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RideAnS [48]
3 years ago
5

Maria had 4 full cement, blocks & one that was 5/9 the normal size, If each full block weight 3 8/9 pounds, What is the Weig

ht blocks maria has? (20 POINTS!)
Mathematics
2 answers:
nekit [7.7K]3 years ago
8 0
I have no idea I hope you find the awnser though
ANEK [815]3 years ago
5 0

Answer:

Every problem in 501 Math Word Problems has a complete answer ... Nicole jogs for 60 minutes four times a week, lifts weights for 45 ... blocks north, 2 blocks east, 1 block south, 2 blocks east, and 1 block ... If a serving size is 4. 3 -cup, how many servings does the box contain?

Step-by-step explanation:

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Step-by-step explanation:

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The volume of a cone-shaped funnel is 603 cubic centimeters. The radius of the funnel is 8 centimeters. Find the height of the f
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Step-by-step explanation:

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4 0
3 years ago
A hiker is hiking in a valley. The height of the valley is h(x,y)=4x2+y2 where x and y are the east-west and north-south distanc
Ainat [17]

Answer:

A. \frac{\partial{h}}{\partial{t}}=0

Step-by-step explanation:

A. The problems asked for 2 ways to solve it, expanding the equation with the substitution  x(t)=2 cos(t) and y(t)=4 sin(t) to differentiate it . The other way is by chain rule.

Expanding and differentiating:

We start by substituting x(t)=2 cos(t) and y(t)=4 sin(t) in h(x,y)=4x2+y2:

h(x,y)=4x^{2}+y^{2}= 4(2cos(t))^{2}+(4sin(t))^{2}\\h(x,y)=4(4cos^{2}(t))+(16sen^{2}(t))\\h(x,y)=16cos^{2}(t)+16sen^{2}(t)=16(sen^{2}(t)+cos^{2}(t))\\h(x,y)=16

So, in the path that the hiker chose:

\frac{\partial{h}}{\partial{t}}=0

Chain rule:

We start differentiating h(x,y) using chain rule as follows:

\frac{\partial{h}}{\partial{t}}= \frac{\partial{h}}{\partial{x}}\frac{\partial{x}}{\partial{t}}+\frac{\partial{h}}{\partial{y}}\frac{\partial{y}}{\partial{t}}

Now, it´s easy to find all these derivatives:

\frac{\partial{h}}{\partial{x}}=8x\\\frac{\partial{x}}{\partial{t}}=-2sin(t)\\\frac{\partial{h}}{\partial{y}}=2y\\\frac{\partial{y}}{\partial{t}}=4cos(t)

Now we replace them in the chain rule, with the replacement x=2cos(t) and y=4sin(t) in the x,y that are left and we operate everything:

\frac{\partial{h}}{\partial{t}}= 8x(-2sin(t))+2y(4cos(t)

\frac{\partial{h}}{\partial{t}}= 8(2cos(t))(-2sin(t))+2(4sin(t))(4cos(t)

\frac{\partial{h}}{\partial{t}}= -32cos(t)sin(t)+32sin(t)cos(t)

\frac{\partial{h}}{\partial{t}}= 0

This will be our answer

6 0
3 years ago
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