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Travka [436]
3 years ago
15

What does Y=-x^2-3x+6 equal

Mathematics
1 answer:
drek231 [11]3 years ago
6 0
The answer to your question is X=1
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4.16161616 as a mixed number?
Assoli18 [71]

4.16161616=41010101/6250000

 

Showing the work

Rewrite the decimal number as a fraction with 1 in the denominator

4.16161616=4.16161616/1

Multiply to remove 8 decimal places. Here, you multiply top and bottom by 10^8 = 100000000

4.16161616/1×100000000/100000000=416161616/100000000

Find the Greatest Common Factor (GCF) of 416161616 and 100000000, if it exists, and reduce the fraction by dividing both numerator and denominator by GCF = 16,

416161616÷16/100000000÷16=26010101/6250000

Simplify the improper fraction,

=4 1010101/6250000

In conclusion,

4.16161616=4 1010101/6250000

8 0
3 years ago
Find the measure of angle D and E Angle A = 90 degrees Angle B = 90 degrees Angle C= Angle D Angle E = 90 degrees ​
Kobotan [32]

Answer:

45

Step-by-step explanation:

all angles have to add up to a total of 360, so the other 3 angles equal 270 and there's two more angles that need to split 90° equally

5 0
4 years ago
Write an inequality to represent the graph
Elis [28]

Answer:

its C. i hope this helps good luck and don't forget to thank and the brainly

Step-by-step explanation:

5 0
4 years ago
Find the product of z1 and z2, where z1 = 2(cos 70° + i sin 70°) and z2 = 4(cos 200° + i sin 200°)
vazorg [7]

Answer:

-8i

Step-by-step explanation:

To multiply numbers is polar form

z1 = r1 ( cos theta 1 + i sin theta 1)

z2 = r2 ( cos theta 2 + i sin theta 2)

z1*z2 = r1*r2 (cos (theta1+theta2) + i sin (theta1+theta2)

z1 = 2(cos 70° + i sin 70°)

z2 = 4(cos 200+ i sin 200)

z1z2 = 2*4 (cos (70+200) + i sin (70+200)

z1z2 = 8 (cos(270) + i sin (270))

       = 8 (0 + i (-1))

       =-8i

5 0
4 years ago
Read 2 more answers
A high school coach wants to buy new shirts for the 25 members of the track team. The coach must spend less than $300 on the shi
aksik [14]
25 less than or equal too 300,not really enough information
3 0
3 years ago
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