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3 years ago

5

The formula for the volume of a square pyramid is v= (b squared •h) ÷3, where b is in the length of one side of the square base

and h is the height of the pyramid. Find the length of a side of the base of a square pyramid that has a height of 3 inches and a volume of 25 cubic inches.
I just need someone to confirm my answer to be correct, also can you explain the answer for me to make sure i understand why i am wrong if im wrong

1 answer:

Verdich [7]3 years ago

5 0

I got 5 what did you get

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If your friend Doug is on a balcony and asks you to throw him a donut, how many stories up can he be and you would still feel co

Katena32 [7]

The building that I would imagine would be 40 stories, about 404 feet tall perhaps. If he were to throw the donut from that distance until the ground, catching it would be impossible. So no, I am not confident in catching the donut all the way there.

8 0

3 years ago

For A = {a,b}, determine A x P(A).

alex41 [277]

Answer:

Since A={a,b}, then $P(A)=\{\emptyset, \{a,b\}, \{a\}, \{b\}\}$ . Remember that the Cartesian Product of two sets A, B is defined by

$A\times B=\{(x,y): x\in A \text{ and } y\in B\}$ . Then

$A\times P(A)=\{(a,\emptyset),(a,\{a,b\}),(a,\{a\}), (a,\{b\}), (b,\emptyset),(b,\{a,b\}),(b,\{a\}), (b,\{b\})\}$

6 0

3 years ago

What are the coordinates of the centroid of a triangle with vertices A(−3, 1) , B(1, 6) , and C(5, 2) ? Enter your answer in the

kirza4 [7]

<u>ANSWER</u>: The centroid is (1,3)

<u>Explanation:</u>

The centroid is the intersection of the medians of the triangle.

So we need to find the equation of any two of the medians and solve simultaneously.

Since the median is the straight line from one vertex to the midpoint of the opposite side, we find the midpoint of any two sides.

We find the midpoint of AC using the formula;

$N=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$

$N=(\frac{-3+5}{2}, \frac{1+2}{2})$

$N=(\frac{2}{2}, \frac{3}{2})$

$N=(1, \frac{3}{2})$

The equation of the median passes through $B(1,6)$ and $N=(1, \frac{3}{2})$ .

This line is parallel to the y-axis hence has equation

$x=1$ -------first median.

We also find the midpoint M of BC.

$M=(\frac{1+5}{2}, \frac{6+2}{2})$

$M=(\frac{6}{2}, \frac{8}{2})$

$M=(3, 4)$

The slope of the median, AM is

$Slope_{AM}=\frac{4-1}{3--3}$

$Slope_{AM}=\frac{3}{6}$

$Slope_{AM}=\frac{1}{2}$

The equation of the median AM is given by;

$y-y_1=m(x-x_1)$

We use the point M and the slope of AM.

$y-4=\frac{1}{2}(x-3)$

$2y-8=(x-3)$

$2y-x=-3+8$

$2y-x=5$ -------Second median

We now solve the equation of the two medians simultaneously by putting $x=1$ in to the equation of the second median.

$2y-1=5$

$2y=5+1$

$2y=6$

$y=3$

Hence the centroid has coordinates $(1,3)$

7 0

3 years ago

Read 2 more answers

1/10( x + 130) = -2(3-X)<br><br> I need help

Artist 52 [7]

Answer:

x = 10

Step-by-step explanation:

1/10(x + 130) = -2(3 - x)

x + 130 = -20(3-x)

x + 130 = -60 + 20x

x - 20x = -60 - 130

-19x = -190

x = 10

5 0

3 years ago

together,sarah's two cats have the same mass as her dog,11 kg.the mass of one cat is 1500 g greater than the mass of the other.

Alika [10]

Let (a) be the first cat
let (b) be the second cat
a +b=11kg-----(1)
a=b+1500g⇒(1.5 kg)-------(2)
so now u have a system of two unknowns
a=11-b
substitute a in (2)
11-b=b+1.5
11-1.5=2b
2b=9.5
b=9.5/2=4.75
subs. b in (1)
a+b=11
a+4.75=11
a=11-4.75=6.25
so cat (a) is 6.25 kg
and (b) is 4.75 kg
hope I helped

6 0

3 years ago