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xz_007 [3.2K]
4 years ago
9

Sketch the region bounded by the curves, and visually estimate the location of the centroid. (Do this on paper. Your instructor

may ask you to turn in this work.) Then find the exact coordinates of the centroid. y = ex y = 0, x = 0, x = 4

Mathematics
1 answer:
DerKrebs [107]4 years ago
3 0

So the Answers are in the following attachments

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18/27=1/3 9/15<4/4 3/4<2/3 5/6>11/12 which is a true statement?
hram777 [196]

Answer:

The correct answer us 9/15<4/4

Can I please have a brainliest? I want to rank up :)

6 0
3 years ago
PLEASEEEEEEE HEEELLLLPPPPP MEEE
Aleks04 [339]

Answer:

The second option

Step-by-step explanation:

For this problem, there are multiple ways to solve. The easiest here is the cross method. For ax²+bx+c, you find 2 numbers that multiply to ac and add to b, which you replace b with.

ac in this case is -8×3, which is -24. b in this case is 23.

You can go through the factor pairs of -24 to find the ones that add to b. These pairs are (1 and -24, -1 and 24, 2 and -12, -2 and 12, 3 and -8, -3 and 8, 4 and -6, -4 and 6). The only pair that adds to 23 is 24 and -1.

You sub this into the original equation, making 3x²+24x-x-8.

From there, you factorise, making 3x(x+8)-x(x+8).

This can be factorised further to (3x-x)(x+8) by collecting like terms.

**This question involves factorising quadratics, which you may wish to revise. I'm always happy to help!

3 0
3 years ago
Help!!!!!!!!!!!!!!!!​
kaheart [24]

Answer:

\cos\alpha\sin\alpha(\cos\alpha-\sin\alpha)

Step-by-step explanation:

First, simplify each term:

\sin\left(\dfrac{\pi}{2}+\alpha\right)=\cos \alpha\\ \\\cos \left(\dfrac{\pi}{2}+\alpha\right)=-\sin \alpha\\ \\\cos \left(\alpha-\dfrac{3\pi}{2}\right)=-\sin \alpha\\ \\\sin \left(\dfrac{3\pi}{2}+\alpha\right)=-\cos \alpha

Then given expression is equivalent to

\cos ^3\alpha+(-\sin \alpha)^3-(-\sin \alpha)+(-\cos \alpha)\\ \\=\cos ^3\alpha-\sin^3 \alpha+\sin \alpha-\cos \alpha\\ \\=(\cos\alpha-\sin\alpha)(\cos^2\alpha+\cos\alpha\sin\alpha+\sin^2\alpha)-(\cos\alpha-\sin\alpha)\\ \\=(\cos\alpha-\sin\alpha)(1+\cos\alpha\sin\alpha-1)\ \ [\cos^2\alpha+\sin^2\alpha=1]\\ \\=\cos\alpha\sin\alpha(\cos\alpha-\sin\alpha)

6 0
3 years ago
What are all the names to a polygon that has 2 pairs of parallel sides 2 pairs of sides equal in length and 2 acute and 2 obtuse
tester [92]

Answer:

a trapezoid

Step-by-step explanation:

A trapezoid is a convex quadrilateral. A trapezoid has at least on pair of parallel sides. the parallel sides are called bases while the non parallel sides are called legs

Characteristics of a trapezoid

It has 4 vertices and edges

If both pairs of its opposite sides of a trapezoid are parallel, it becomes a parallelogram

Area of a trapezoid =   x (sum of the lengths of the parallel sides) x height

Perimeter =  sum of lengths of sides of a trapezoid

5 0
3 years ago
Dominic is cutting a rectangular hole in his paper, from top left corner to the bottom right corner. The rectangle will be 4.5 c
Readme [11.4K]

Answer:

Step-by-step explanation:

So initially you have a square piece of metal, 15" x 15". You cut out squares from the corners of each side with side length x.

 

So you'll have a sheet that looks something like this now (assuming the formatting works):

 

   _____

__|       |__

|               |

|__        __|

   |____|

 

Your original side length was 15, but it has now been reduced by x on each side, so you're now going to have a new side length of 15 - 2x

 

When you fold the box up, the box will have a height of just x.

 

So you have a base with lengths (15 - 2x) and (15 - 2x), and a height which is just x.

 

V = l*w*h

 

So

 

V = (15 - 2x)(15 - 2x)(x)

 

= 225x - 60x^2 + 4x^3

 

Hope this helps

8 0
3 years ago
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