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Olin [163]
2 years ago
13

What is the surface area of the regular pyramid given below?

Mathematics
1 answer:
frosja888 [35]2 years ago
3 0

Answer:  

Total Surface Area of the given regular pyramid = 153 units²

Step-by-step explanation:

Base length of the regular pyramid, l  = 9 units

Base width of the regular pyramid, w = 9 units

Height of the regular pyramid, h = 7 units

Perimeter of base = 3 × 9

                              = 27 units

\text{Area of base = }\frac{1}{\times Base\times Height\\\\Area = \frac{1}{2}\times 9\times 7=21.5\:\:units^2

\text{Now, Total Surface Area = }\frac{1}{2}\times p\times l+A\\\\\implies\text{Total Surface Area = }\frac{1}{2}\times 27\times 9+31.5

Hence, Total Surface Area of the given regular pyramid = 153 units²

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Eriko cuts out a paper quadrilateral with two angles that each measure 60° and two angles that each measure 120°. Only two of th
Whitepunk [10]

Answer:

The quadrilateral is Parallelogram.

Step-by-step explanation:

A quadrilateral is a four-sided regular polygon.

A parallelogram is a quadrilateral.

The opposite sides of a parallelogram are parallel and equal in length.

The opposite angles of a parallelogram are also equal.

In this case the quadrilateral is defined as follows:

  • two angles that each measure 60°
  • two angles that each measure 120°
  • Only two of the sides have a length of 3.2 inches.

Consider the quadrilateral ABCD below.

The quadrilateral ABCD is Parallelogram.

8 0
3 years ago
Okay can someone pls help me i literally am so confused
masha68 [24]
View answer attached below

5 0
3 years ago
Please help me with this question!! :)
Gre4nikov [31]

Answer:

D. The last 26/8, 3-32, -Pi, -4 2/3

Step-by-step explanation:

26/8= 3.25

3-32=3.17

-Pi= -3.14

-4 2/3= -4.67

6 0
9 months ago
Ian tosses a bone up in the air for his dog, Spot. The height, h, in feet, that Spot is above the ground at the time t seconds a
kirill115 [55]

Answer:

12 feet per second.

Step-by-step explanation:

Please consider the complete question.

Ian tosses a bone up in the air for his dog, Spot. The height, h, in feet, that Spot is above the ground at the time t seconds after she jumps for the bone can be representedh(t)=-16t^2+20t.

What is Spot's average rate of ascent, in feet per second, from the time she jumps into the air to the time she catches the bone at t=1/2?  

We will use average rate of change formula to solve our given problem.

\text{Average rate of change}=\frac{f(b)-f(a)}{b-a}

\text{Average rate of change}=\frac{h(\frac{1}{2})-h(0)}{\frac{1}{2}-0}

\text{Average rate of change}=\frac{-16\cdot(\frac{1}{2})^2+20\cdot \frac{1}{2}-(-16\cdot(0)^2+20\cdot (0))}{\frac{1}{2}-0}

\text{Average rate of change}=\frac{-16\cdot\frac{1}{4}+10-(0)}{\frac{1}{2}}

\text{Average rate of change}=\frac{-4+10}{\frac{1}{2}}

\text{Average rate of change}=\frac{6}{\frac{1}{2}}

\text{Average rate of change}=\frac{2\cdot 6}{1}

\text{Average rate of change}=12

Therefore, Spot's average rate of ascent is 12 feet per second.

3 0
3 years ago
What is 9/10 divided by 2/5
MakcuM [25]
\frac{9}{10}:\frac{2}{5}=\frac{9}{10}*\frac{5}{2}= \boxed{\frac{9}{4}}
3 0
3 years ago
Read 2 more answers
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