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Vlad1618 [11]
3 years ago
8

An expression is shown below

Mathematics
1 answer:
mestny [16]3 years ago
5 0
The expression is a rational expression and the product is (x^6/9)

Brainliest my answer if it helps you out?
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You are playing a game using the spinners shown.
Otrada [13]

Answer and Step-by-step explanation:

<u>You should pick Spinner A.</u> There is a high chance of getting red since there are three pieces of the spinner that has red on it, while Spinner B only has 2 instances of red.

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

7 0
3 years ago
Please help me !
miv72 [106K]

List out the multiples of 8

8, 16, 24, 32, 40, 48, 56, 64, 72, 80

I'm going to highlight in bold the values between 20 and 50

8, 16, 24, 32, 40, 48, 56, 64, 72, 80

So the favorable outcomes, aka the outcomes we want, are: {24, 32, 40, 48}

These four values are all divisible by 8. Also, these values are between 20 and 50.

7 0
3 years ago
X^2+X=56<br> Help please!!!!
stich3 [128]
The value of x is either +7 or -8
3 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
Round off 174 to the nearest hundred​
shepuryov [24]

Answer: 200

Step-by-step explanation:

174 rounded to the nearest hundred is:

200

When rounding to the nearest hundred, like we did with 174 above, we use the following rules:

A) We round the number up to the nearest hundred if the last two digits in the number are 50 or above.

B) We round the number down to the nearest hundred if the last two digits in the number are 49 or below.

C) If the last two digits are 00, then we do not have to do any rounding, because it is already to the hundred.

pls mark brainliest!

4 0
3 years ago
Read 2 more answers
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