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3 years ago

Genevieve deposited $400 into her bank account. The equation A(t) = 400(1.07) can be used to calcula

Mathematics

1 answer:

UNO [17]3 years ago

4 0

Answer: 7%

Step-by-step explanation: I did the test.

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You deposit $3000 into a bank account that pays 1.25% annual interest, compounded semi-annually. What is the final balance after

mojhsa [17]

Answer:

$3,153.32

Step-by-step explanation:

Given that

The deposited amount is $3,000

The annual interest rate is 1.25%

So, the semi-annual interest rate is 1.25% ÷2 = 0.625%

And, the time period = 4 × 2 = 8

We need to find out the final balance

So,

As we know that

Future value = Present value × (1 + rate of interest)^number of years

= $3,000 × (1 + 0.625%)^8

= $3,153.32

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3 years ago

If a single one of these can be answered, can someone help me? I put my book somewhere. I’ll mark brainliest to actual answers :

Alisiya [41]

16. Anaerobic means Without oxygen

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3 years ago

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Select the correct answer.<br> Σ(-)),<br> Which expression gives the same result as t=0

Monica [59]

Answer:

Step-by-step explanation:

since 5 is constant so you can take it out from the summation

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3 years ago

CAN SOMEONE HELP ME PLEASE ASAP!?

VikaD [51]

Answer:

40°

Step-by-step explanation:

∠AOE and ∠COD are vertical angles meaning they have the same value.

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3 years ago

What is the form of the two squares identity?

zvonat [6]

Answer:

Step-by-step explanation:

Consider all options:

A. False, because

$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\\ \\(ab-cd)^2+(ac+bd)^2=a^2b^2-2abcd+c^2d^2+a^2c^2+2abcd+b^2d^2=\\ =a^2b^2+c^2d^2+a^2c^2+b^2d^2\\ \\a^2c^2+a^2d^2+b^2c^2+b^2d^2\neq a^2b^2+c^2d^2+a^2c^2+b^2d^2$

B. False, because

$(a^2-b^2)(c^2+d^2)=a^2c^2+a^2d^2-b^2c^2-b^2d^2\\ \\(ac-bd)^2-(ad+bc)^2=a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2=\\=a^2c^2+b^2d^2+a^2d^2+b^2c^2\\ \\a^2c^2+a^2d^2-b^2c^2-b^2d^2\neq a^2c^2+b^2d^2+a^2d^2+b^2c^2$

C. False, because

$(a^2+b^2)(c^2-d^2)=a^2c^2-a^2d^2+b^2c^2-b^2d^2\\ \\(ac+bd)^2-(ad+bc)^2=a^2c^2+2abcd+b^2d^2-a^2d^2-2abcd-b^2c^2=\\=a^2c^2+b^2d^2-a^2d^2-b^2c^2\\ \\a^2c^2-a^2d^2+b^2c^2-b^2d^2\neq a^2c^2+b^2d^2-a^2d^2-b^2c^2$

D. True, because

$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\\ \\(ac-bd)^2-(ad+bc)^2=a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2=\\=a^2c^2+b^2d^2+a^2d^2+b^2c^2\\ \\a^2c^2+a^2d^2+b^2c^2+b^2d^2= a^2c^2+b^2d^2+a^2d^2+b^2c^2$

6 0

3 years ago