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V125BC [204]
3 years ago
10

BRAINLIEST HELPPPPPO

Mathematics
1 answer:
Zarrin [17]3 years ago
6 0

Question 5

To use the quadratic formula, we need to rewrite the equation so that all the terms are on one side: by subtracting (x+13) from both sides, we get that x^2-x-6=0. The quadratic formula tells us that the roots of this are \frac{-b \pm \sqrt{b^2-4ac}}{2a}, where a=1, b=-1, and c=-6. This is equal to \frac{-(-1)\pm \sqrt{(-1)^2-4(1)(-6)}}{2(1)} = \frac{1 \pm 5}{2}, which gives us the two roots -2 and 3.

Question 7

If you look at the attached graph, you will see that the graph of x^2-x-6 intersects the x-axis at the points (-2,0) and (3,0).

Question 8

The roots are the same. Both finding the x-intercepts and solving for the roots of a quadratic using the quadratic formula will give the values of x for which the function value is zero.

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Please help!! Thanks
Troyanec [42]
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3x-5y=10<br> -3x-4y=35<br> Find the Solution set . Solve the system using addition method ?
trapecia [35]
Hello: 
<span>3x-5y=10 .... (1)
-3x-4y=35 ..... (2) 
</span><span>using addition method (1)+(2) :-9y=45        y=-6
</span>subst in (1) :  3x-5(-6)=10
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