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4 years ago

PLEASE answer 5, 8, 9, and 11. please also show work if possible. THANK YOU

Mathematics

2 answers:

Igoryamba4 years ago

4 0

Answer:For those problems USE PHOTOMATH it is your algebra savior!

Step-by-step explanation:

soldier1979 [14.2K]4 years ago

3 0

So GCF is greatest common factor. This means- what is the biggest number that goes into both 4a^4 and 6a^2. Well, 2 goes into both 4 and 6! And a^2 goes into both of those a values. So we take out 2a^2 and put the rest in parenthesis. By taking out that value, we will divide the expression by it. We get 2a^2(2a^2 + 3)
For 11, I’m not sure what the table is. I think it’s a 4 square box where you put 5w and 3 on one side and 2w and 7 on the adjacent side. Then you multiply each value together. Let me know if you need more help with that

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An equation for the depreciation of a car is given by y = A(1 – r)t , where y = current value of the car, A = original cost, r =

avanturin [10]

Answer:

Step-by-step explanation:

We know that the equation for the depreciation of the car is:

$y=A(1-r)^t$

Where y is the current cost, A is the original cost, r is the rate of depreciation, and t is the time, in years.

We are told that the value of the car now is half of what it originally cost. So:

$y=\frac{1}{2}A$

Substitute this for y:

$\frac{1}{2}A=A(1-r)^t$

We also know that the rate of depreciation is 10% or 0.1. Substitute 0.1 for r:

$\frac{1}{2}A=A(1-0.1)^t$

So, let's solve for t. Divide both sides by A:

$\frac{1}{2}=(1-0.1)^t$

Subtract within the parentheses:

$\frac{1}{2}=(0.9)^t$

Take the natural log of both sides:

$\ln(\frac{1}{2})=\ln((0.9)^t})$

Using the properties of logarithms, we can move the t to the front:

$\ln(\frac{1}{2})=t\ln(0.9)$

Divide both sides by ln(0.9):

$t=\frac{\ln(\frac{1}{2})}{\ln(0.9)}$

Use a calculator. So, the car is approximately:

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3 years ago

1. On the last 5 quizzes, Marco has earned a total of 21 out of 25 possible points. If he has a perfect

Nonamiya [84]

The number of points Marco needs to bring his average up to 90% is 15points.

If there's a possible 25 possible points, and Marco has taken 5 quizzes;

Evidently, each quiz has weighs 5 points.

Mathematically;

$\frac{21 + 5x}{5(5 + x)} \times 100\% = \frac{90}{100}$

2100 + 500x = 2250 + 450x

50x = 150

x = 3.

Therefore, he needs 5x points = 15 points to bring his average up to 90%.

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egoroff_w [7]

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