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3 years ago

What is the volume of one cube with 1/2 inch sides

Mathematics

1 answer:

Anit [1.1K]3 years ago

8 0

Answer:

0.125 cubic inches

Step-by-step explanation:

The volume of a cube is just the length of a side length cubed. 0.5^3=0.125 in^3. Hope this helps!

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Jaheed is making puch for a party. The recipe calls for 3.5 liters of fruit juice. Write a formula to help jaheed determine the

Nimfa-mama [501]

$\pi x^{2} \frac{x}{y}$

5 0

3 years ago

In Exercise,find the balance in the account after 10 years.

djverab [1.8K]

Answer: The balance in the account after 10 years is $3374.65

Step-by-step explanation:

The exponential equation for growth [ compounded continuously] is

$y=pe^{rt}$

, where P= Present value

r= growth rate ( in decimal)

t= time (years)

By considering the given information , we have

p=$2500 r = % =0.03 and t= 10

Substitute all the values in the above equation , we get

$y=2500e^{0.03(10)}$

$y=2500e^{0.3}$

$y=2500(1.3498588)=3374.64701894\approx3374.65$ [Round to the nearest cent]

Therefore, the balance in the account after 10 years is $3374.65

3 0

2 years ago

Can someone help me with this?

Mandarinka [93]

Answer:

Step-by-step explanation:

Area of rectangle = length * width

= (x + 4) (5x)

= x *5x + 4 *5x

= 5x² + 20x

Perimeter of rectangle = 2*(length + width)

= 2*(x + 4 + 5x)

= 2*(6x + 4)

= 2*6x + 2 *4

= 12x + 8

4 0

3 years ago

BRAINLIEST TO CORRECT

Helga [31]

I think it’s b
have a nice day

5 0

2 years ago

Read 2 more answers

Pratap puri rowed 10 miles down a river in 2 hours, but the return trip took him 2 and one half hours. Find the rate pratap can

N76 [4]

Answer:

Rate of Pratap in still water is 4.5 miles/hour and rate of current is 0.5 miles/hour.

Step-by-step explanation:

Pratap Puri rowed 10 miles down a river in 2 hours, but the return trip took him 2.5 hours.

We know that, $Speed = \frac{Distance}{Time}$

So, the speed of Pratap with the current will be: $(\frac{10}{2})miles/hour = 5$ miles/hour

and the speed of Pratap against the current will be: $(\frac{10}{2.5})miles/hour = 4$ miles/hour.

Suppose, the rate of Pratap in still water is $x$ and the rate of current is $y$ .

So, the equations will be........

$x+y= 5 .............................. (1)\\ \\ x-y=4 .............................. (2)$

Adding equation (1) and (2) , we will get......

$2x=9\\ \\ x=\frac{9}{2}= 4.5$

Now, plugging this $x=4.5$ into equation (1), we will get.....

$4.5+y=5\\ \\ y=5-4.5 =0.5$

Thus, Pratap can row at 4.5 miles per hour in still water and the rate of the current is 0.5 miles/hour.

6 0

3 years ago