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Alenkasestr [34]
3 years ago
7

A Boxerville Express commuter train leaves the station every 30 minutes. When Theo arrives at the station, a woman tells him she

has been waiting for 12 minutes and has not yet seen a train. If the variable T stands for the amount of time Theo will have to wait, which of the following units could apply to this variable?
A. passengers
B. trains
C. minutes
D. miles per hour​
Mathematics
2 answers:
sladkih [1.3K]3 years ago
8 0
I believe the answer would be D
Dmitrij [34]3 years ago
4 0
Answer: D

The answe is just simply D because of the minutes when it adds up is almost the same as MPH
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What do I put? I am having so much trouble with this.
jek_recluse [69]
12. \frac{3}{4} ,  \frac{9}{12}

13. \frac{4}{10} ,  \frac{4}{100}

16. \frac{3}{9} is already in simplest form.
8 0
3 years ago
A snail travels 10 cm in 4 minutes. At this rate: How long will it take the snail to travel 24 cm? minutes How far does the snai
vovangra [49]

Answer:

1) 9 min 36 sec (9.6min)

2) 15 cm

Step-by-step explanation:

10 cm -->4 min

1 cm --> 0.4 min

24cm --> 9.6 min

15 cm--> 6min

3 0
3 years ago
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What is the product (2x-1)(x+4)
Ostrovityanka [42]
2x(x+4)-1(x+4)
2x^2+8x-x-4
2x^2+7x-4
8 0
3 years ago
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Find the area of the part of the plane 3x 2y z = 6 that lies in the first octant.
gavmur [86]

The area of the part of the plane 3x 2y z = 6 that lies in the first octant  is  mathematically given as

A=3 √(4) units ^2

<h3>What is the area of the part of the plane 3x 2y z = 6 that lies in the first octant.?</h3>

Generally, the equation for is  mathematically given as

The Figure is the x-y plane triangle formed by the shading. The formula for the surface area of a z=f(x, y) surface is as follows:

A=\iint_{R_{x y}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y(1)

The partial derivatives of a function are f x and f y.

\begin{aligned}&Z=f(x)=6-3 x-2 y \\&=\frac{\partial f(x)}{\partial x}=-3 \\&=\frac{\partial f(y)}{\partial y}=-2\end{aligned}

When these numbers are plugged into equation (1) and the integrals are given bounds, we get:

&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{(-3)^{2}+(-2)^2+1dxdy} \\\\&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{14} d x d y \\\\&=\sqrt{14} \int_{0}^{2}[y]_{0}^{3-\frac{3}{2} x} d x d y \\\\&=\sqrt{14} \int_{0}^{2}\left[3-\frac{3}{2} x\right] d x \\\\

&=\sqrt{14}\left[3 x-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3.2-\frac{3}{2} \cdot \frac{1}{2} \cdot 3^{2}\right] \\\\&=3 \sqrt{14} \text { units }{ }^{2}

In conclusion,  the area is

A=3 √4 units ^2

Read more about the plane

brainly.com/question/1962726

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5 0
1 year ago
A pilot flew a 400-mile flight in 2.5 hours flying into the wind . Flying the same rate and with the same wind speed, the return
k0ka [10]

Answer:

20

Step-by-step explanation:

pilot flew a 400-mile flight in 2.5 hours flying into the wind . Flying the same rate and with the same wind speed, the return trip took only 2hours with a tailwind

The speed can be defined as the ratio distance over time, but in this case the pilot is flying at the speed of the plane as well of speed of the wind. We can denote speed of plane as "s" and speed of the wind as "k"

Then the speed of the plane=(s-k) this is because it is moving in the opposite direction of the wind.

But From the question, the plane distance=400mile and the corresponding time =2.5 h

Then the speed of the plane=(s-k) =400mile/2.5

=160mph

By the time it return from trip, the speed the plane flew with is (s+k) because the wind is at the rear.

From the question, the plane distance=400mile and the corresponding time =2 h

Then (s+k)= (400/2)

(s+k)=200mph

8 0
3 years ago
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