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koban [17]
3 years ago
12

Plz answer 1 and 2 for me plz plz plz

Mathematics
1 answer:
OLEGan [10]3 years ago
7 0
A 20 is at 70%
A 23 is at 100%
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Write this number in standard form 12 ten thousands,8 thousands,14 hundreds, 7 ones
sweet-ann [11.9K]
128,147 !!!!!!!!!!!!!
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Can someone pls give answer to number 4
katen-ka-za [31]

Step-by-step explanation:

Supplement= 180°- 13°

167° Answer

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2 years ago
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Liam wants to save $6,265 to purchase a car. if $4,000 is deposited into an account at 7.5% interest, compounded monthly, how ma
timurjin [86]
To solve for time taken we use the compound formula:
A=P(1+r/100*n)^(n*t)
where:
A=future value
P=principle
r=rate
t=time
n=number of terms

6265=4000(1+7.5/(12*100))^(12t)
solving for t we get
1.56625=1.00625^12t
12t=72.0136
t=6.00 years
Answer: 6 years
5 0
3 years ago
$53,000 is placed in an investment account that grows at a fixed rate of 2% (compound growth) per year. how much is in the accou
ella [17]

Answer:

$57,369

Step-by-step explanation:

We have been given that an amount of $53,000 is placed in an investment account that grows at a fixed rate of 2% (compound growth) per year. We are asked to find the amount in the account after 4 years.

To solve our given problem we will use compound interest formula.\

A=P(1+\frac{r}{n})^{nt}, where,

A = Final amount after t years,

P = Principal amount,

r = Annual interest rate in decimal form,

n = Number of times interest is compounded per year,

t = Time in years.

Let us convert our given rate in decimal form.

2\%=\frac{2}{100}=0.02

Upon substituting our given values in compound interest formula we will get,

A=\$53,000(1+\frac{0.02}{1})^{1*4}

A=\$53,000(1+0.02)^{4}

A=\$53,000(1.02)^{4}

A=\$53,000*1.08243216

A=\$57368.90448\approx \$57,369

Therefore, an amount of $57,369 will be in the account after 4 years.

3 0
3 years ago
In Exploration 5.4.2 Question 2, what conclusion can you make about the value of the derivative at
givi [52]

The value of the derivative at the maximum or minimum for a continuous function must be zero.

<h3>What happens with the derivative at the maximum of minimum?</h3>

So, remember that the derivative at a given value gives the slope of a tangent line to the curve at that point.

Now, also remember that maximums or minimums are points where the behavior of the curve changes (it stops going up and starts going down or things like that).

If you draw the tangent line to these points, you will see that you end with horizontal lines. And the slope of a horizontal line is zero.

So we conclude that the value of the derivative at the maximum or minimum for a continuous function must be zero.

If you want to learn more about maximums and minimums, you can read:

brainly.com/question/24701109

4 0
2 years ago
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