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3 years ago

What is the soultion of 25>-3(4n-3)

Mathematics

1 answer:

fredd [130]3 years ago

8 0

Answer:

n > -4/3

Step by step

Step by step solution :

Step 1 :

Equation at the end of step 1 :

25 - (0 - 3 • (4n - 3)) > 0

Step 2 :

Step 3 :

Pulling out like terms :

3.1 Pull out like factors :

12n + 16 = 4 • (3n + 4)

Equation at the end of step 3 :

4 • (3n + 4) > 0

Step 4 :

4.1 Divide both sides by 4

4.2 Divide both sides by 3

n+(4/3) > 0

Solve Basic Inequality :

4.3 Subtract 4/3 from both sides

n > -4/3

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An article is purchased for Rs.5000 andsold for Rs.4500.Find loss percent

bixtya [17]

Answer:

10% loss

Step-by-step explanation:

the difference = Rs. 5000 -Rs. 4500 = Rs. 500

Rs. 500/Rs. 5000*100%= 10%

Since it was reduction in price, it is 10% loss

3 0

3 years ago

I need help with this my guys !

luda_lava [24]

Answer:

1/2

Step-by-step explanation:

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Rise=1

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Rise/run=1/2

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3 years ago

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What is 0.00277 scientific notation?

Harlamova29_29 [7]

2.77 × 10 ^ ( -3 ) is the answer

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3 years ago

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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

1 year ago

$2.25

DiKsa [7]

2.50 that is the answer

8 0

2 years ago