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Solnce55 [7]
3 years ago
9

HELP HELP HELP HELP HELP

Mathematics
1 answer:
Vadim26 [7]3 years ago
5 0

Answer:

0+(-2)+9=7

Step-by-step explanation:

You start at 0 then go down by 2 or minus by , Then add by 9 then you end up at 7.

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I need some help please :)​
const2013 [10]
ADC = DAB
CAD = BDA
Answer: DAB, BDA
8 0
3 years ago
Can someone help?????
dexar [7]

Answer:

JL = 56

NK = 28

Step-by-step explanation:

Since JL  = MK

5x + 1 = 8x - 32

8x - 5x = 32 + 1

3x = 33

x = 11

so

JL = 5x + 1 = 5(11) + 1 = 55 + 1 = 56

NK = 1/2MK = 1/2JL = 56/2 = 28

6 0
2 years ago
Circumference involves u its or squate units
kodGreya [7K]

Circumference has the answer in square units.

5 0
3 years ago
20 grams is 60% of what weight
Assoli18 [71]

Answer:12

Step-by-step explanation: 0.6 x 20= 12

7 0
3 years ago
Suppose that $3000 is placed in an account that pays 16% interest compounded each year. Assume that no withdrawals are made from
Papessa [141]

Answer:

a) $3480

b) $4036.8

Step-by-step explanation:

The compound interest formula is given by:

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

Where A(t) is the amount of money after t years, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per year and t is the time in years for which the money is invested or borrowed.

Suppose that $3000 is placed in an account that pays 16% interest compounded each year.

This means, respectively, that P = 3000, r = 0.16, n = 1

So

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

A(t) = 3000(1 + \frac{0.16}{1})^{t}

A(t) = 3000(1.16)^{t}

(a) Find the amount in the account at the end of 1 year.

This is A(1).

A(t) = 3000(1.16)^{t}

A(1) = 3000(1.16)^{1} = 3480

(b) Find the amount in the account at the end of 2 years.

This is A(2).

A(2) = 3000(1.16)^{2} = 4036.8

4 0
3 years ago
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