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2 years ago

You need 20 pounds of grass seed to plant grass inside the baseball diamond below. About how many pounds of grass seed

Mathematics

1 answer:

taurus [48]2 years ago

6 0

Answer:

13 pounds of grass.

Step-by-step explanation:

You need 20 pounds of grass seed to plant inside the baseball diamond.

The Baseball Diamond is a square of side = 90 ft

The Softball Diamond is a square of side = 60 ft

The ratio of the size of the Baseball Diamond to that of Softball Diamond is:

90: 60 = 3:2

If 3 represents 20 pounds, then 2 represents;

$\frac{2}{3}$ × 20 = 13.33333333 pounds = 13 pounds (rounded to nearest pound)

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The formula for the volume V of a cylinder is V = πr2h, where r is the radius of the base and h is the height of the cylinder. S

lana [24]

Answer:

Part A) $h=V/(\pi r^{2})$

Part B) $h=4\ cm$

Step-by-step explanation:

Part A)

we have the formula of the volume of a cylinder

$V=\pi r^{2}h$

Solve for h

That means ----> isolate the variable h

Divide both side by πr²

$V/(\pi r^{2})=\pi r^{2}h/(\pi r^{2})$

Simplify

$V/(\pi r^{2})=h$

Rewrite

$h=V/(\pi r^{2})$

Part B) Find the height of a cylinder with a volume of 36π cm3 and a base with a radius of 3 cm

we have

$V=36\pi\ cm^3$

$r=3\ cm$

substitute in the formula and solve for h

$h=36\pi/(\pi 3^{2})$

$h=4\ cm$

8 0

2 years ago

In ABC, AB=5, BC=9, and AC=8. choose the angle measures in order from greatest to least.

shepuryov [24]

angle measures in order from greatest to least.
<BAC, <ABC, <ACB

hope it help.

6 0

2 years ago

Read 2 more answers

Which of the following could be the equation for a parabola that opens to the left with vertex (-17,2)?

Sholpan [36]

A sideways opening parabola is in the form $x= y^{2}$ , so we know from the process of elimination that it will either be b or c. Next we have to realize that if the parabola opens to the left it is a negative parabola, just like if a parabola opens upside down it is a negative parabola. So the one that has the negative out front is b.

3 0

2 years ago

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Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$