The last answer is the correct one.
If it does not make sense ask for clarification.
If it does not make sense ask for clarification.
I REQUIRE ASSISTANCE!
1 answer:
You might be interested in
By solving the system of equations, we can see that the solutions are:
(A, B) = (4, 9)
(C, D) = (1, -6).
<h3>How to solve the system of equations?</h3>Here we have the following system of equations:
x^2 - y = 7
y - 5x = -11
If we isolate the variable y in both equations, we get:
y = x^2 - 7
y = -11 + 5x
Now we can equate these two to get:
x^2 - 7 = -11 + 5x
now we have a quadratic equation, this can be rewritten as:
x^2 - 7 + 11 - 5x = 0
x^2 - 5x + 4 = 0
Using the quadratic formula, we will see that the solutions are:
Solving that we get:
So the two solutions for x are:
x = (5 + 3)/2 = 4
x = (5 - 3)/2 = 1
Evaluating the second equation in these x-values we get:
y = -11 + 5*4 = -11 + 20 = 9
y = -11 + 5*1 = -6
Then we have the coordinate pairs: (4, 9) (on the first quadrant) and (1, -6) on the fourth quadrant.
Learn more about systems of equations:
#SPJ1
Answer: Anything between 0 and 10, excluding both endpoints.
In terms of symbols we can say 0 < w < 10 where w is the width.
===================================================
Explanation:
You could do this with two variables, but I think it's easier to instead use one variable only. This is because the length is dependent on what you pick for the width.
w = width
2w = twice the width
2w-5 = five less than twice the width = length
So,
- width = w
- length = 2w-5
which lead to
area = length*width
area = (2w-5)*w
area = 2w^2-5w
area < 150
2w^2 - 5w < 150
2w^2 - 5w - 150 < 0
To solve this inequality, we will solve the equation 2w^2-5w-150 = 0
Use the quadratic formula. Plug in a = 2, b = -5, c = -150
Ignore the negative solution as it makes no sense to have a negative width.
The only practical root is w = 10.
If w = 10 feet, then the area = 2w^2-5w results in 150 square feet.
----------------------
Based on that root, we need to try a sample value that is to the left of it.
Let's say we try w = 5.
2w^2 - 5w < 150
2*5^2 - 5*5 < 150
25 < 150 ... which is true
This shows that if 0 < w < 10, then 2w^2-5w < 150 is true.
Now try something to the right of 10. I'll pick w = 15
2w^2 - 5w < 150
2*15^2 - 5*15 < 150
375 < 150 ... which is false
It means w > 10 leads to 2w^2-5w < 150 being false.
Therefore w > 10 isn't allowed if we want 2w^2-5w < 150 to be true.