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3 years ago

8

The graph represents an object that is shot upwards from a tower and then falls to the ground. The independent variable is time

in seconds and the dependent variable is the object’s height above the ground in meters.

1 answer:

elixir [45]3 years ago

3 0

Answer:

maybe pemdas or choose c lol

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Convert 0.0855 to a fraction. Move numbers to the blanks to complete the fraction.

Elis [28]

Answer:

a

Step-by-step explanation:

no

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3 years ago

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What is equivalent to 2 (m-4) + 1

natima [27]

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2m-7

Step-by-step explanation:

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3 years ago

1. Use the data points about the distance and elevations of the camps to graph the data points where the x-coordinate is the tot

Vadim26 [7]

You should plot the total distance traveled from the base camp on the x-coordinate while the elevation should be plotted on the y-coordinate.

<h3>What is a graph?</h3>

A graph simply refers to a type of chart which is commonly used to graphically represent data on both the vertical and horizontal lines of a cartesian coordinate (x-coordinate and y-coordinate).

<h3>How to plot this graph?</h3>

In this scenario, you would plot the total distance traveled from the base camp on the x-coordinate while the elevation of the camps would be plotted on the y-coordinate as shown in the image attached below.

Read more on graphs here: brainly.com/question/25799000

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2 years ago

Solve the following initial-value problem, showing all work, including a clear general solution as well as the particular soluti

Vikki [24]

Answer:

General Solution is $y=x^{3}+cx^{2}$ and the particular solution is $y=x^{3}-\frac{1}{2}x^{2}$

Step-by-step explanation:

$x\frac{\mathrm{dy} }{\mathrm{d} x}=x^{3}+3y\\\\Rearranging \\\\x\frac{\mathrm{dy} }{\mathrm{d} x}-3y=x^{3}\\\\\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{3y}{x}=x^{2}$

This is a linear diffrential equation of type

$\frac{\mathrm{d} y}{\mathrm{d} x}+p(x)y=q(x)$ ..................(i)

here $p(x)=\frac{-2}{x}$

$q(x)=x^{2}$

The solution of equation i is given by

$y\times e^{\int p(x)dx}=\int e^{\int p(x)dx}\times q(x)dx$

we have $e^{\int p(x)dx}=e^{\int \frac{-2}{x}dx}\\\\e^{\int \frac{-2}{x}dx}=e^{-2ln(x)}\\\\=e^{ln(x^{-2})}\\\\=\frac{1}{x^{2} } \\\\\because e^{ln(f(x))}=f(x)]\\\\Thus\\\\e^{\int p(x)dx}=\frac{1}{x^{2}}$

Thus the solution becomes

$\tfrac{y}{x^{2}}=\int \frac{1}{x^{2}}\times x^{2}dx\\\\\tfrac{y}{x^{2}}=\int 1dx\\\\\tfrac{y}{x^{2}}=x+c$ $y=x^{3}+cx^{2$

This is the general solution now to find the particular solution we put value of x=2 for which y=6

we have $6=8+4c$

Thus solving for c we get c = -1/2

Thus particular solution becomes

$y=x^{3}-\frac{1}{2}x^{2}$

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4 years ago

Translate point P(3,3) left 1 unit and up 4 units. What are the coordinates of the image P’?

ANTONII [103]

The coordinates would be (2,7) you just have to move 1 left 4 up

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3 years ago