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PolarNik [594]
4 years ago
12

Convert the fraction 50/100 to a decimal.

Mathematics
1 answer:
evablogger [386]4 years ago
4 0

Answer:

0.5

Step-by-step explanation:

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Multiply: (5 ft 6 in) 2.4
sergey [27]

to make it easier, convert 5 feet to x inches:

5*12=60

60+6=66

So, 66 inches.

Multiply 2.4 by 66:

2.4*66=158.4

So in inches, the answer is 158.4

In feet and inches, 13 feet and 2.4 inches.

4 0
3 years ago
Step by step what is 2z+7z-z+5
MArishka [77]
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3 0
4 years ago
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Kelly works at her own nail salon. Today, she did 9 manicures and received $36.00 overall in tips.
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Y=9x+36 because she did 9 manicures x is how much is she got paid+ her $36 in tips:)

3 0
3 years ago
Using y=mx+b find the slope for<br> 7) 8x + 3y = −9
malfutka [58]

Answer:

-8/3

Step-by-step explanation:

8x + 3y = - 9

Let's make y the subject of the formula.

y = - 9/3 - 8x/3

Rearrange

y = - 8x/3 - 3

The slope is the coefficient of x which is - 8/3

If you don't understand, please leave a comment

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7 0
3 years ago
If a polynomial function f(x) has roots -8, 1, and 6i, what must also be a root of f(x)?
Naddik [55]

Answer:

it must also have the root : - 6i

Step-by-step explanation:

If a polynomial is expressed with real coefficients (which must be the case if it is a function f(x) in the Real coordinate system), then if it has a complex root "a+bi", it must also have for root the conjugate of that complex root.

This is because in order to render a polynomial with Real coefficients, the binomial factor  (x - (a+bi)) originated using the complex root would be able to eliminate the imaginary unit, only when multiplied by the binomial factor generated by its conjugate: (x - (a-bi)). This is shown below:

(x-(a+bi))*(x-(a-bi))=\\(x-a-bi)*(x-a+bi)=\\([x-a]-bi)*([x-a]+bi)=\\(x-a)^2-(bi)^2=\\(x-a)^2-b^2(-1)=\\(x-a)^2+b^2

where the imaginary unit has disappeared, making the expression real.

So in our case, a+bi is -6i (real part a=0, and imaginary part b=-6)

Then, the conjugate of this root would be: +6i, giving us the other complex root that also may be present in the real polynomial we are dealing with.

5 0
3 years ago
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