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4 years ago

You have a pile of 20 dimes AND quarters which total $2.75. How many quarters do you have?

1 answer:

5 0

you have five quarters you didn't read the question right i had it before in quarters you have 5 = 1.25 en dimes you have 15= 1.50

1.25+1.50=2.75

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Please help! I am stuck!

Mrac [35]

the answer is D the last one,

the reason is just that any negative number is automatically smaller than a positive number. so -2.8 is smaller than even the smallest decimal possible as long as that decimal is positive. 1.4 is actually about 3 time more that -2.8

(both are still freezing cold)

5 0

3 years ago

Find the length of the missing side. Please help!

Dvinal [7]

Answer:

$b=32$ or $b=\sqrt{1025}$

Step-by-step explanation:

$a^2+b^2=c^2\\20^2+b^2=25^2\\b^2=25^2-20^2\\\sqrt{b^2=1025}\\$

$b=32$ or $b=\sqrt{1025}$

8 0

3 years ago

Use the zero product property to find the solutions to the equation 2x2 + x – 1 = 2.

FromTheMoon [43]

$2 {x}^{2} + x - 1 = 2$

Subtract sides -2

$2 {x}^{2} + x - 1 - 2 = 2 - 2$

$2 {x}^{2} + x - 3 = 0$

$(x - 1)(2x + 3) = 0$

_________________________________

$x - 1 = 0$

$x = 1$

##############################

$2x + 3 = 0$

$x = - \frac{3}{2} \\$

_________________________________

Done♥️♥️♥️♥️♥️

3 0

3 years ago

A closed box with a square base is to have a volume of 171 comma 500 cm cubed. The material for the top and bottom of the box co

Zepler [3.9K]

Answer:

$C(x)=\dfrac{20x^3+1715000}{x}\\$Minimum cost, C(35)=\$29,400$

The dimensions that will lead to minimum cost of the box are a base length of 35 cm and a height of 140 cm.

Step-by-step explanation:

Volume of the Square-Based box=171,500 cubic cm

Let the length of a side of the base=x cm

Volume $=x^2h$

$x^2h=171,500\\h=\dfrac{171500}{x^2}$

The material for the top and bottom of the box costs $10.00 per square centimeter.

Surface Area of the Top and Bottom $=2x^2$

Therefore, Cost of the Top and Bottom $=\$10X2x^2=20x^2$

The material for the sides costs $2.50 per square centimeter.

Surface Area of the Sides=4xh

Cost of the sides=$2.50 X 4xh =10xh

$\text{Substitute h}$=\dfrac{171500}{x^2} $into 10xh\\Cost of the sides=10x(\dfrac{171500}{x^2})=\dfrac{1715000}{x}$

Therefore, total Cost of the box

$= 20x^2+\dfrac{1715000}{x}\\C(x)=\dfrac{20x^3+1715000}{x}$

To find the minimum total cost, we solve for the critical points of C(x). This is obtained by equating its derivative to zero and solving for x.

$C'(x)=\dfrac{40x^3-1715000}{x^2}\\\dfrac{40x^3-1715000}{x^2}=0\\40x^3-1715000=0\\40x^3=1715000\\x^3=1715000\div 40\\x^3=42875\\x=\sqrt[3]{42875}=35$

Recall that:

$h=\dfrac{171500}{x^2}\\Therefore:\\h=\dfrac{171500}{35^2}=140cm$

The dimensions that will lead to minimum costs are base length of 35cm and height of 140cm.

Therefore, the minimum total cost, at x=35cm

$C(35)=\dfrac{20(35)^3+1715000}{35}=\$29,400$

8 0

3 years ago

The data shows the weights of 2 groups of house cats in pounds.

lesantik [10]

I would say the answer is C

5 0

3 years ago