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sp2606 [1]
3 years ago
10

the student of Turner Middle School are going on a field trip. there are 540 students attending. A bus can hold 45 students. How

many buses are needed for the field trip?
Mathematics
2 answers:
AnnZ [28]3 years ago
8 0
Divide 45 from 540. You get 12. The answer wold be 12 busses.
KiRa [710]3 years ago
7 0
All you need to do is divide the number of students by the number of students a bus can hold.

540 / 45 = 12. 
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Here is triangle ABC. Triangle XYZ is similar to ABC with scale factor ¼,center A
wlad13 [49]

Answer:

1

Step-by-step explanation:

4 0
2 years ago
PLS HURRY GIVING BRAINLIEST
hoa [83]

Answer:

<u>220.5 sq feet</u>

Step-by-step explanation:

Given : Bev has to cut her grandma's grass this weekend

Six-sided polygon that includes two isosceles right triangles, one with height and base of 15 feet, the other height and base of 4 feet, and one rectangle measuring 25 feet by 4 feet.

To Find : Area of Polygon

Solution:

Area of triangle = (1/2) bh

= (1/2) * 15 * 15

= 112.5   sq feet

= (1/2) * 4 * 4

=  8   sq feet

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3 0
2 years ago
Read 2 more answers
If You choose a card at random, what is the possibility of choosing an ace or a heart?
natima [27]
RE-POSTING MY PREVIOUS ANSWER
***************************************************************
there are 13 hearts in every deck as well as 4 aces making 17 cards. However, you are counting the ace of hearts twice so you are trying to choose 16 cards out of 52.
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16 / 52 = 4 / 13.

8 0
3 years ago
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Plzzzz help with atleast 1question❤️
velikii [3]

Answer:

4) 3, -2, -8

5) -7

6) E

7) D

8) 5, -1, 34

Step-by-step explanation:

4) There are at least two prominent points on line EF: (0, -4) and (6, 5). From there, we can find the slope-intercept formula of it.

The slope: (5 + 4) / (6 - 0) = 9 / 6 = 3 / 2

The intercept: (0, -4)

So, we have y = 3/2x - 4.

-3/2x + y + 4 = 0

Multiply every term by 2.

-3x + 2y + 8 = 0.

But the question asks for A to be >0! Multiply everything by -1.

3x - 2y - 8 = 0

So, the values of A, B, and C, respectively, are 3, -2, -8.

5) There are at least two prominent points on line GH: (-9, 4) and (-1, 2). From there, we have...

The slope: (4 - 2) / (-9 + 1) = 2 / -8 = -1/4

y = -1/4x + b

2 = (-1/4 * -1) + b

2 = (1/4) + b

b = 1.75; or 1 and 3/4; or 7/4.

Then, we have our intercept: 7/4.

We have an equation in slope-intercept form: y = -1/4x + 7/4

1/4x + y - 7/4 = 0

Multiply everything by 4.

x + 4y - 7 = 0.

The value of C is -7.

6) Lines that are parallel to the y-axis are vertical lines, which means that the x-values never change. We are looking for an equation where x = 3. That is code letter E.

7) Lines perpendicular to the y-axis are horizontal lines, which means that the y-values never change. We are looking for an equation where y = -5. That is code letter D (y + 5 = 0; y = -5).

8) A line parallel to 5x - 8y + 12 = 0 would have the same slope as it.

5x - 8y + 12 = 0

-8y = -5x - 12

y = 5/8x + 3/2

So, the line should have a slope of 5/8.

3 = (5/8) * (-2) + b

3 = -10/8 + b

3 = -5/4 + b

b = 3 + 5/4

b = 4 and 1/4; 17/4.

y = 5/8x + 17/4

-5/8x + y - 17/4 = 0

Multiply all terms by 8.

-5x + y - 34 = 0

Multiply all terms by -1.

5x - y + 34 = 0.

The values of A, B, and C, respectively, are 5, -1, 34.

I REALLY hope this helps because my brain kinda hurts now XD Have a great day!

3 0
3 years ago
Two Tibetan monks are having a race along a straight path. They both start at the same time, and the race ends in a tie. Prove t
Triss [41]
This is so provided that the velocity changes continuously in which case we can apply the mean value theorem. 
<span>Velocity (v) is the derivative of displacement (x) : </span>
<span>v = dx/dt </span>
<span>Monk 1 arrives after a time t* and Monk 2 too. </span>
<span>Name v1(t) and v2(t) their respective velocities throughout the trajectory. </span>
<span>Then we know that both average velocities were equal : </span>
<span>avg1 = avg2 </span>
<span>and avg = integral ( v(t) , t:0->t*) / t* </span>
<span>so </span>
<span>integral (v1(t), t:0->t*) = integral (v2(t), t:0->t*) </span>
<span>which is the same of saying that the covered distances after t* seconds are the same </span>
<span>=> integral (v1(t) - v2(t) , t:0->t*) = 0 </span>
<span>Thus, name v#(t) = v1(t) - v2(t) , then we obtain </span>
<span>=> integral ( v#(t) , t:0->t*) = 0 </span>
<span>Name the analytical integral of v#(t) = V(t) , then we have </span>
<span>=> V(t*) - V(0) = 0 </span>
<span>=> V(t*) = V(0) </span>
<span>So there exist a c in [0, t*] so that </span>
<span>V'(c) = (V(t*) - V(0)) / (t* - 0) (mean value theorem) </span>
<span>We know that V(0) = V(t*) = 0 (covered distances equal at the start and finish), so we get </span>
<span>V'(c) = v#(c) = v1(c) - v2(c) = 0 </span>
<span>=> v1(c) = v2(c) </span>
<span>So there exist a point c in [0, t*] so that the velocity of monk 1 equals that of monk 2. </span>
8 0
3 years ago
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