<span>60
Sorry, but the value of 150 you entered is incorrect. So let's find the correct value.
The first thing to do is determine how large the Jefferson High School parking lot was originally. You could do that by adding up the area of 3 regions. They would be a 75x300 ft rectangle, a 75x165 ft rectangle, and a 75x75 ft square. But I'm lazy and another way to calculate that area is take the area of the (300+75)x(165+75) ft square (the sum of the old parking lot plus the area covered by the school) and subtract 300x165 (the area of the school). So
(300+75)x(165+75) - 300x165 = 375x240 - 300x165 = 90000 - 49500 = 40500
So the old parking lot covers 40500 square feet. Since we want to double the area, the area that we'll get from the expansion will also be 40500 square feet. So let's setup an equation for that:
(375+x)(240+x)-90000 = 40500
The values of 375, 240, and 90000 were gotten from the length and width of the old area covered and one of the intermediate results we calculated when we figured out the area of the old parking lot. Let's expand the equation:
(375+x)(240+x)-90000 = 40500
x^2 + 375x + 240x + 90000 - 90000 = 40500
x^2 + 615x = 40500
x^2 + 615x - 40500 = 0
Now we have a normal quadratic equation. Let's use the quadratic formula to find its roots. They are: -675 and 60. Obviously they didn't shrink the area by 675 feet in both dimensions, so we can toss that root out. And the value of 60 makes sense. So the old parking lot was expanded by 60 feet in both dimensions.</span>
Answer:
range of f(x) = [-4, -2) ∪ [2, 8)
a+b+c+d = -4
Step-by-step explanation:
The graph is attached. The range is the vertical extent of the function. It is defined at f(0) = -4 and f(2) = 2.
The limits f(2-) and f(4-) are -2 and 8, respectively, so the graph has open circles there. These are the ends of the two half-open intervals that make up the range of the function.
The portion of the graph in the domain [4, 7) is included in the range [2, 8), so no special treatment is needed for that piece of the function.
It’s 4.2 because it’s and given and you add then yeah you get the point
B. The product of two irrational numbers if rational
Example: The square root of 2 times the square root of 3, is the square root of 6, which is still irrational.
Hey There!
Remember, first find the gcf of 4,3 in this problem it is 12.
Steps to solve
1.6
- 3
2.6
- 3
= 2
So, The answer is <span>2
</span>
Hope This Helps :)
