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3 years ago

What is the relationship between CE and AD? I’ll give brainly

Mathematics

1 answer:

Step2247 [10]3 years ago

4 0

It’s the second one because it can’t be half but it can be 1/4 and if you do the measurements you can tell it is the second one

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Need help please help

lilavasa [31]

Answer:

27 of the pets are dogs.

Step-by-step explanation:

Real easy way to do this, treat 45 (the total) as 100. So, do 45/100 then multiply the result of that by 60, and you end up with 27.

45/100=0.45

0.45*60=27

7 0

3 years ago

3. What are the factors of 9x2 - 30x + 25?

vaieri [72.5K]

{(3x)}^{2}-2(3x)(5)+{5}^{2}

−2(3x)(5)+5

{(3x-5)}^{2}

4 0

3 years ago

Find the distance between the given point (2,-2) and (-4,-7)

tatiyna

Answer:

-11

Step-by-step explanation:

that's my answer please follow me

7 0

3 years ago

The braking distance, in feet of a car a Travling at v miles per hour is given.

irakobra [83]

The braking distance is the distance the car travels before coming to a stop after the brakes are applied

a. The braking distances are as follows;

The braking distance at 25 mph, is approximately 63.7 ft.

The braking distance at 55 mph, is approximately 298.35 ft.

The braking distance at 85 mph, is approximately 708.92 ft.

b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s

Reason:

The given function for the braking distance is D = 2.6 + v²/22

a. The braking distance if the car is going 25 mph is therefore;

25 mph = 36.66339 ft./s

$D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.$

At 25 mph, the braking distance is approximately 63.7 ft.

At 55 mph, the braking distance is given as follows;

55 mph = 80.65945 ft.s

$D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.$

At 55 mph, the braking distance is approximately 298.35 ft.

At 85 mph, the braking distance is given as follows;

85 mph = 124.6555 ft.s

$D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.$

At 85 mph, the braking distance is approximately 708.92 ft.

b. The speed of the car when the braking distance is 450 feet is given as follows;

$450 = 2.6 + \dfrac{v^2}{22}$

v² = (450 - 2.6) × 22 = 9842.8

v = √(9842.2) ≈ 98.211 ft./s

The car was moving at v ≈ 98.211 ft./s

Learn more here:

brainly.com/question/18591940

8 0

2 years ago

Pls help Given: △ABC, CM⊥ AB, BC = 5, AB = 7 CA = 4 sqrt(2) Find: CM

babymother [125]

Answer:

The general plan is to find BM and from that CM. You need 2 equations to do that.

Step One

Set up the two equations.

(7 - BM)^2 + CM^2 = (4*sqrt(2) ) ^ 2 = 32

BM^2 + CM^2 = 5^2 = 25

Step Two

Subtract the two equations.

(7 - BM)^2 + CM^2 = 32

BM^2 + CM^2 = 25

(7 - BM)^2 - BM^2 = 7 (3)

Step three

Expand the left side of the new equation labeled (3)

49 - 14BM + BM^2 - BM^2 = 7

Step 4

Simplify And Solve

49 - 14BM = 7 Subtract 49 from both sides.

-49 - 14BM = 7 - 49

- 14BM = - 42 Divide by - 14

BM = -42 / - 14

BM = 3

Step Five

Find CM

CM^2 + BM^2 = 5^2

CM^2 + 3^2 = 5^2 Subtract 3^2 from both sides.

CM^2 = 25 - 9

CM^2 = 16 Take the square root of both sides.

sqrt(CM^2) = sqrt(16)

CM = 4 < Answer

Step-by-step explanation:

6 0

3 years ago