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3 years ago

How do use a net to find the surface area of a square Pyramid

Mathematics

1 answer:

Gekata [30.6K]3 years ago

7 0

Find the area of each shape first like first the middle square then the triangles and multiply the triangles by 4

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Maggie is creating a cushion for a circular stool. The stool has a diameter of 14 inches. If she needs fabric to extend 2 inches

Marianna [84]

Answer:

396 in.^2

Step-by-step explanation:

The cushion is shaped like a cylinder with 14 inch diameter and 2 inch height.

We need to find the surface area of the cylinder.

diameter = 14 in.

radius = diameter/2 = 7 in.

height = 2 in.

A = 2(pi)r^2 + 2(pi)rh

A = 2(3.14159)(7 in.)^2 + 2(3.14159)(7 in.)(2 in.)

A = 395.84 in.^2

A = 396 in.^2

8 0

2 years ago

What is the area of the figure shown delow?

Mariulka [41]

Answer:

160mm2 is the correct answer

Step-by-step explanation:

7 0

2 years ago

Please help me branniest

Goshia [24]

Answer:

the answer is C

Step-by-step explanation:

7 0

3 years ago

Prove 2√(x) + 1/√(x + 1) <= 2√(x+1) for all x in [0,inf)

EleoNora [17]

Start by multiplying each side of the inequality by $\sqrt{x + 1}$ and simplifying:

$2 \sqrt x + \frac{1}{\sqrt{x+1}} \leq 2 \sqrt{x + 1}$
$(2 \sqrt x + \frac{1}{\sqrt{x+1}})(\sqrt{x + 1}) \leq (2 \sqrt{x + 1})(\sqrt{x + 1})$
$2 \sqrt{x(x + 1)} + 1 \leq 2(x + 1)$
$2 \sqrt{x^2 + x} + 1 \leq 2x + 2$
$2 \sqrt{x^2 + x} \leq 2x + 1$
$\sqrt{x^2 + x} \leq x + \frac{1}{2}$

From here, we can square both sides to get

$x^2 + x \leq (x + \frac{1}{2})^2$
$x^2 + x \leq x^2 + x + \frac{1}{4}$
$0 \leq \frac{1}{4}$ , which is always true.

3 0

2 years ago

A number decreased by the sum of the number and seven. what is the algebraic expression?

Aleks04 [339]

The answer would be x-(7+x)

5 0

2 years ago

Read 2 more answers