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3 years ago

I need help don't know no links

Mathematics

1 answer:

DaniilM [7]3 years ago

3 0

Answer:

Slope = -1

Step-by-step explanation:

Hope it helps you.......

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What is the percent of change from 75 to 14 rounded to nearest percent??

zhuklara [117]

- 81 %

Step-by-step explanation:

To find the percentage change;

([Current – Initial] / Initial) * 100

([14 - 75] / 75) * 100

(-61 / 75) * 100

(-0.81333) * 100

= - 81.3333 %

To the nearest percentage;

= - 81 %

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3 0

2 years ago

You could win a giant jar of M&Ms, if only you could guess how many are in the jar. You find a smallerjar such that the gian

sleet_krkn [62]

Answer:

There are 7360 M&Ms in larger jar.

Step-by-step explanation:

We are given the following in the question:

A giant jar is such that it is exactly four times times the smaller jar in dimensions.

If r and h is the radius and height of the cylindrical jar respectively, then the dimensions of the larger are:

$R = 4r\\H=4h$

Number of M&Ms in smaller jar = 115

Volume of smaller jar =

$v = \dfrac{1}{3}\pi r^2 h$

Volume of larger jar =

$V =\dfrac{1}{3}\pi R^2 H=\dfrac{64}{3}\pi r^2 h$

Number of M&Ms in larger jar:

$v = \dfrac{1}{3}\pi r^2 h\rightarrow 115\\\\V=\dfrac{64}{3}\pi r^2 h\rightarrow 115\times 64\\\\V=\dfrac{64}{3}\pi r^2 h\rightarrow 7360$

Thus, there are 7360 M&Ms in larger jar.

6 0

3 years ago

A car dealership sold 850 cars last year. This year the dealership sold 715 cars. What is the percent decrease, to the nearest p

Bumek [7]

The answer is 16%

Hope this helps :)

3 0

2 years ago

Read 2 more answers

Wayne is 12 mi due north of Jefferson, and Oakland is due west of Jefferson. Wayne is 28 mi from Oakland. How far apart are Jeff

bija089 [108]

The anwer is 233,000 and that is the anwer so i need a thank you

3 0

3 years ago

Find the solution of the given initial value problem. ty' + 2y = sin t, y π 2 = 9, t > 0 y(t) =

Helen [10]

For the ODE

$ty'+2y=\sin t$

multiply both sides by t so that the left side can be condensed into the derivative of a product:

$t^2y'+2ty=t\sin t$

$\implies(t^2y)'=t\sin t$

Integrate both sides with respect to t :

$t^2y=\displaystyle\int t\sin t\,\mathrm dt=\sin t-t\cos t+C$

Divide both sides by $t^2$ to solve for y :

$y(t)=\dfrac{\sin t}{t^2}-\dfrac{\cos t}t+\dfrac C{t^2}$

Now use the initial condition to solve for C :

$y\left(\dfrac\pi2\right)=9\implies9=\dfrac{\sin\frac\pi2}{\frac{\pi^2}4}-\dfrac{\cos\frac\pi2}{\frac\pi2}+\dfrac C{\frac{\pi^2}4}$

$\implies9=\dfrac4{\pi^2}(1+C)$

$\implies C=\dfrac{9\pi^2}4-1$

So the particular solution to the IVP is

$y(t)=\dfrac{\sin t}{t^2}-\dfrac{\cos t}t+\dfrac{\frac{9\pi^2}4-1}{t^2}$

$y(t)=\dfrac{4\sin t-4t\cos t+9\pi^2-4}{4t^2}$

6 0

2 years ago