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3 years ago

Please help find the percent of increase?!

Mathematics

1 answer:

alex41 [277]3 years ago

7 0

Take how much was added, divide by how much you originally had. 100-80=20. 20/80= 25%

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Graph the linear equation to -2x+7y=-21

mr Goodwill [35]

Answer:

This equation is in standard form, so to make graphing easier you need to put it in y=mx+b form or slope, intercept form.

Step-by-step explanation:

I have attached a photo of what the graph would look like.

However, in the future you should know how to turn this into a slope intercept equation.

-2x+7y=-21

+2x +2x

7y=2x-21

divide the entire equation by 7

y= 2/7x-3

That is a full equation in slope intercept form.

Hopefully this helps!

8 0

3 years ago

Read 2 more answers

For the high school basketball game, it costs $6 for every 3 tickets. complete the table below showing the cost and the number o

zhenek [66]

Every 3 tickets cost $6

Tickets Cost
3 $6
6 $12
9 $18
12 $24
15 $30
18 $36
21 $42
24 $48
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8 0

3 years ago

The coordinates of a particle in the metric xy-plane are differentiable functions of time t with:

frez [133]

Answer:

The rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12 is $-7.6\:\frac{m}{s}$ .

Step-by-step explanation:

This is an example of a related rate problem. A related rate problem is a problem in which we know one of the rates of change at a given instant $\frac{dx}{dt}$ and we want to find the other rate $\frac{dy}{dt}$ at that instant.

We know the rate of change of x-coordinate and y-coordinate:

$\frac{dx}{dt}=-2\:\frac{m}{s} \\\\\frac{dy}{dt}=-8\:\frac{m}{s}$

We want to find the rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12.

The distance of a point (x, y) and the origin is calculated by:

$s=\sqrt{x^2+y^2}$

We need to use the concept of implicit differentiation, we differentiate each side of an equation with two variables by treating one of the variables as a function of the other.

If we apply implicit differentiation in the formula of the distance we get

$s=\sqrt{x^2+y^2}\\\\s^2=x^2+y^2\\\\2s\frac{ds}{dt}= 2x\frac{dx}{dt}+2y\frac{dy}{dt}\\\\\frac{ds}{dt}=\frac{1}{s}(x\frac{dx}{dt}+y\frac{dy}{dt})$

Substituting the values we know into the above formula

$s=\sqrt{9^2+12^2}=15$

$\frac{ds}{dt}=\frac{1}{15}(9(-2)+12(-8))\\\\\frac{ds}{dt}=\frac{1}{15}\left(-18-96\right)\\\\\frac{ds}{dt}=\frac{1}{15}(-114)=-\frac{38}{5}=-7.6\:\frac{m}{s}$