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3 years ago

-2s(5st+3t) need help with problem

1 answer:

7 0

The answer to this problem is -10x^2t-6st to get this answer or this problem you must use the product rule formula x^a x^b=x^a=b

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a fleece sitting on the very top of the minute hand on a grandfather clock. the minute hand measures 6 inches long. how far does

lawyer [7]

The minute hand makes 24 revolutions in one day, so the distance traveled is
.. 24*2π*r
.. = 24*2*π*(6 in) = 288π in ≈ 904.78 inches

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4 years ago

Find the smallest number by which 6615 is to be multiplied by to get a perfect square

Ipatiy [6.2K]

Answer:

3x 5

Step-by-step explanation:

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2 years ago

A department store advertised that for every $25 you spend at the store,

Semmy [17]

His claim is resonable 154/25=6 r4 6x2=12

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3 years ago

Read 2 more answers

Which of the following situations WOULD NOT represent a binomial application? A. Choosing a card randomly from a standard deck a

SVETLANKA909090 [29]

Answer:

Choosing a card randomly and noting its suit

Step-by-step explanation:

Choosing a card randomly and noting its suit

This is because binomial distributions only work for bernoulli trials (a trail in which there are only two outcomes)

4 0

3 years ago

Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 21 in. by 12 in. by

a_sh-v [17]

Answer:

Dimension of the box is $16.1\times 7.1\times 2.45$

The volume of the box is 280.05 in³.

Step-by-step explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the box?

Solution :

Let h be the height of the box which is the side length of a corner square.

According to question,

A sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.

The length of the box is $L=21-2h$

The width of the box is $W=12-2h$

The volume of the box is $V=L\times W\times H$

$V=(21-2h)\times (12-2h)\times h$

$V=(21-2h)\times (12h-2h^2)$

$V=252h-42h^2-24h^2+4h^3$

$V=4h^3-66h^2+252h$

To maximize the volume we find derivative of volume and put it to zero.

$V'=12h^2-132h+252$

$0=12h^2-132h+252$

Solving by quadratic formula,

$h=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$h=\frac{-(-132)\pm\sqrt{132^2-4(12)(252)}}{2(12)}$

$h=\frac{132\pm72.99}{24}$

$h=2.45,8.54$

Now, substitute the value of h in the volume,

$V=4h^3-66h^2+252h$

When, h=2.45

$V=4(2.45)^3-66(2.45)^2+252(2.45)$

$V\approx 280.05$

When, h=8.54

$V=4(8.54)^3-66(8.54)^2+252(8.54)$

$V\approx -170.06$

Rejecting the negative volume as it is not possible.

Therefore, The volume of the box is 280.05 in³.

The dimension of the box is

The height of the box is h=2.45

The length of the box is $L=21-2(2.45)=16.1$

The width of the box is $W=12-2(2.45)=7.1$

So, Dimension of the box is $16.1\times 7.1\times 2.45$

6 0

4 years ago