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OlgaM077 [116]
2 years ago
14

Can someone help me out pls.

Mathematics
1 answer:
Eduardwww [97]2 years ago
3 0

Ok ill answer the question but do you want me to evaluate it or find it to the simplest form ?

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PLEASE HELP ME<br> ( I WILL MARK BRAINLIEST )
AlekseyPX
The numbers are 15, 17 and 19
8 0
1 year ago
HELP! ASAP!
Novosadov [1.4K]

Step-by-step explanation:

Part A:

So the height is going to be x when you fold the sides up. So that's one part of the volume but for the width it was going to be 4 but since two corners were cut out with the length x the new width is going to be (4-2x). The same thing applies for the length which should be 8 inches but since two corners were removed with the length x it's now (8-2x)

v = x(4-2x)(8-2x)

Part B:

The volume can be graphed although there must be a domain restriction since the height, width, or length cannot be negative. So let's look at each part of the equation

so for the x in front it must be greater than 0 to make sense

for the (4-2x), the x must be less than 2 or else the width is  negative.

for the (8-2x) the x must be less than 4 or else the length is negative

so the domain is going to be restricted to 0 < x < 2 so all the dimensions are greater than 0

By using a graphing calculator you can see the maximum of the given equation with the domain restrictions is 0.845 which gives a volume of 12.317  

5 0
2 years ago
PLEASE HELP QUICK
son4ous [18]
A should be the correct answer

7 0
2 years ago
Read 2 more answers
Solve the system of equations by graphing.<br> X+ y = 7<br> 2x - 4y=8
Sidana [21]

Answer:

(6,1)

Step-by-step explanation:

after switching the equations from y=mx+b, graph both equations. you will see that the point of intersection is (6,1)

6 0
3 years ago
G with the definition of covariance, prove cov[(ay − b),(cy − d)] = accov(x, y ), where x, y are random variables and a, b, c, d
____ [38]

By definition of covariance,


\mathrm{Cov}(X,Y)=\mathbb E[(X-\mathbb E[X])(Y-\mathbb E[Y])]


\mathrm{Cov}(X,Y)=\mathbb E[XY-\mathbb E[X]Y-X\mathbb E[Y]+\mathbb E[X]\mathbb E[Y]]=\mathbb E[XY]-\mathbb E[X]\mathbb E[Y]


We have


\mathbb E[(aX-b)(cY-d)]=\mathbb E[acXY-adX-bcY+bd]

=ac\mathbb E[XY]-ad\mathbb E[X]-bc\mathbb E[Y]+bd


\mathbb E[aX-b]=a\mathbb E[X]-b


\mathbb E[cY-d]=c\mathbb E[Y]-d


\mathbb E[aX-b]\mathbb E[cY-d]=ac\mathbb E[X]\mathbb E[Y]-ad\mathbb E[X]-bc\mathbb E[Y]+bd


Putting everything together, we find the covariance reduces to


\mathrm{Cov}(aX-b,cY-d)=ac(\mathbb E[XY]-\mathbb E[X]\mathbb E[Y])=ac\mathrm{Cov}(X,Y)


as desired.

5 0
2 years ago
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