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taurus [48]
3 years ago
12

Factor this expression completely. Write the factored expression in the space provided.

Mathematics
1 answer:
icang [17]3 years ago
5 0
<h3>I'll teach you how to solve y^2 - y - 12</h3>

--------------------------------------------------

y^2 - y - 12

Break the expression into groups:

(y^2+3y)+(-4y-12)

Factor out y from y^2+37:

y^2+37 = y(y+3)

Factor out -4 from -4y-12:

-4y-12 = -4(y+3)

y(y+3)-4y-12

Factor out common term y+3:

(y+3)(y-4)

Your Answer Is (y+3)(y-4)

plz mark me as brainliest :)

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Although five of John's answers were incorrect, 90% of his answers were correct. How many questions did John answer in all?
fiasKO [112]

Answer:

John answered 50 questions.

Step-by-step explanation:

Although five of John's answers were incorrect, 90% of his answers were correct.

This means that 5 is 100 - 90% = 10% of the total number of questions, n. So

0.1n = 5

n = \frac{5}{0.1}

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John answered 50 questions.

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2 years ago
Find CD if C(0,3) and D(4,7)
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3 years ago
1. Make sure to respond with complete sentences! Show all work please:)
tino4ka555 [31]

Explanation:

The numerator of the rational exponent will be the product of the exponents inside and outside the radical: 5·7=35. The denominator of the rational exponent will be the index of the radical: 6. Then the equivalent expression is x^(35/6)

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A point located at (6, -4) is reflected over the x-axis. What are the coordinates of the image?
frutty [35]
It will be (6,4) all u do is slip it over t he x-axis
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3 years ago
Read 2 more answers
A lidless box is to be made using 2m^2 of cardboard find the dimensions of the box that requires the least amount of cardboard
Jlenok [28]
1.8, Problem 37: A lidless cardboard box is to be made with a volume of 4 m3 . Find the dimensions of the box that requires the least amount of cardboard. Solution: If the dimensions of our box are x, y, and z, then we’re seeking to minimize A(x, y, z) = xy + 2xz + 2yz subject to the constraint that xyz = 4. Our first step is to make the first function a function of just 2 variables. From xyz = 4, we see z = 4/xy, and if we substitute this into A(x, y, z), we obtain a new function A(x, y) = xy + 8/y + 8/x. Since we’re optimizing something, we want to calculate the critical points, which occur when Ax = Ay = 0 or either Ax or Ay is undefined. If Ax or Ay is undefined, then x = 0 or y = 0, which means xyz = 4 can’t hold. So, we calculate when Ax = 0 = Ay. Ax = y − 8/x2 = 0 and Ay = x − 8/y2 = 0. From these, we obtain x 2y = 8 = xy2 . This forces x = y = 2, which forces z = 1. Calculating second derivatives and applying the second derivative test, we see that (x, y) = (2, 2) is a local minimum for A(x, y). To show it’s an absolute minimum, first notice that A(x, y) is defined for all choices of x and y that are positive (if x and y are arbitrarily large, you can still make z REALLY small so that xyz = 4 still). Therefore, the domain is NOT a closed and bounded region (it’s neither closed nor bounded), so you can’t apply the Extreme Value Theorem. However, you can salvage something: observe what happens to A(x, y) as x → 0, as y → 0, as x → ∞, and y → ∞. In each of these cases, at least one of the variables must go to ∞, meaning that A(x, y) goes to ∞. Thus, moving away from (2, 2) forces A(x, y) to increase, and so (2, 2) is an absolute minimum for A(x, y).
5 0
3 years ago
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