode is 5 and 7.

Step-by-step explanation: the mean is 6.1 because the number add up to 61 and there are 10 numbers 61/10=6.1

The mode is 5 and 7 because they occur most frequently

3 0

3 years ago

Helppppp!!!! Ty ✨:)))))))))

LiRa [457]

Answer:

12/7 or 1 5/7

Step-by-step explanation:

6 * 2 = 12

so

12/7 = 1 5/7

8 0

3 years ago

Read 2 more answers

A piece of cardboard is 15 inches by 30 inches. A square is to be cut from each corner and the sides folded up to make an open-t

solmaris [256]

Answer:

Maximum volume = 649.519 cubic inches

Step-by-step explanation:

A rectangular piece of cardboard of side 15 inches by 30 inches is cut in such that a square is cut from each corner. Let x be the side of this square cut. When it was folded to make the box the height of box becomes x, length becomes (30-2x) and the width becomes (15-2x).

Volume is given by

V = $V = Length\times Width\times Height\\V = (30 - 2x)(15-2x)x= 4x^3-90x^2+450x\\So,\\V(x) = 4x^3-90x^2+450x$

First, we differentiate V(x) with respect to x, to get,

$\frac{d(V(x))}{dx} = \frac{d(4x^3-12x^2+9x)}{dx} = 12x^2 - 180x +450$

Equating the first derivative to zero, we get,

$\frac{d(V(x))}{dx} = 0\\\\12x^2 - 180x +450 = 0$

Solving, with the help of quadratic formula, we get,

$x = \displaystyle\frac{5(3+\sqrt{3})}{2}, \frac{5(3-\sqrt{3})}{2}$ ,

Again differentiation V(x), with respect to x, we get,

$\frac{d^2(V(x))}{dx^2} = 24x - 180$

At x =

$\displaystyle\frac{5(3-\sqrt{3})}{2}$ ,

$\frac{d^2(V(x))}{dx^2} < 0$

Thus, by double derivative test, the maxima occurs at

x = $\displaystyle\frac{5(3-\sqrt{3})}{2}$ for V(x).

Thus, largest volume the box can have occurs when $x = \displaystyle\frac{5(3-\sqrt{3})}{2}}$ .

Maximum volume =

$V(\displaystyle\frac{5(3-\sqrt{3})}{2}) = (30 - 2x)(15-2x)x = 649.5191\text{ cubic inches}$

8 0

3 years ago