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3 years ago

Help!! please if can i need answer

Mathematics

1 answer:

leva [86]3 years ago

8 0

The answer is C

I think it is

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If I currently have a 77% in my course, and I have final that's worth 30%, how would a score of exactly 0% or 100% on the final

Neporo4naja [7]

Answer:

If you score a 0% on your exam, you get 53.9% as your overall grade
If you score 100% on your exam, you get 83.9% as your overall grade

====================================================

Explanation:

Your final is worth 30% of your grade, so the other 70% is what you've already done so far

The expression below

0.30x + 0.70*0.77

represents what your final grade would be in decimal form. We multiply the weights 0.30 and 0.70 with each of their corresponding grade value.

The x represents the decimal form of the final exam score.

------------

Let's say you get a 0% on the final. That would mean x = 0 and

0.30x + 0.70*0.77

0.30*0 + 0.70*0.77

0.539

So you'd end up with a 53.9% as the overall grade.

------------

Now let's say you get 100% on the final. We plug in x = 1 since it's the decimal form of 100%

So,

0.30x + 0.70*0.77

0.30*1 + 0.70*0.77

0.839

You would end up with a 83.9% as the overall grade.

4 0

3 years ago

The solution of 4(x + 1) = 4(2 – x)

Galina-37 [17]

4x+4=8-4x

4x+4x=8-4

8x=4

x=4/8

x=1/2

5 0

3 years ago

Read 2 more answers

Please help with this AP Calculus question !

melomori [17]

Answer:

C. $\displaystyle \frac{cos(x)}{x} - ln(x)sin(x)$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

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

Trig Derivatives

Logarithmic Derivatives

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f(x) = ln(x)cos(x)$

Step 2: Differentiate

Derivative Rule [Product Rule]: $\displaystyle f'(x) = \frac{d}{dx}[ln(x)]cos(x) + ln(x)\frac{d}{dx}[cos(x)]$
Logarithmic Derivative: $\displaystyle f'(x) = \frac{1}{x}cos(x) + ln(x)\frac{d}{dx}[cos(x)]$
Trig Derivative: $\displaystyle f'(x) = \frac{1}{x}cos(x) + ln(x)[-sin(x)]$
Simplify: $\displaystyle f'(x) = \frac{cos(x)}{x} - ln(x)sin(x)$