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3 years ago

What is the surface area of the square pyramid below?

Mathematics

1 answer:

Reika [66]3 years ago

6 0

The surface area of square pyramid is the sum of the area of all its sides.

If l is the slant height and base is s each then surface area of pyramid is given by:

Surface area = s2+2xsx l

From the given figure in the question:

S=6 cm ,l= 8cm.

Substituting these values in surface area formula we have:

Surface area = 62+2x6x8

Surface area = 36+96

Surface area =132cm2

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salantis [7]

Answer:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

We are given the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}$

When we directly plug in x = 0, we see that we would have an indeterminate form:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}$

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}$

Plugging in x = 0 again, we would get:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}$

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}$

Substitute in x = 0 once more:

$\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}$

And we have our final answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

6 0

3 years ago

Ms. Jones had $40 to purchase groceries. She bought 4 packs of rice and 3 boxes of granola bars. The packs of rice cost $1.09 ea

Nesterboy [21]

Answer:

so to know full price of granola bars and rice packs; you would multiply 1.09 × 4 for the rice which you get 4.36 then for granola bars you do 4.29 × 3 and you get 12.87 then you add 12.87 + 4.36 = 17.23 then you subtract this from the $$ Mr.Jones has which is 40 so its 40-17.23 and you get 22.77

7 0

2 years ago

Graph the inequality y > 3x + 3

natita [175]

Answer:

see below

Step-by-step explanation:

7 0

2 years ago

A square has an area of 100 in ^2. What is the perimeter of the square

almond37 [142]

Area of a square is its side length squared...

A=s^2 solving for s...

s=√A

The perimeter of a square is 4 times its side length...

P=4s, and since s=√A

P=4√A, and we are told that A=100 in^2 so

P=4√100

P=4*10

P=40 in

6 0

3 years ago

Circumference of a circle use 22/7 for pie 10 1/2

Fittoniya [83]

Answer:

Step-by-step explanation:

formula- radius X 2 = diameter x 3.14 | 22/7 = circumference

10 1/2 x 22/7 = 33

4 0

2 years ago