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3 years ago

What is the value of x? Enter your answer in the box. x =

Mathematics

1 answer:

ioda3 years ago

5 0

Answer:

The answer of your question is: x = 5

Step-by-step explanation:

It's and isosceles triangle then ∠ A = ∠C

73 = 8y - 7

8y = 73 + 7

8y = 80

y = 80 / 8

y = 10

∠ C = 8(10) -7

∠ C = 73°

The sum of the internal angles in a triangle equals 180°

∠A + ∠B + ∠C = 180°

73° + (6x + 4) + 73° = 180° Substitution

6x + 4 = 180 - 73 - 73

6x =34 - 4

6x = 30

x = 30 / 6

x = 5

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6x/5 - x = x/15 - 10/3

Vadim26 [7]

Answer:

x=-25

6x/5 - x = x/15 - 10/3

x/5 = x/15 - 10/3

3x = x - 50

3x - x = -50

2x = -50

x = -50/2

X=-25

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2 years ago

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Ymorist [56]

Answer:

Hello! answer: y = 122

Step-by-step explanation:

360 - 116 = 244 244 ÷ 2 = 122 therefore y = 122

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3 years ago

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Hank is taking a walking tour of a park. The route he takes is shown on the map above. How far is the fountain from the entrance

Colt1911 [192]

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3 years ago

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goldenfox [79]

This should be right.

Part A:

-mk - 110 > 75

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m < 185/-k (Fraction)

m < - 185/k (Fraction)

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2c - 8f + 24

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2 years ago

Find the derivative.

krek1111 [17]

Answer:

$\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}$

General Formulas and Concepts:

Algebra I

Terms/Coefficients

Expanding/Factoring

Calculus

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f(x) = \frac{\sqrt{x}}{e^x}$

Step 2: Differentiate

Derivative Rule [Quotient Rule]: $\displaystyle f'(x) = \frac{(\sqrt{x})'e^x - \sqrt{x}(e^x)'}{(e^x)^2}$
Basic Power Rule: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}(e^x)'}{(e^x)^2}$
Exponential Differentiation: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{(e^x)^2}$
Simplify: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{e^{2x}}$
Rewrite: $\displaystyle f'(x) = \bigg( \frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x \bigg) e^{-2x}$
Factor: $\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}$

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

Unit: Differentiation

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2 years ago