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3 years ago

PLEASE HELP PLEAAE HURRY ANS NO LINKS PLEASE I REALLY NEED THE ANSWER

Mathematics

2 answers:

natita [175]3 years ago

7 0

Answer:

-11

Step-by-step explanation:

-4 + (6 + 8) / (-2)

-4 + (14) / (-2)

-4 + -7

-4 - 7

-11

Hopefully this helped!

Brainliest please?

Harlamova29_29 [7]3 years ago

3 0

6+8=14÷(-2)=-7-4=-11

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3 years ago

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2 years ago

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3 years ago

Douglas invests money in two simple interest accounts. He invests three times as much in an account paying 14% as he does in an

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Douglas invested $ 1300 altogether

Solution:

Let x represent the amount invested in the account paying 14% interest.

Let y represent the amount invested in the account paying 5% interest

He invests three times as much in an account paying 14% as he does in an account paying 5%

Which means,

x = 3y

The simple interest is given by formula:

$S.I = \frac{p \times n \times r}{100}$

Where,

"p" is the principal

"n" is the number of years

"r" is the rate of interest

He earns $152.75 in interest in one year from both accounts combined

Therefore,

Combined S.I = 152.75

n = 1 year

Considering the account earning 14% interest:

$S.I = \frac{3y \times 14 \times 1}{100}\\\\S.I = 0.42y$

Considering the account earning 5% interest:

$S.I = \frac{y \times 5 \times 1}{100}\\\\S.I = 0.05y$

Since, Combined S.I = 152.75

Therefore,

0.42y + 0.05y = 152.75

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Divide both sides by 0.47

y = 325

Therefore,

x = 3y

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Amount invested together = x + y = 975 + 325 = 1300

Thus he invested $ 1300 altogether

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3 years ago