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3 years ago

Erica invested $200 last year.

Mathematics

1 answer:

Semenov [28]3 years ago

7 0

Investment = $200

Gain = $200 × .2 = 40

Total investment value = 200 + 40 = 240

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BRAINLIEST FOR CORRECT ANSWER! <br> What is the transformation rule for a translation 3 units right?

Butoxors [25]

Answer:Translation happens when we move the image without changing anything in it. ...

Rotation is when we rotate the image by a certain degree. ...

Reflection is when we flip the image along a line (the mirror line). ...

Dilation is when the size of an image is increased or decreased without changing its shape.

5 0

3 years ago

Suppose that, in 1995, Germany produced 4.719 million computers. The total world production that year totaled 80 million compute

Reptile [31]

Answer:

To the nearest tenth

Percentage= 5.9%

Step-by-step explanation:

In 1995, Germany produced 4.719 million computers.

In that same year, the total computer production of the whole world was 80 million.

The percentage of the world production contributed by Germany

=Number produced by Germany/Number produced by the whole world * 100

The percentage of the world production contributed by Germany

=( 4.719/80)*100

The percentage of the world production contributed by Germany

=5.89875%

To the nearest tenth

= 5.9%

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3 years ago

What is the radius of a sphere with the volume of 113.04 in.³

Nimfa-mama [501]

3 is the answer! Have a nice day!

7 0

3 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago

258,197-64,500 (show me how to round this and I will give brainliest!)

Mars2501 [29]

There are a variety of methods of performing this subtraction, and corresponding different methods of regrouping.

If you're taught (as I was) to do the subtraction right-to-left, then the first regrouping you need to do is when you try to subtract 500 from 100. You must regroup the 8 thousands and 1 hundred to 7 thousands and 11 hundreds. Then, when you subtract 5 hundreds, you end with 7 thousands and 6 hundreds.

The next regrouping you need to do is when you try to subtract 6 ten-thousands from 5 ten-thousands. You must regroup the 2 hundred-thousands and 5 ten-thousands to 1 hundred-thousand and 15 ten-thousands. Then, when you subtract 6 ten-thousands, you end with 1 hundred-thousand and 9 ten-thousands.

The end result is

... 258,197 - 64,500 = 193,697

_____

If you use an abacus or soroban or similar tool to help you keep track of the numbers, you were likely taught to do the subtraction left-to-right. In this case, the first regrouping comes when you want to subtract 6 ten-thousands. Practitioners of this method know that -6 = -10 +4, so the number represented on the tool becomes (2-1) hundred-thousands and (5+4) ten-thousands plus the rest of the initial number, or 198,197 after subtracting the 6 ten-thousands.

The subtraction proceeds until you find you need to subtract 500 from 100. At this point, the tool is representing the partial result as 194,197. Again, if you practice this method, you know that -5 = -10 +5, so you reduce the thousands digit by 1 (to 3) and add 5 to the hundreds digit to get 193,697.

_____

An attempt is made to show the regroupings in the attachment. In each case there are two of them. However, working left-to-right, the result of the first subtraction of 6 ten-thousands is 19 ten-thousands, so you never actually write down anything else. Of course, if you're using an abacus or soroban, you don't write down anything—you simply change the position of the beads on the tool.

4 0

4 years ago