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Ksju [112]
3 years ago
12

It says ... Which expression is equivalent to the following complex fraction?.

Mathematics
1 answer:
Sergeeva-Olga [200]3 years ago
4 0
-2/x + 5/y ÷ 3/y - 2/x

(-2/x * y/y) + (5/y * x/x) ÷ (3/y * x/x) - (2/x * y/y)
-2y/xy + 5x/xy ÷ 3x/xy - 2y/xy

(-2y + 5x)/xy ÷ (3x - 2y)/xy
(-2y + 5x)/xy * xy/(3x - 2y)
-2y + 5x / 3x - 2y  answer is the first choice

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I had 3 apples. I add 2 more apples. How many apples do I have?
navik [9.2K]

Answer:

5 lol

Step-by-step explanation:

6 0
3 years ago
M(x)=x^2-3x<br> N(x)=x-5<br><br> MN(-1)= ?<br><br> A.18<br> B.-20<br> C.-22<br> D. -24
schepotkina [342]

Answer:

d. -24

Step-by-step explanation:

M(-1)=(-1)^2-3(-1)

M(-1)=1+3

M(-1)=4

N(-1)=-1-5

N(-1)=-6

MN(-1)=(4)(-6)

MN(-1)=-24

3 0
4 years ago
Group of 7 people spend 33.50 movie tickets 6.50 adults and 3.50 for child ticket. Which system of equation could be used to fin
kramer
The answer is
x + y = 7
6.50x + 3.50y = 33.50

x - the number of adult tickets
y - the number of child tickets

<span>Group of 7 people: x + y = 7
</span> 6.50 adults and 3.50 for child ticket cost <span>33.50: 6.50x + 3.50y = 33.50

So, we have two equations:
</span>x + y = 7
6.50x + 3.50y = 33.50
7 0
3 years ago
Read 2 more answers
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
Which expression makes the equation true?
N76 [4]
1. C. 

The rule of powers is that when a parenthesis is raised to a power, you multiply the two numbers. Since it is being raised to the second power, it needs to have a 10 in the parenthesis. That would make 20. 

2. B

Firstly 10^2 = 100, so we know that the number must be 10. That leaves us with just A and B. Then we know the same thing from the previous equation, that they need to be multiplied together. 
8 0
3 years ago
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