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3 years ago

Need a GED Math test tutor. Comment if I can give you my number to help

Mathematics

1 answer:

love history [14]3 years ago

3 0

Answer:

sure I'll help, 59.99 per hour. it's a pretty solid deal.

Step-by-step explanation:

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What is the solution set for 10−2n=4, given the replacement set {1,2,3,4}?

Pachacha [2.7K]

The correct solution is n=3

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2 years ago

What area of the clock is not covered with clock face in square inches

Kay [80]

Hello again,
We know that the clock has rectangular shapes, and the clock face is circle.
Area of the clock is:
15×8=120 square inches
To find the area of the circle:
πr^2. However, we did not know the radius yet, but we had circumference of the circle which is 6 inches
C=2πr
6=2πr
Divided 2 from both side
3=πr
Divide π or 3.14 from both side
r≈1 inches
Area of the clock face is:
1×1×3.14
≈3.14.
To find the area of the clock that does not covered with clock face take:
120-3.14=116.86 square inches. Hope it help!

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=f%28x%29%3D%281-0.08%29%5E%7B%5Cfrac%7B1%7D%7B12%7D%20%7D%20%5E%7B%2812t%29%7D" id="TexFormula

oksano4ka [1.4K]

Answer:

See Below.

Step-by-step explanation:

We have:

$\displaystyle f(x)=(1-0.08)^{\frac{1}{12}(12t)}$

First, we can subtract within the parentheses:

$f(x)=(0.92)^{\frac{1}{12}(12t)}$

By the properties of exponents:

$f(x)=((0.92)^\frac{1}{12})^{12t}$

Approximate. Use a calculator:

$f(x)\approx (0.993)^{12t}$

Notes:

0.993 is only an approximation, hence the approximately equal sign.

I'm not given the context of the problem, but it's simpler to just simplify in the exponent like so (the fractions cancel):

$\displaystyle f(x)=(1-0.08)^{\frac{1}{12}(12t)}=(0.92)^t$

Full Problem:

The value of Sara's car decreases at a rate of 8% per year.

We will use the exponential decay formula with a set time, given by:

$f(x)=a(r)^{x/d}$

Where a is the initial value, r is the rate, x is the time that has passed (dependent on d), and d is the amount of time for one decrease.

For this problem, we can ignore the initial value.

And since the value decreases at a rate of 8% per year, r = 0.92 (we acquire this from 1 - 0.08).

Part 1) Per Month:

Since it decreases per month, d = 12.

$f(x)=(0.92)^{x/12}$

Approximate:

$f(x)=((0.92)^{1/12})^x\approx(.993)^x$

In this case, x is measured in months.

Part 2) Per Week:

Since it decreases per week, d = 52.

$f(x)=(0.92)^{x/52}$

Approximate:

$f(x)=((0.92)^1/52)^x\approx (.998)^x$

In this case, x is measured in weeks.

Part 3) Per Day:

So, d = 365.

$f(x)=(0.92)^{x/365}$

Simplify:

$f(x)=((0.92)^{1/365})^x\approx(.999)^x$

In this case, x is measured in days.

Part 4)

So, as d increases, our r increases as well.

Therefore, the smaller the time interval (from months to weeks to days), the higher our rate of decrease is.

8 0

3 years ago

Write an inequality that could be used to solve the following problem. Do not solve. Sal's Pizza charges $14 for a large cheese

Novosadov [1.4K]

He can get 2 toppings on his pizza because 20-14=6 and 2*3=6

3 0

4 years ago

What is this and how do you do it ? thank you

mars1129 [50]

Answer:

$568.38

Step-by-step explanation:

Michaels starts the day with $782.88 in his account.

Then he uses $55 from his debit cars to pay for car repairs. Subtract 55 from 782.88 to get $727.88 in the account after the repairs.

After that, he uses $89.50 for a clothing purchase. Subtract $89 from $727.88 to get 638.88, then subtract another 50 cents to get $638.38 after the clothing purchase.

Lastly, he withdraws $70 from his account. Subtract 70 from $638.38 to get a final answer of $568.38 at the end of the day.

7 0

3 years ago

Need a GED Math test tutor. Comment if I can give you my number to help​

Need a GED Math test tutor. Comment if I can give you my number to help