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3 years ago

How do you divide $306 into the ratio of 9:5:3? Can someone give me some advice on how to solve this problem?

1 answer:

4 0

In order to divide $306 into the ratio of 9 to 5 to 3, first you have to make each section of the ratio represented by a variable, and put it into an equation.

9x + 5x + 3x = 306

Our next step is to simplify the left side of the equation by combining all of the like terms, the ones that contain variables.

17x = 306

Finally, to solve this equation, you have to divide both sides by 17 to isolate the variable x on the left side of the equation.

x = 18

However, this is not our answer as it isn't in a ratio format and doesn't really make sense. To find our ratio, we have to multiply each of our initial numbers (9, 5, and 3) by our variable, x.

9 * 18 = 162
5 * 18 = 90
3 * 18 = 54

You can verify that these numbers are correct because if you add them together you get 306.

Your final ratio is 162:90:54.

Hope this helps! :)

If you have a separate specific question about pre-tax prices and PMed me the link, I'd be happy to help you.

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Answer:

Step-by-step explanation:

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3 years ago

Which polynomial function has a leading coefficient of 1 and roots 21 and 31 with multiplicity 1?

schepotkina [342]

Answer:

Which i think is the first one, there may just be a typing error.

Step-by-step explanation:

A polynomial of order n has the following format:

$f(x) = a(x - x_{0})(x - x_{1})...(x - x_{n-1})$

In which a is the leading coefficient, $x_{0}, x_{1},..., x_{n-1}$ are the roots.

If a root appears m times, they are said to have multiplicity m.

Leading coefficient of 1 and roots 21 and 31 with multiplicity 1

$f(x) = 1(x - 21)(x - 31)$

So the correct answer is:

$f(x) = 1(x - 21)(x - 31)$

Which i think is the first one, there may just be a typing error.

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3 years ago

Please answer asap no trolls

Svetradugi [14.3K]

Answer:

It would probably be 15 actually

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3 years ago

Read 2 more answers

If x and y are both negative when is x-y positive

jarptica [38.1K]

Hmm

let's try some numbers

we have 3 scenarios
x>y
x=y
x<y

x>y
x=-1 and y=-2
x-y=-1-(-2)=-1+2=1
it's positive when x>y

x=y
x=-1, y=-1
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nope

x<y
x=-2 and y=-1
-2-(-1)=-2+1=-1
nope

x-y is positive when 0>x>y

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3 years ago

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Perform the indicated operation and express the result as a simplified complex number. I need help with 35

skelet666 [1.2K]

Given:-

$\frac{3+4i}{2-i}$

To find:-

The simplified form.

At first we take conjucate and multiply below and above.

The conjucate is,

$2+i$

So now we multiply. we get,

$\frac{3+4i}{2-i}\times\frac{2+i}{2+i}$

Now we simplify. so we get,

$\frac{3+4i}{2-i}\times\frac{2+i}{2+i}=\frac{6+3i+8i+4(i)^2}{2^2-i^2}$

We know the value of,

$i^2=-1$

Substituting the value -1. we get,

$\begin{gathered} \frac{6+3i+8i+4(i)^2}{2^2-i^2}=\frac{6+3i+8i+4(-1)_{}^{}}{2^2-(-1)^{}} \\ \text{ =}\frac{6-4+11i}{2+1} \\ \text{ =}\frac{2+11i}{3} \end{gathered}$

So now we split the term to bring it into the form a+ib. so we get,

$\frac{2+11i}{3}=\frac{2}{3}+i\frac{11}{3}$

So the required solution is,

$\frac{2}{3}+i\frac{11}{3}$

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1 year ago