So when X is 0, the area of the pool is 1200. That makes A and C impossible... then you can plug in numbers...say the boardwalk is 1 ft wide, add 2 to each dimension and plug in 1 for X. (You're adding 2 because there is a tile on either side of each dimension, including corners).

32x42=1344=4+140+1200

Therefore the answer is B.

8 0

3 years ago

Please answer I’ll rate brainlyest

12345 [234]

Answer:

$5P3*6C4=60*15=900$

Step-by-step explanation:

$5P3$ refers to the permutations of 5 items taken 3 at a time. To evaluate this, we use factorials as follows;

$5P3=\frac{5!}{(5-3)!}$

The factorial of an integer n is evaluated as;

$n!=n*(n-1)*(n-2)*(n-3).......3*2*1$

Using this concept, the above expression can now be simplified as follows;

$5P3=\frac{5!}{2!}=\frac{5*4*3*2!}{2!}=5*4*3=60$

Therefore, the permutations of 5 items taken 3 at a time is 60.

The next expression, $6C4$ refers to the combinations of 6 items taken 4 at a time. The simplification utilizes similar concepts of permutations since we shall be involving factorials;

$6C4=\frac{6!}{4!(6-4)!}=\frac{6!}{4!*2!}=\frac{6*5*4!}{4!*2*1}=\frac{6*5}{2*1}=15$

Therefore, the combinations of 6 items taken 4 at a time is 15.

The final step is to evaluate the product;

$5P3*6C4=60*15=900$

3 0

3 years ago

A number divided by 58 is close to 30.

Rom4ik [11]

X/58 (is roughly) 30
x/58 = 30
Multiply 58 on both sides.
x = 1,740

So it has to be c. 1,843

5 0

4 years ago

Read 2 more answers

PLS HELP ME!!! The figure is made up of a cone and a hemisphere. To the nearest whole number, what is the approximate volume of

Pachacha [2.7K]

Hello!

The figure is made up of a cone and a hemisphere. To the nearest whole number, what is the approximate volume of this figure? Use 3.14 to approximate π . Enter your answer in the box. cm³

Data: (Cone)

h (height) = 12 cm

r (radius) = 4 cm (The diameter is 8 being twice the radius)

Adopting: $\pi \approx 3.14$

V (volume) = ?

Solving: (Cone volume)

$V = \dfrac{ \pi *r^2*h}{3}$

$V = \dfrac{ 3.14 *4^2*\diagup\!\!\!\!\!12^4}{\diagup\!\!\!\!3}$

$V = 3.14*16*4$

$\boxed{V = 200.96\:cm^3}$

Note: Now, let's find the volume of a hemisphere.

Data: (hemisphere volume)

V (volume) = ?

r (radius) = 4 cm

Adopting: $\pi \approx 3.14$

If: We know that the volume of a sphere is $V = 4* \pi * \dfrac{r^3}{3}$ , but we have a hemisphere, so the formula will be half the volume of the hemisphere $V = \dfrac{1}{2}* 4* \pi * \dfrac{r^3}{3} \to \boxed{V = 2* \pi * \dfrac{r^3}{3}}$