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2 years ago

Help please!

Mathematics

1 answer:

ki77a [65]2 years ago

7 0

Answers:

5 two-pointers
11 three-pointers

=====================================================

Explanation:

x = number of two-pointers made

2x+1 = number of three-pointers made

This is because of the fact that she made one more than twice as many three-pointers than two-pointers.

In other words,

number of three pointers = 2*(number of two pointers) + 1

number of three pointers = 2x + 1

-------------------

If she made x two-pointers, then she gets 2x points from just those made shots.

If Nikki made 2x+1 three-pointers, then she gets 3(2x+1) = 6x+3 points from just the stuff beyond the arc.

In total, she earns (2x)+(6x+3) = 8x+3 points from both types of shots.

-------------------

Set this equal to the stated total 43 points and solve for x.

8x+3 = 43

8x = 43-3

8x = 40

x = 40/8

x = 5

Nikki made 5 two-pointers and 2x+1 = 2*5+1 = 11 three-pointers

--------------------

Check:

A = points from two-pointers only = 2*5 = 10 points

B = points from three-pointers only = 3*11 = 33 points

A+B = total points = 10+33 = 43 points

The answer has been confirmed.

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3 years ago

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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 $16.1\times 7.1\times 2.45$

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