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Scilla [17]
3 years ago
6

Which multiplication problem is represented by the model

Mathematics
1 answer:
Ivenika [448]3 years ago
3 0

Answer:

It's the second one. Press 'Ask For Help' if you need more help on problems

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What is the answer x2 = 400.
Westkost [7]

Answer:

200

Step-by-step explanation:

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Use the graph of the function to find its domain and range. Write the domain and range in interval notation
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Step-by-step explanation:

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2 years ago
PLEASE HELP! I WILL GIVE BRAINLIEST! Determine whether the statement is always, sometimes, or never true.
hammer [34]

Answer:

13 True

14 Sometimes

15 Never

Step-by-step explanation:

All equilateral triangles must have 60 degree angles

Sometimes a acute triangle can have 2 acute angles

Obtuse angles take up majority of the mesuares in the triangle. All triangles must have 3 angles equal up to 180 degrees

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In the figure below, angle y and angle x form vertical angles. Angle x forms a straight line with the 50° angle and the 40° angl
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Answer: The answer is B

Step-by-step explanation:

4 0
3 years ago
$6000 are invested in a bank account at an interest rate of 10 percent per year.
Licemer1 [7]

The amount for the investment of $6000 will be a.$6369  b. $6090  and c.$6030.

<h3>What is compound interest?</h3>

Compound interest is the interest levied on the interest. The formula for the calculation of compound interest is given as:-

A=P[1+\dfrac{r}{n}]^{nt}

a) The amount in the bank after 6 years if interest is compounded annually.

A=P[1+\dfrac{r}{1}]^{t}\\\\\\A=6000[1+\dfrac{0.01}{1}]^{  6}

A= $6369

b) The amount in the bank after 6 years if interest is compounded quarterly.

A=P[1+\dfrac{r}{4}]^{4t}\\\\\\A=6000[1+\dfrac{0.01}{4}]^{4\times 6}

A= $6090

c ) The amount in the bank after 6 years if interest is compounded monthly.

A=P[1+\dfrac{r}{12}]^{4t}\\\\\\A=6000[1+\dfrac{0.01}{12}]^{12\times 6}

A=$6030

Hence the amount for the investment of $6000 will be a.$6369  b. $6090  and c.$6030.

To know more about Compound interest follow

brainly.com/question/24924853

#SPJ1

7 0
2 years ago
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