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3 years ago

14

Trayvion measured his basketball rim. The radius of the rim is \large 7\frac{1}{2} inches. What is the circumference of the rim?

1 answer:

fenix001 [56]3 years ago

7 0

Answer:

<em>(b) 47.10 inch.</em>

Step-by-step explanation:

C = 2πr

C ≈ 2 × 3.14 × 7.5 = 47.1 inches

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How many times greater is the value of 3 in 300 than the value of the 3 in 30

Vikki [24]

Well it sounds like you mean smaller, not greater. 3/300 = 1/100 while 3/30 = 1/10. So 3 out of 300 is 10 times SMALLER than 3 out of 30.

3 0

3 years ago

Read 2 more answers

In the diagram, the measurements that are labeled are known, while the other measurements are unknown. Which measurement of acut

andrew-mc [135]

As the sine rule states,

A/sina = B/sinb = C/sinc .

in the diagram, there are two identified sides and if you use the sine rule, you can find the opposed angles easily.

there are:

side a, with angle â .
side b, with angle b.

so the answer is C.

if you input these into the sine rule, it would be:

a/ sin a = b/sin b

8 0

3 years ago

The formula for the perimeter of a rectangle is p=2(1+w) solve for w.

Dvinal [7]

P=2*(l+w)
p=2l+2w
2w=p - 2l
w= (p-2l)/2

3 0

3 years ago

The length of a 200 square foot rectangular vegetable garden is 4feet less than twice the width. Find the length and width of th

inna [77]

Answer:

Length = 18.099 ft

Width = 11.049 ft

Step-by-step explanation:

let the length of the field be x ft

and the width be y ft

as per the condition given in problem

x=2y-4 -----------(A)

Also the area is given as 200 sqft

Hence

xy=200

Hence from A we get

y(2y-4)=200

taking 2 as GCF out

2y(y-2)=200

Dividing both sides by 2 we get

$y(y-2)=100$

$y^2-2y=100$

subtracting 100 from both sides

$y^2-2y-100=0$

Now we solve the above equation with the help of Quadratic formula which is given in the image attached with this for any equation in form

$ax^2+bx+c=0$

Here in our case

a=1

b=-1

c=-100

Putting those values in the formula and solving them for y

$y=\frac{-(-2)+\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

$y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

Solving first

$y=\frac{2+\sqrt{4+400}{2}$

$y=\frac{2+\sqrt{404}{2}$

$y=\frac{2+20.099}{2}$

$y=\frac{22.099}{2}$

$y=11.049$

Solving second one

$y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

$y=\frac{2-\sqrt{4+400}{2}$

$y=\frac{2-\sqrt{404}{2}$

$y=\frac{2-20.099}{2}$

$y=\frac{-18.99}{2}$

$y=-9.045$

Which is wrong as the width can not be in negative

Our width of the field is

y=11.099

Hence the length will be

x=2y-4

x=2(11.049)-4

x=22.099-4

x=18.099

Hence our length x and width y :

Length = 18.099 ft

Width = 11.049 ft

4 0

3 years ago

Which is the product of (3)(7) and why?

irina [24]

Answer:

21

Step-by-step explanation:

8 0

3 years ago