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mote1985 [20]
3 years ago
14

A plane flies 720 miles with the wind in 3 hours. The return trip against the wind takes 4 hours. What is the speed of the wind

and the speed of the plane in still air?
Mathematics
1 answer:
Mandarinka [93]3 years ago
4 0
Let p and w represent the speed of the plane and the speed of the wind, respectively.
.. speed = distance/time
.. p +w = 720/3 = 240
.. p -w = 720/4 = 180

Add the two equations to eliminate w.
.. 2p = 420
.. p = 210
.. w = p -180 = 30

The speed of the wind is 30 mph.
The speed of the plane in still air is 210 mph.
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Point K is on line segment JL. Given JK = 5x + 7, JL = 2x+8, and KL = 4,
kati45 [8]

Answer:

The numerical length of JL is 6 units

Step-by-step explanation:

Here, we want to determine the numerical length of JL

Mathematically;

JL = JK + KL

2x + 8 = 5x + 7 + 4

2x + 8 = 5x + 11

5x -2x = 8-11

3x = -3

x = -1

But JL = 2x + 8

JL = 2(-1) + 8 = -2 + 8 = 6

3 0
3 years ago
Can someone please help quick!
Jet001 [13]

Answer:

the answer is 8

Step-by-step explanation:

1/2(8 + 4) = 6

0.5(8 + 4) = 6

(4 + 2) = 6

6 = 6

7 0
3 years ago
the performance score of 10 adults is recorded, and the results are 83 87 90 92 93 100 104 111 115 121 find the standard deviati
luda_lava [24]

Answer:

The standard deviation of the data set is \sigma = 12.7906.

Step-by-step explanation:

The Standard Deviation is a measure of how spread out numbers are. Its symbol is σ (the greek letter sigma)

To find the standard deviation of the following data set

\begin{array}{cccccccc}83&87&90&92&93&100&104&111\\115&121&&&&&&\end{array}

we use the following formula

                                             \sigma = \sqrt{ \frac{ \sum{\left(x_i - \overline{X}\right)^2 }}{n-1} }

Step 1: Find the mean \left( \overline{X} \right).

The mean of a data set is the sum of the terms divided by the total number of terms. Using math notation we have:

                                     Mean = \frac{Sum ~ of ~ terms}{Number ~ of ~ terms}

Mean = \frac{Sum ~ of ~ terms}{Number ~ of ~ terms}=\frac{83+87+90+92+93+100+104+111+115+121}{10} \\\\Mean = \frac{996}{10} =\frac{498}{5}=99.6

Step 2: Create the below table.

Step 3: Find the sum of numbers in the last column to get.

\sum{\left(x_i - \overline{X}\right)^2} = 1472.4

Step 4: Calculate σ using the above formula.

\sigma = \sqrt{ \frac{ \sum{\left(x_i - \overline{X}\right)^2 }}{n-1} }       = \sqrt{ \frac{ 1472.4 }{ 10 - 1} } \approx 12.7906

3 0
3 years ago
The point (16, 12) lies on the terminal side of an angle. find cos, tan, and sin
frez [133]
The terminal side means that it's 16 to the right and 12 up. Which means that the triangle made from this has the sides of 16 and 12 and the hypotenuse is 20 units long. Theta, in this case, would have 16 as it's adjacent and 12 as it's opposite.
Therefore, using the Pythagorean identities:
<span>sin(\theta) = \frac{12}{20} = \boxed{\frac{3}{5}} \\\ cos(\theta) = \frac{16}{20} = \boxed{\frac{4}{5}} \\\ tan(\theta) = \frac{12}{16} = \boxed{\frac{3}{4}} </span>
8 0
3 years ago
NO LINKS OR FILES!
Archy [21]

(a) If the particle's position (measured with some unit) at time <em>t</em> is given by <em>s(t)</em>, where

s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}

then the velocity at time <em>t</em>, <em>v(t)</em>, is given by the derivative of <em>s(t)</em>,

v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}

(b) The velocity after 3 seconds is

v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}

(c) The particle is at rest when its velocity is zero:

\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|

In interval notation, this happens for <em>t</em> in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

\displaystyle \int_0^8 |v(t)|\,\mathrm dt

By definition of absolute value, we have

|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)

In part (d), we've shown that <em>v(t)</em> > 0 when -√11 < <em>t</em> < √11, so we split up the integral at <em>t</em> = √11 as

\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt

and by the fundamental theorem of calculus, since we know <em>v(t)</em> is the derivative of <em>s(t)</em>, this reduces to

s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}

7 0
2 years ago
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