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Roman55 [17]
3 years ago
15

Pyramid A is a square pyramid with a base side length of 18 inches and a height of 9 inches. Pyramid B has a volume of 3,136 cub

ic inches. How many times bigger is the volume of pyramid B than pyramid A? Give your answer as a percent. Provide an explanation and proof for your answer to receive full credit.
Mathematics
1 answer:
seraphim [82]3 years ago
8 2

Answer:

322.63%

Step-by-step explanation:

Volume of square pyramid = 1/3a²h

a = base side length ; h = height

Volume of pyramid A :

a = 18 ; h = 9

V = 1/3*18²*9

V = 18² * 3

V = 972 in³

Volume of pyramid B = 3136 in³

Volume of pyramid B / Volume of pyramid A

(3136 / 972) * 100% = 322.63%

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coldgirl [10]

Answer:

x=3

Step-by-step explanation:

Use pemdas!

First use distributive property to get rid of the parenthesis, by multiplying 2 by 3x and -4. Then combine like terms and use inverse operations to simplify each side until you get it down to just x! Hope this helped! :)

3 0
3 years ago
160% as a decimal Please help
kolezko [41]

Answer:

160/100=1.6

Step-by-step explanation:

6 0
3 years ago
Given interest of $10,510 at 12 percent for 30 days one can calculate the principal as:
gtnhenbr [62]

Answer:

$1,065,597.22

Step-by-step explanation:

Simple interest = Principal * Rate * Time/100

10,510 = P * 12 * 30/365*100

10,510*365*100 = 360P

383,615,000 = 360P

Note that the time was converted to years by dividing by 365

P = 383,615,000/360

P = $1,065,597.22

Hence the principal is $1,065,597.22

7 0
3 years ago
2x + y = −3 −2y = 6 + 4x Write each equation in slope-intercept form.
Zanzabum

slope intercept form is y=mx+b

so solve for y


2x+y=-3

minus 2x both sides

y=-2x-3


-2y=6+4x

divide both sides by -2

y=-3-2x

y=-2x-3

6 0
3 years ago
A geometric sequence is defined by the equation an = (3)3 − n.
Delvig [45]
PART A

The geometric sequence is defined by the equation

a_{n}=3^{3-n}

To find the first three terms, we put n=1,2,3

When n=1,

a_{1}=3^{3-1}

a_{1}=3^{2}

a_{1}=9
When n=2,

a_{2}=3^{3-2}
a_{2}=3^{1}

a_{2}=3

When n=3

a_{3}=3^{3-3}

a_{3}=3^{0}
a_{1}=1
The first three terms are,

9,3,1

PART B

The common ratio can be found using any two consecutive terms.

The common ratio is given by,
r= \frac{a_{2}}{a_{1}}
r = \frac{3}{9}

r = \frac{1}{3}

PART C

To find
a_{11}

We substitute n=11 into the equation of the geometric sequence.

a_{11} = {3}^{3 - 11}

This implies that,

a_{11} = {3}^{ - 8}

a_{11} = \frac{1}{ {3}^{8} }

a_{11}=\frac{1}{6561}
4 0
3 years ago
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