Find the area of the region between the curves y=12-x^2 and y=x^2-6

spayn [35]

The answer is 4
The area is length X width
It is also a square meaning all sides are equal
So with this is mind we can ask the question what times what equals 16 and is the same number
This would be 4 since 4 times 4 is sixteen

4 0

3 years ago

Dose someone know hope to do this can someone help

Art [367]

The point where they intersect

5 0

2 years ago

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Which of the following represents the zeros of f(x) = 2x3 − 5x2 − 28x + 15?

likoan [24]

$\mathrm{Use\:the\:rational\:root\:theorem}$

$a_0=15,\:\quad a_n=2$

$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:3,\:5,\:15,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\:2$

$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:3,\:5,\:15}{1,\:2}$

$-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3$

$\mathrm{Compute\:}\frac{2x^3-5x^2-28x+15}{x+3}\mathrm{\:to\:get\:the\:rest\:of\:the\:eqution:\quad }2x^2-11x+5$

$=\left(x+3\right)\left(2x^2-11x+5\right)$

$Factor: 2x^2-11x+5$

$2x^2-11x+5=\left(2x^2-x\right)+\left(-10x+5\right)$

$=x\left(2x-1\right)-5\left(2x-1\right)$

$2x^3-5x^2-28x+15=\left(x+3\right)\left(x-5\right)\left(2x-1\right)$

$\left(x+3\right)\left(x-5\right)\left(2x-1\right)=0$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:}$

thus zeros of f(x) is

$x=-3,\:x=5,\:x=\frac{1}{2}$

5 0

2 years ago

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What is the M.A.D of 1,1,1,1,5?

algol [13]

I think it is 3.6 .......

8 0

2 years ago

Mason gave the waiter a $14.58 tip, which was 15 percent of the dinner bill. What was the amount of the dinner bill before he ad

lapo4ka [179]

Answer:

21.87

Step-by-step explanation:

8 0

3 years ago