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KIM [24]
3 years ago
10

The base of the square pyramid shown has an area of 576 square units. If the slant height of the pyramid is 20 units, what is th

e height of the pyramid?
18 units


13.4 units


16 units


31.2 units
Mathematics
1 answer:
Andreas93 [3]3 years ago
5 0

Answer:

Option C - 16 units

Step-by-step explanation:

Since the base is a square and it has an area of 576 sq.units, thus, if the side length of the base is x, then;

x² = 576

x = √576

x = 24

To get the height, we will divide this side length by 2 and use Pythagoreas Theorem.

Since, we are told that the pyramid has a slant height of 20 units,

So using pythagoras theorem, we can find the height of the pyramid as;

h² = 20² - (24/2)²

h² = 400 - 144

h² = 256

h = √256

h = 16 units

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A rectangular chalkboard is 1.7 m long and 0.9 m wide.
IRINA_888 [86]

Answer:

A=1.53\ m^2

Step-by-step explanation:

The length of a chalkboard, l = 1.7

The width of a chalkboard, b = 0.9 m

We need to find the area of the chalkboard. The formula for the area of a rectangle is given by :

A=L\times B\\\\A=1.7\times 0.9\\\\A=1.53\ m^2

So, the area of the chalkboard is equal to 1.53\ m^2.

3 0
2 years ago
What is 124/87 SOMEONE?
Rufina [12.5K]

Answer:

1.53

Step-by-step explanation:

5 0
3 years ago
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An art class costs $45 for materials and then
Rudiy27
45+10x the x stands for how many classes you take
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3 years ago
Given the parent function f(x) = 8x + 9, rewrite the function after the following transformation: reflect over the x-axis. Group
kari74 [83]

Given:

The parent function is

f(x)=8x+9

To find:

The function after the reflection over the x-axis.

Solution:

We know that, if a function f(x) reflected over x-axis to get the function g(x), then

g(x)=-f(x)

Putting f(x)=8x+9, we get

g(x)=-(8x+9)

g(x)=-8x-9

The function after the reflection over the x-axis is g(x)=-8x-9.

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3 0
3 years ago
Can you guys give me a serious answer please?
grin007 [14]

Answer:

800 bacteria

Step-by-step explanation:

Answer of this question depend upon number of hours at the end of day

Let Take Full Day of 24 hrs

Colby bacteria will doubled 12 times ( doubled once in 2hrs => 24/2 = 12)

Colby bacteria at end of day = 50 * 2¹²

Jaquan bacteria will doubled 8 times ( doubled once in 3hrs => 24/3 = 8)

Let say Jaquan bacteria at start  of Day = J

Jaquan bacteria at end of day = J * 2⁸

Equating Both

J * 2⁸ = 50 * 2¹²

=> J = 50 * 2⁴

=> J = 50 * 16

=> J = 800

Jaquan will start  with 800 Bacteria

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