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3 years ago

5

What function is a quadratic function

2 answers:

Nikolay [14]3 years ago

5 0

The answer is d or the last one

alina1380 [7]3 years ago

3 0

I think it's the first one

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9 boys can paint a basketball court in 3 hours how long should 1 boy take working at the same rate

GaryK [48]

1 boy can work about 0.3 hours
i think that’s the answer,but good luck

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3 years ago

Find the sale price. The original price of a remote control car is $70. The car is on sale for 20% off.

Mrac [35]

$56
Explanation:
70(20%)=14
70-14=56

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3 years ago

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If elephant grass grows 2 1/3 inches a day, how many inches will it grow in 8 days?

geniusboy [140]

2 1/3 * 8 = 16 8/3 or 18 2/3 inches in 8 days.

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3 years ago

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Okay someone help me please what is ten times negative three ? ):

mr_godi [17]

Answer:

-30

Step-by-step explanation:

10 x 3 = 30

10 x -3= -30

please mark me brainliest

two negatives makes a positive

two positives markets a negative

and

a positive and a negative makes a negative

5 0

3 years ago

Read 2 more answers

If 47400 dollars is invested at an interest rate of 7 percent per year, find the value of the investment at the end of 5 years f

Anni [7]

Answer:

Part A) Annual $\$66,480.95$

Part B) Semiannual $\$66,862.38$

Part C) Monthly $\$67,195.44$

Part D) Daily $\$67,261.54$

Step-by-step explanation:

we know that

The compound interest formula is equal to

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

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

so

Part A) Annual

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=1$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{1})^{1*5}$

$A=\$47,400(1.07)^{5}$

$A=\$66,480.95$

Part B) Semiannual

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=2$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{2})^{2*5}$

$A=\$47,400(1.035)^{10}$

$A=\$66,862.38$

Part C) Monthly

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=12$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{12})^{12*5}$

$A=\$47,400(1.0058)^{60}$

$A=\$67,195.44$

Part D) Daily

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=365$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{365})^{365*5}$

$A=\$47,400(1.0002)^{1,825}$

$A=\$67,261.54$

8 0

3 years ago