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oee [108]
3 years ago
11

A ball drops from the top of a tower. At the same time a rocket is launched from a different level of the tower.

Mathematics
1 answer:
ExtremeBDS [4]3 years ago
5 0

Answer:a

Step-by-step explanation:

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SVETLANKA909090 [29]

Answer: I believe it's nonlinear

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3 years ago
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Please give me the answer to number 2 you don’t need to explain it just give me the answer
Yakvenalex [24]

Hello from MrBillDoesMath!

Answer:   Perimeter = 17x + 4

Discussion:

(So you are trying to sneak #2 out of me too..... );

Perimeter = sum of lengths of sides. Starting at the bottom of the figure and proceeding counterclockwise.

Perimeter =  (3x +2) + (3x+2) + (4x-1) + (4x-1) + (3x + 2) .

Separate and combine x's and constants:

Perimeter = (3x + 3x+4x+4x+3x) + (2 + 2 -1 -1 + 2) =

                = 17x + 4

You answer has the wrong constant value ( you wrote 17x + 1. The constant should be 4)


Thank you,

MrB

8 0
3 years ago
What is 23.7 minus 11.82
larisa86 [58]

Answer:

11.88

Step-by-step explanation:

4 0
3 years ago
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Let ????C be the positively oriented square with vertices (0,0)(0,0), (2,0)(2,0), (2,2)(2,2), (0,2)(0,2). Use Green's Theorem to
bonufazy [111]

Answer:

-48

Step-by-step explanation:

Lets call L(x,y) = 10y²x, M(x,y) = 4x²y. Green's Theorem stays that the line integral over C can be calculed by computing the double integral over the inner square  of Mx - Ly. In other words

\int\limits_C {L(x,y)} \, dx + M(x,y) \, dy =  \int\limits_0^2\int\limits_0^2 (M_x - L_y ) \, dx \, dy

Where Mx and Ly are the partial derivates of M and L with respect to the x variable and the y variable respectively. In other words, Mx is obtained from M by derivating over the variable x treating y as constant, and Ly is obtaining derivating L over y by treateing x as constant. Hence,

  • M(x,y) = 4x²y
  • Mx(x,y) = 8xy
  • L(x,y) = 10y²x
  • Ly(x,y) = 20xy
  • Mx - Ly = -12xy

Therefore, the line integral can be computed as follows

\int\limits_C {10y^2x} \, dx + {4x^2y} \,dy = \int\limits_0^2\int\limits_0^2 -12xy \, dx \, dy

Using the linearity of the integral and Barrow's Theorem we have

\int\limits_0^2\int\limits_0^2 -12xy \, dx \, dy = -12 \int\limits_0^2\int\limits_0^2 xy \, dx \, dy = -12 \int\limits_0^2\frac{x^2y}{2} |_{x = 0}^{x=2} \, dy = -12 \int\limits_0^22y \, dy \\= -24 ( \frac{y^2}{2} |_0^2) = -24*2 = -48

As a result, the value of the double integral is -48-

3 0
4 years ago
F(x)=-x^2-5 find the range
lina2011 [118]

x =  \frac{ - b}{2a}  \\  =   \frac{0}{2(1)}  = 0 \\ f(0) = (0)^{2}  - 5 \\  = 0 - 5 =  - 5
The vertex is at (0,-5), therefore the range goes from (-infinity, - 5]
4 0
3 years ago
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