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3 years ago

Find the Surface Area of the cylinder

Mathematics

2 answers:

marusya05 [52]3 years ago

5 0

It’s C! A=2πrh+2πr2 ovi cause the in is even if you keep adding to the same area

Evgen [1.6K]3 years ago

4 0

Answer:

the answer is D according to my scorrs

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U=ak/b solve for this equation a

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A=ub/k. the way you get this is by symplfying both sides of the equation then isolate the variable. hope this helped :)

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3 years ago

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Keith bought a tree for his backyard. The graph shows the height of the tree after a certain number of years.

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Answer:18

Step-by-step explanation:

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2 years ago

If the gradient of the tangent to

Marizza181 [45]

Answer:

Point A(9, 3)

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Algebra I

Coordinates (x, y)
Functions
Function Notation
Terms/Coefficients
Anything to the 0th power is 1
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

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:

Step 1: Define

Identify

$\displaystyle y = \sqrt{x}$

$\displaystyle y' = \frac{1}{6}$

Step 2: Differentiate

[Function] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y = x^{\frac{1}{2}}$
Basic Power Rule: $\displaystyle y' = \frac{1}{2}x^{\frac{1}{2} - 1}$
Simplify: $\displaystyle y' = \frac{1}{2}x^{-\frac{1}{2}}$
[Derivative] Rewrite [Exponential Rule - Rewrite]: $\displaystyle y' = \frac{1}{2x^{\frac{1}{2}}}$
[Derivative] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y' = \frac{1}{2\sqrt{x}}$

Step 3: Solve

Find coordinates of A.

x-coordinate

Substitute in y' [Derivative]: $\displaystyle \frac{1}{6} = \frac{1}{2\sqrt{x}}$
[Multiplication Property of Equality] Multiply 2 on both sides: $\displaystyle \frac{1}{3} = \frac{1}{\sqrt{x}}$
[Multiplication Property of Equality] Cross-multiply: $\displaystyle \sqrt{x} = 3$
[Equality Property] Square both sides: $\displaystyle x = 9$