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3 years ago

Help please (●´⌓`●)

Mathematics

2 answers:

Triss [41]3 years ago

8 0

Answer: $39ft^2$

Step-by-step explanation:

This is a rectangle missing a piece which looks like a triangle.

Find the area of the rectangle and subtract the area of the triangle.

Rectangle:

$A=l*w\\A=(9ft)(5ft)\\A=45ft^2$

Triangle:

$A=\frac{hb}{2}\\ A=\frac{(3ft)(4ft)}{2}\\ A= \frac{12ft^2}{2}\\ A=6ft^2$

Now subtract the area of the triangle from the area of the rectangle.

$45ft^2-6ft^2=39ft^2$

Sonbull [250]3 years ago

3 0

The answer is 39ft squared

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The quadratic equation ax^2+32x+c=0 has exactly one solution. If a+c=130, and a>c, find the ordered pair (a,c).

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Need help and explain please!!

lukranit [14]

Answer:

$x=-4\text{ and } x=3$

Step-by-step explanation:

We are given the second derivative:

$g''(x)=(x-3)^2(x+4)(x-6)$

And we want to find its inflection points.

To do so, we will first determine possible inflection points. These occur whenever g''(x) = 0 or is undefined.

Next, we will test values for the intervals. Inflection points occur if and only if the sign changes before and after the point.

So first, finding the zeros, we see that:

$0=(x-3)^2(x+4)(x-6)\Rightarrow x=-4, 3, 6$

So, we can draw the following number-line:

<----(-4)--------------(3)----(6)---->

Now, we will test values for the intervals x < -4, -4 < x < 3, 3 < x < 6, and x > 6.

Testing for x < -4, we can use -5. So:

$g^\prime^\prime(-5)=(-5-3)^2(-5+4)(-5-6)=704>0$

Since we acquired a positive result, g(x) is concave up for x < -4.

For -4 < x < 3, we can use 0. So:

$g^\prime^\prime(0)=(0-3)^2(0+4)(0-6)=-216$

Since we acquired a negative result, g(x) is concave down for -4 < x < 3.

And since the sign changed before and after the possible inflection point at x = -4, x = -4 is indeed an inflection point.

For 3 < x < 6, we can use 4. So:

$g^\prime^\prime(4)=(4-3)^2(4+4)(4-6)=-16$

Since we acquired a negative result, g(x) is concave down for 3 < x < 6.

Since the sign didn't change before and after the possible inflection point at x = 3 (it stayed negative both times), x = -3 is not a inflection point.

And finally, for x > 6, we can use 7. So:

$g^\prime^\prime(7)=(7-3)^2(7+4)(7-6)=176>0$