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3 years ago

What is the average acceleration of a southbound train that slows down from 15 m/s to 8.6 m/s in 1.2 s?

Mathematics

1 answer:

horrorfan [7]3 years ago

5 0

Answer:

The average acceleration of train is $5.34\ m/s^2$

Step-by-step explanation:

We have,

A train that slows down from 15 m/s to 8.6 m/s in 1.2 s. It means that 15 m/s is its initial velocity and 8.6 m/s is its final velocity.

It is required to find the average acceleration of the train.

$a=\dfrac{v-u}{t}\\\\a=\dfrac{(8.6-15)\ m/s}{1.2\ s}\\\\a=-5.34\ m/s^2$

The average acceleration of train is $5.34\ m/s^2$ . Negative signs shows that the train is decelerating.

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Can y’all give me the area of each figure

KIM [24]

Answer:

1. 208 in^2

Step-by-step explanation:

1. We can break the shape up into a rectangle in the middle and 2 triangles on either side of said rectangle.

The dimensions of the rectangle are 8 in by 20 in, and we only know one leg of the triangle as well as the hypotenuse.

If we know one leg and the hypotenuse we can use the pythagorean theormed to sovle for the other side and get 6 in.

So we have

(8 * 20) + 2((1/2)(6)(8))

160 + 48

208 in^2

5 0

3 years ago

Help with exercise

GalinKa [24]

The probability that a randomly selected chocolate bar will have between 200 and 220 calories is 3.97%

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

The zscore is given a:

z = (raw score - mean) / standard deviation

mean = 225, standard deviation o = 10.

a) For x = 200:

z = (200 - 225) / 200 = -0.125

For x = 220:

z = (220 - 225) / 200 = -0.025

P(-0.125 < z < -0.025) = P(z < -0.025) - P(z < -0.125) = 0.4880 - 0.4483 = 3.97%

b) For x = 190:

z = (190 - 225) / 200 = -0.175

P(z < -0.025) = P(z < -0.175) = 0.4286

The probability that a randomly selected chocolate bar will have between 200 and 220 calories is 3.97%

Find out more on equation at: brainly.com/question/2972832

#SPJ1

7 0

2 years ago

ZE is the angle bisector of AngleYEX and the perpendicular bisector of Line segment G F. Line segment G X is the angle bisector

frozen [14]

Answer:

Point A is the center of the circle that passes through points E, F, and G and the center of the circle that passes through points X, Y, and Z.

Step-by-step explanation:

A is the intersection of angle bisectors, so is the incenter of triangle EFG. It is also the intersection of the perpendicular bisectors of the sides of triangle EFG, so is the circumcenter.

The altitudes at X, Y, and Z are perpendicular to sides EF, EG, and FG, and pass through the incenter, so X, Y, Z are points on the incircle.

A is the center of circles through E, F, and G, and through X, Y, and Z.

6 0

4 years ago

Read 2 more answers

A rectangular park of length 60 m breadth 50 m encloses w volleyball court of length 18 m and breadth 10 m. Find the area of the

Zanzabum

Given

A rectangular park of length 60 m breadth 50 m encloses with volleyball court of length 18 m and breadth 10 m.

To find:

The area of the park excluding the court at the rate of Rs 110 per square meter.

Solution:

Area of a rectangle is:

$Area=length \times breadth$

Area of whole park is:

$A_1=60 \times 50$

$A_1=3000$

Area of volleyball court is:

$A_2=18 \times 10$

$A_2=180$

Now, the area of the park excluding the court is:

$A=A_1-A_2$

$A=3000-180$

$A=2820$

Therefore, the area of the park excluding the court is 2820 square meter.

4 0

3 years ago

50 POINTS!!! In rectangle ABCD, AB = 6 cm, BC = 8 cm, and DE = DF. The area of triangle DEF is one-fourth the area of rectangle

aalyn [17]

Answer:

$EF=4\sqrt{3}$

Step-by-step explanation:

In rectangle ABCD, AB = 6, BC = 8, and DE = DF.

ΔDEF is one-fourth the area of rectangle ABCD.

We want to determine the length of EF.

First, we can find the area of the rectangle. Since the length AB and width BC measures 6 by 8, the area of the rectangle is:

$A_{\text{rect}}=8(6)=48\text{ cm}^2$

The area of the triangle is 1/4 of this. Therefore:

$\displaystyle A_{\text{tri}}=\frac{1}{4}(48)=12\text{ cm}^2$

The area of a triangle is half of its base times its height. The base and height of the triangle is DE and DF. Therefore:

$\displaystyle 12=\frac{1}{2}(DE)(DF)$

Since DE = DF:

$24=DF^2$

Thus:

$DF=\sqrt{24}=\sqrt{4\cdot 6}=2\sqrt{6}=DE$

Since ABCD is a rectangle, ∠D is a right angle. Then by the Pythagorean Theorem:

$(DE)^2+(DF)^2=(EF)^2$

Therefore:

$(2\sqrt6)^2+(2\sqrt6)^2=EF^2$

Square:

$24+24=EF^2$

Add:

$EF^2=48$

And finally, we can take the square root of both sides:

$EF=\sqrt{48}=\sqrt{16\cdot 3}=4\sqrt{3}$

6 0

3 years ago

Read 2 more answers