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ira [324]
2 years ago
10

Y = x + 2 y = 2x - 5 what is x and y

Mathematics
1 answer:
lana66690 [7]2 years ago
7 0

Answer:

x= 7 y= 9 (7,9)

Step-by-step explanation:

solve by using system if equations. multiply top answer by 2, giving u 2y=2x+4 and y=2x-5 subtract the equations. y=9 (the x cancels out) then put 9 in for y. 9=x+2. subtract 2. giving u 7. and now x =7

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What is the square root of 18
kifflom [539]

Answer:

4.24

Step-by-step explanation:

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3 0
3 years ago
Identify the name and label the parts of the following polynomial: − 2x^4+24^2-10
lianna [129]
-2x^4 + 24x^2 - 10

u have 3 terms.....and since u are only dealing with 1 variable with the highest exponent being 4, then what u have here is a 4th degree trinomial.
U have a lead coefficient of -2.
Ur constant term is -10
And ur middle term (24x^2) has a degree of 2

a cubic binomial.....a binomial has 2 terms and it being cubic means the highest term has a degree of 3
example would be : x^3 - 4

how many constants can a polynomial have ? I am not sure about this one...I wanna say 1 because u can simplify it if it has more then 1...but I am not 100% sure on this one

3x^2 + 6xy -10x^5 + y^6 - 10x^3y^5

when u have a polynomial with more then 1 variable, such as this one, the degree is not the highest exponent, it is the highest term.....-10x^3y^5...u add the exponents....so this term has a degree of 8, and it is the highest one in this problem....so this is an 8th degree polynomial with 5 terms


7 0
3 years ago
Which are the solutions of the quadratic equation? x2 = –5x – 3 –5, 0 StartFraction negative 5 minus StartRoot 13 EndRoot Over 2
boyakko [2]

<u>Given</u>:

The quadratic equation is x^{2}=-5 x-3

We need to determine the solutions of the quadratic equation.

<u>Solution</u>:

Let us solve the equation to determine the value of x.

Adding both sides of the equation by 5x and 3, we get;

x^{2}+5 x+3=0

The solution of the equation can be determined using quadratic formula.

Thus, we get;

x=\frac{-5 \pm \sqrt{5^{2}-4 \cdot 1 \cdot 3}}{2 \cdot 1}

x=\frac{-5 \pm \sqrt{25-12}}{2 }

x=\frac{-5 \pm \sqrt{13}}{2 }

Thus, the two roots of the equation are x=\frac{-5 + \sqrt{13}}{2 } and x=\frac{-5- \sqrt{13}}{2 }

Hence, the solutions of the equation are x=\frac{-5 + \sqrt{13}}{2 } and x=\frac{-5- \sqrt{13}}{2 }

6 0
3 years ago
Read 2 more answers
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