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3 years ago

A purse contains 21 coins consisting of nickels and dimes. Find the number of coins of each kind if the total value is $1.65.

Mathematics

1 answer:

MaRussiya [10]3 years ago

4 0

the value of the dimes is 10(21 - 5)

hope this helps

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34÷3=? dcvsfregbrfdewqsdcfvgrbtrfdcfvgb

castortr0y [4]

Answer: 11.33333333333333

8 0

3 years ago

Read 2 more answers

2.3 to 1 significant figure

azamat

☽------------❀-------------☾

Hi there!

~

2.3 to 1 significant figure.

= 2

❀Hope this helped you!❀

☽------------❀-------------☾

3 0

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 $16.1\times 7.1\times 2.45$

6 0

3 years ago

Pls I’m about to cry...

marissa [1.9K]

FORMULA:

Volume of cylinder = πr²h
Volume of sphere = 4/3πr³

ANSWER:

Volume of capsule = 4/3πr³ + πr²h

πr²(4/3r + h)

3.14 × 2²(8/3 + 5.6)
3.14 × 4(2.6 + 5.6)
12.56(8.2)
102.99 or 103 rounded.

Now, Density = 0.7mg/mm³

0.7/103
0.00679mg

7 0

3 years ago

PLEAS HELP ASAP :)))))

Jet001 [13]

Answer:

18 inches

Step-by-step explanation:

54*1/3=18(inches)

8 0

3 years ago