Answer:
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Answer:
Option C is correct.
Step-by-step explanation:
y = x^2-x-3 eq(1)
y = -3x + 5 eq(2)
We can solve by substituting the value of y in eq(2) in the eq(1)
-3x+5 = x^2-x-3
x^2-x+3x-3-5=0
x^2+2x-8=0
Now factorizing the above equation
x^2+4x-2x-8=0
x(x+4)-2(x+4)=0
(x-2)(x+4)=0
(x-2)=0 and (x+4)=0
x=2 and x=-4
Now finding the value of y by placing value of x in the above eq(2)
put x =2
y = -3x + 5
y = -3(2) + 5
y = -6+5
y = -1
Now, put x = -4
y = -3x + 5
y = -3(-4) + 5
y = 12+5
y =17
so, when x=2, y =-1 and x=-4 y=17
(2,-1) and (-4,17) is the solution.
So, Option C is correct.
Do you mean
![3 \sqrt[3]{125}](https://tex.z-dn.net/?f=3%20%5Csqrt%5B3%5D%7B125%7D%20)
or
![\sqrt[3]{125}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B125%7D%20)
remember that
![\sqrt[n]{x^m}=x^\frac{m}{n}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%5Em%7D%3Dx%5E%5Cfrac%7Bm%7D%7Bn%7D)
resolve the
![\sqrt[3]{125}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B125%7D)
part first
![\sqrt[3]{125}=](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B125%7D%3D)
![\sqrt[3]{5^3}=](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B5%5E3%7D%3D)
so
![3 \sqrt[3]{125}=3*5=15](https://tex.z-dn.net/?f=3%20%5Csqrt%5B3%5D%7B125%7D%3D3%2A5%3D15%20)
or
![\sqrt[3]{125} =5](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B125%7D%20%3D5)
not sure which one you mean
2.8-(-5.1)
-(-5.1)=5.1
2.8+5.1= 7.9
Is this what you were asking?
