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mihalych1998 [28]
3 years ago
15

I need help on my math hw please and thanks!!

Mathematics
1 answer:
jarptica [38.1K]3 years ago
3 0
Attached solutions and work.

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Please answer seriously thanks !! will give brainliest
deff fn [24]

Answer:

Step-by-step explanation:

21) CD -  chord   (segment)

22)FG - line

23) EC  - segment -  diameter

24) AB - segment - radius

25) H - Point

26) A - Point

4 0
2 years ago
Four boys went to the amusement park. They paid $123 altogether to get into the park. About how much did each boy pay?
andre [41]

Answer:

30.75

Step-by-step explanation:

by dividing the total by 4 you get your answer to be 30 dollars and 75 cents per boy

4 0
3 years ago
Read 2 more answers
If an angle of 110º bisected, find the measure of each angle formed?
bekas [8.4K]

Answer:

55°

Step-by-step explanation:

if an angle is bisected, it get divided into two equal angles .

so, each resulting angle will be half of the angle which is bisected.

hence, measure of each angle formed is

110° / 2 = 55°

3 0
3 years ago
Read 2 more answers
80 gallons turned into kilograms
zubka84 [21]
80 gallons turned into kilograms is 302.8 kilograms.
Explanation:
1 gallon = 3.785 kg
80 * 3.785 kg = 302. 8
Another explanation:
80 gallons = 302.8 liters
1 liter = 1 kilogram
302.8 * 1 kg = 302.8 kg
8 0
2 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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