All categories

3 years ago

The axis of the sphere below is 16 units in length. What is the surface area of the sphere?.

Mathematics

2 answers:

coldgirl [10]3 years ago

8 0

Answer:

256 $\pi$ units²

Step-by-step explanation:

~apex

Nataly_w [17]3 years ago

5 0

There are enough information's already given in the question. Based on those information's that answer can be easily deduced.
Radius of the sphere = 8 units
Then
Surface area of the sphere = 4 *pi * r^2
= 4 * pi * (8)^2
= 4 * pi * 64
= 256pi
I hope that this answer has come to your help.

You might be interested in

I need help solving this

max2010maxim [7]

Answer:

sorry cant help u

Step-by-step explanation:

im bad at this

5 0

2 years ago

Evaluat the exression when X is equal to3 X+1

Gemiola [76]

Just substitute 3 for x, and it’s 3+1=4. Your answer is 4!

4 0

3 years ago

A total of 30 percent of the geese included in a certain migration study were male. If some of the geese migrated during the stu

Serggg [28]

<h2>Answer:</h2>

(B). $\frac{7}{12}$

<h2>Step-by-step explanation:</h2>

In the question,

Let the total number of geese be = 100x

Number of Male geese = 30% = 30x

Number of Female Geese = 70x

Let us say 'kx' geese migrated from these geese.

Number of migrated Male geese = 20% of kx = kx/5

Number of migrated Female geese = 4kx/5

So,

Migration rate of Male geese is given by,

$\frac{(\frac{kx}{5})}{30x}$

Migration rate of Female geese is given by,

$\frac{(\frac{4kx}{5})}{70x}$

So,

The ratio of Migration rate of Male geese to that of Female geese is given by,

$\frac{\left[\frac{(\frac{kx}{5})}{30x}\right]}{\left[\frac{(\frac{4kx}{5})}{70x}\right]}=\frac{350}{4\times 150}=\frac{7}{12}$

Therefore, the ratio of the rate of migration of Male geese to that of Female geese is,

$\frac{7}{12}$

Hence, the correct option is (B).

6 0

3 years ago

Pls helppp math work bleh

joja [24]

Answer:

q = -2

Step-by-step explanation:

when solving 12q >= -24 you get

q >= -2 (12 divides 24 by 2)

therefore q is greater or equal to -2.

3 0

2 years ago

A gold mine has two elevators, one for equipment and another for the miners. The equipment elevator descends 4 feet per second.

Ad libitum [116K]

Let $d_m,\ d_e$ represent the depth of the miner and equipments elevator, respectively. We know that equipment elevator descends with a velocity of $v_e = 4$ feet per second, and the miners one descends with a velocity of $v_m = 15$ feet per second.

If we start counting time ( $t=0$ ) when the equipment elevator begins to descend, after $t$ seconds its depth will be

$d_e(t) = v_et = 4t$

On the other hand, the miner elevator follows the same rule, but we have to use its velocity, and remember that it starts with a delay of thirty seconds:

$d_m(t) = v_m(t-30) = 15(t - 30)$

Now, we have to wait the 30 seconds of delay, and then another 14 seconds. This means that we want to know the positions of both elevators when $t = 44$ . Let's plug this value into the two equations:

$d_e(44) = 4\cdot 44 = 176$

$d_m(44) = 15(44-30) = 15\cdot 14 = 210$

So, the equipment elevator is 176 feet deep, and the miner elevator is 210 feet deep, and thus this is the deeper one.

7 0

3 years ago