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3 years ago

On the last history quiz, Knesha answered 0.75 of the 20 question correctly How many questions did Kanesha answer correctly?

1 answer:

5 0

.75 is 75% of the test, and 75 of 20 is equal to 1/4. So 20/4 is equal to 5, meaning 5 questions were wrong, and 15 were answered correctly.

I hope this was not too confusing for you.

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If g (x) = 1/x then [g (x+h) - g (x)] /h

lys-0071 [83]

Answer:

$\dfrac{-1}{x(x+h)}, h\ne 0$

Step-by-step explanation:

If $g(x) = \dfrac{1}{x}$ , then $g(x+h) = \dfrac{1}{x+h}$ . It follows that

$\begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \cdot [g(x+h) - g(x)] \\&= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\end{aligned}$

Technically we are done, but some more simplification can be made. We can get a common denominator between 1/(x+h) and 1/x.

$\begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\\&=\frac{1}{h} \left(\frac{x}{x(x+h)} - \frac{x+h}{x(x+h)} \right) \\ &=\frac{1}{h} \left(\frac{x-(x+h)}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{x-x-h}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{-h}{x(x+h)}\right) \end{aligned}$

Now we can cancel the h in the numerator and denominator under the assumption that h is not 0.

$= \dfrac{-1}{x(x+h)}, h\ne 0$

5 0

3 years ago

PLEASE HELP!!!!

Artyom0805 [142]

Answer:

The total cost of producing 6 widgets is $231.

Step-by-step explanation:

5 0

3 years ago

Simplify the following: 7-3[(n^3+8n)/(-n)+9n^2]

Pachacha [2.7K]

If you would like to simplify 7 - 3[(n^3 + 8n) / (-n) + 9n^2], you can do this using the following steps:

7 - 3[(n^3 + 8n) / (-n) + 9n^2] = 7 - 3[(-n^2 - 8) + 9n^2] = 7 - 3[-n^2 - 8 + 9n^2] = 7 - 3[ - 8 + 8n^2] = 7 - 3[8n^2 - 8] = 7 - 24n^2 + 24 = - 24n^2 + 31

The correct result would be - 24n^2 + 31.

7 0

3 years ago

Each chicken eats about 6 pounds of chicken feed in a month.how many pounds of chicken feed will be left after 3 months? Write a

NARA [144]

Dear Vegafred77, use 3x6=18 since a chicken eats 6 lbs of chicken.

6 0

3 years ago

F(x) = e^-x . Find the equation of the tangent to f(x) at x=-1

natima [27]

Answer:

The equation of the tangent line is given by the following equation:

$\displaystyle y - \frac{1}{e} = \frac{-1}{e} \bigg( x - 1 \bigg)$

General Formulas and Concepts:

Algebra I

Point-Slope Form: y - y₁ = m(x - x₁)

x₁ - x coordinate
y₁ - y coordinate
m - slope

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]:
$\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Rule [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:

*Note:

Recall that the definition of the derivative is the slope of the tangent line.

Step 1: Define

Identify given.

$\displaystylef(x) = e^{-x} \\x = -1$

Step 2: Differentiate

[Function] Apply Exponential Differentiation [Derivative Rule - Chain Rule]:
$\displaystyle f'(x) = e^{-x}(-x)'$
[Derivative] Rewrite [Derivative Rule - Multiplied Constant]:
$\displaystyle f'(x) = -e^{-x}(x)'$
[Derivative] Apply Derivative Rule [Derivative Rule - Basic Power Rule]:
$\displaystyle f'(x) = -e^{-x}$