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3 years ago

Sike this system of equations by graphing . First graph the equations and then type the solution y=7/2x+3 y=-x-6

Mathematics

1 answer:

sergij07 [2.7K]3 years ago

3 0

Answer:

I’ve got the graphs for both equations below.

With y = 7/2x + 3

y = -x-6

I used substitution with simultaneous equations and got x = -2 and y = -4 :)

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3 years ago

Y=sqrt(x)(8x-5) find the derivative

garri49 [273]

Answer:

$\displaystyle y' = \frac{24x - 5}{2\sqrt{x}}$

General Formulas and Concepts:

Algebra I

Exponentials [Fractions] - Are radicals
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Product Rule: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Step-by-step explanation:

Step 1: Define

$\displaystyle y = \sqrt{x}(8x - 5)$

Step 2: Differentiate

$\displaystyle f(x) = \sqrt{x}, \ g(x) = (8x - 5)$

Product Rule: $\displaystyle y' = \frac{d}{dx}[\sqrt{x}] \cdot (8x - 5) + \sqrt{x} \cdot \frac{d}{dx}[(8x - 5)]$
Rewrite: $\displaystyle y' = \frac{d}{dx}[x^{\frac{1}{2}}] \cdot (8x - 5) + \sqrt{x} \cdot \frac{d}{dx}[(8x - 5)]$
Basic Power Rule: $\displaystyle y' = \frac{1}{2}x^{\frac{1}{2} - 1} \cdot (8x - 5) + \sqrt{x} \cdot 1 \cdot 8x^{1 - 1}$
Simplify: $\displaystyle y' = \frac{1}{2}x^{-\frac{1}{2}} \cdot (8x - 5) + \sqrt{x} \cdot 1 \cdot 8x^{0}$
Rewrite: $\displaystyle y' = \frac{1}{2x^{\frac{1}{2}}} \cdot (8x - 5) + \sqrt{x} \cdot 1 \cdot 8$
Multiply: $\displaystyle y' = \frac{8x + 5}{2x^{\frac{1}{2}}} + 8\sqrt{x}$
Rewrite: $\displaystyle y' = \frac{8x + 5}{2\sqrt{x}} + 8\sqrt{x}$
Add/Rewrite: $\displaystyle y' = \frac{24x - 5}{2\sqrt{x}}$

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Answer:

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Step-by-step explanation:

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