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3 years ago

10

640 children participated in the local recreation program last year. This year the program saw a 5% decrease in the number of pa

rticipants. How many fewer children participated this year?

1 answer:

denis-greek [22]3 years ago

4 0

32 less children participated

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Find the mean of the following data set.

rosijanka [135]

Answer:

10

Step-by-step explanation:

the sum of 8,5,15,12,10 is 50 and there are 5 numbers so 50 divided by 5 is 10 and it's mean is also 10

hope this helps !

4 0

3 years ago

Read 2 more answers

15. If x=a Sin2t (I+Cos2t) and y=b Cos 2t (1-Cos2t) then find<br>dy/dx at =22/7*4<br>

Sav [38]

By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}$

It looks like we're given

$\begin{cases}x=a\sin(2t)(1+\cos(2t))\\y=b\cos(2t)(1-\cos(2t))\end{cases}$

where <em>a</em> and <em>b</em> are presumably constant.

Recall that

$\cos^2t=\dfrac{1+\cos(2t)}2$

$\sin^2t=\dfrac{1-\cos(2t)}2$

so that

$\begin{cases}x=2a\sin(2t)\cos^2t\\y=2b\cos(2t)\sin^2t\end{cases}$

Then we have

$\dfrac{\mathrm dx}{\mathrm dt}=4a\cos(2t)\cos^2t-4a\sin(2t)\cos t\sin t$

$\dfrac{\mathrm dy}{\mathrm dt}=-4b\sin(2t)\sin^2t+4b\cos(2t)\sin t\cos t$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{4b\cos(2t)\sin t\cos t-4b\sin(2t)\sin^2}{4a\cos(2t)\cos^2t-4a\sin(2t)\cos t\sin t}$

$\implies\boxed{\dfrac{\mathrm dy}{\mathrm dx}=\dfrac ba\tan t}$

where the last reduction follows from dividing through everything by $\cos(2t)\cos^2t$ and simplifying.

I'm not sure at which point you're supposed to evaluate the derivative (22/7*4, as in 88/7? or something else?), so I'll leave that to you.

8 0

3 years ago

What are the coordinates of the endpoints of the midsegment for △JKL that is parallel to KL⎯⎯⎯⎯⎯? Enter your answers in the boxe

algol [13]

The points you are looking for are the midpoints of segments JL and JK.
J(-2, -1), K(4, -5), L(0, -5)

The midpoint of segment JL is

(-2 + 0)/2, (-1 + (-5))/2) = (-2/2, -6/2) = (-1, -3)

The midpoint of segment JK is

(-2 + 4)/2, (-1 + (-5))/2) = (2/2, -6/2) = (1, -3)

Answer: The coordinates are (-1, -3), (1, -3)

6 0

3 years ago

Read 2 more answers

A system of linear equations has been graphed in the diagram. determine a reasonable solution for the system of equations.

Grace [21]

Answer:

(1,-1)

Step-by-step explanation:

The solution to a system of equation means the x and y values if we were to solve both equations (of lines) simultaneously.

That is also known as the intersecting points.

The top-right part is known as the 1st Quadrant in a coordinate system.

Top-left is 2nd Quadrant.

Bottom left is 3rd Quadrant.

Bottom right is 4th Quadrant.

The values of x and y (positive or negative) of each quadrant is shown below:

Quadrant 1: x +ve, y +ve
Quadrant 2: x -ve, y +ve
Quadrant 3: x -ve, y -ve
Quadrant 4: x +ve, y -ve

The intersection point (solution) of the 2 lines shown is in the 4th quadrant, which has x values positive and y values negative.

From the answer choices, we see that 3rd answer choice goes with this -- (1,-1), so this is the correct choice.

4 0

3 years ago

Find the slope of the line that contains (−3,5) and (3,−9).

snow_lady [41]

The slope is this and that

5 0

3 years ago