3 years ago

3yards per second in miles per hour

Mathematics

1 answer:

Amanda [17]3 years ago

3 0

3 yards per second is 6.136364 mph.

You might be interested in

Answer this fast and I’ll make you Brainly what are perimeters of the triangles shown are equal

Zigmanuir [339]

Answer:

12x+15,13x+1

Step-by-step explanation:

ABC=4x+8+4x+4x+7=12x+15

PQR=4x+5x+1+4x=13x+1

4 0

3 years ago

What algebraic expression is equivalent to 8(2x+4) ?

alexandr402 [8]

Answer:

10x + 32

Step-by-step explanation:

BRANLIEST PLZZZ. I will make my day :)))

6 0

3 years ago

A system of equations has no solution. If y = 8x + 7 is one of the equations, which could be the other equation?

adell [148]

Y=8x-7 is the answer. hope this helps!

5 0

3 years ago

Read 2 more answers

What is the value for x? 69 (5x + 2) Enter your answer in the box (9y-3)

Leviafan [203]

Answer:

$5x + 2 \\ x = 5 - 2 \\ x = 3 \\ \\ 9y - 3 \\ y = 9 + 3 \\ y = 12$

4 0

3 years ago

Suppose astronomers built a 50-meter telescope. how much greater would its light-collecting area be than that of the 10-meter ke

drek231 [11]

Answer: Its light-collecting area would be 25 times greater than that of the 10-meter keck telescope.

Step-by-step explanation:

1. To solve this problem you must apply the formula for calculate the area of a circle, which is shown below:

$A=r^{2}\pi$

Where r is the radius of the circle.

2. The diameter of a circle is:

$D=\frac{r}{2}$

Where r is the radius of the circle.

3. Therefore, keeping this on mind, you have that the light-collecting area of a 50-meter keck telescope is:

$A_1=(\frac{50m}{2})^{2}\pi=1963,49m^{2}$

4. And the light-collecting area of a 10-meter keck telescope is:

$A_2=(\frac{10m}{2})^{2}\pi=78.53m^{2}$

5. Divide $A_1$ by $A_2$ , then:

$\frac{1963.49m^{2}}{78.53m^{2}}=25$

6. Therefore, its light-collecting area would be 25 times greater than that of the 10-meter keck telescope.

4 0

4 years ago