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3 years ago

Is 15 yards to 18 yard increase or decrease

2 answers:

8 0

It is increase, because 15 is less than 18. Other way around, if the question is 18 to 15 yards, then that would be decrease, because it is getting smaller.

bulgar [2K]3 years ago

5 0

15 to 18 yards is increasing because it goes up three times

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The grassy area of a playground has a side length of x and a side of width of x + 3 what is the area of the playground?

Kipish [7]

Answer:

Step-by-step explanation:

We assume this playground is a rectangle. The area of a rectangle is of the form: A = lw, where l is the length and w is the width.

Here, the length is l = x and the width is w = x + 3. So, plug these in:

A = lw

A = x(x + 3)

Expand this out:

A = x² + 3x

The answer is thus A.

4 0

3 years ago

Read 2 more answers

What is the perimeter of an equilateral triangle which has an area of 96\sqrt{3}96 <br> 3 ?

IRISSAK [1]

Answer:

Perimeter = 44.67

Step-by-step explanation:

7 0

3 years ago

The diameter of a tire is 2.5 ft. Use this measurement to answer parts a and b. Show all work to receive full credit.

Alex787 [66]

Divide them all together.

8 0

3 years ago

Doug hits a baseball straight towards a 15 ft high fence that is 400 ft from home plate. The ball is hit 2.5ft above the ground

Scorpion4ik [409]

Answer:

$125.4\ \text{m/s}$

Step-by-step explanation:

u = Initial velocity of baseball

$\theta$ = Angle of hit = $30^{\circ}$

x = Displacement in x direction = 400 ft

y = Displacement in y direction = 15 ft

$y_0$ = Height of hit = 2.5 ft

$a_y$ = g = Acceleration due to gravity = $32.2\ \text{ft/s}^2$

t = Time taken

Displacement in x direction

$x=u_xt\\\Rightarrow x=u\cos\theta t\\\Rightarrow t=\dfrac{x}{u\cos\theta}\\\Rightarrow t=\dfrac{400}{u\cos30^{\circ}}\\\Rightarrow t=\dfrac{400}{u\dfrac{\sqrt{3}}{2}}\\\Rightarrow t=\dfrac{800}{u\sqrt{3}}$

Displacement in y direction

$y=y_0+u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow y=y_0+u\sin\theta t+\dfrac{1}{2}a_yt^2\\\Rightarrow 15=2.5+u\sin30^{\circ}(\dfrac{800}{u\sqrt{3}})+\dfrac{1}{2}\times -32.2\times (\dfrac{800}{u\sqrt{3}})^2\\\Rightarrow \dfrac{400}{\sqrt{3}}-\dfrac{10304000}{3u^2}-12.5=0\\\Rightarrow 218.44=\dfrac{10304000}{3u^2}\\\Rightarrow u=\sqrt{\dfrac{10304000}{3\times218.44}}\\\Rightarrow u=125.4\ \text{m/s}$

The minimum initial velocity needed for the ball to clear the fence is $125.4\ \text{m/s}$

5 0

3 years ago

Identify the new perimeter of the square using a scale factor of 1/3

Verizon [17]

Answer:

Step-by-step explanation:

15 × $\frac{1}{3}$ = 5

5 0

3 years ago