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3 years ago

Write an expression for calculation double the product of 4 and 7

Mathematics

1 answer:

Over [174]3 years ago

3 0

Answer:2(4*7)

this is an expression

Step-by-step explanation:

if you want to solve it

you would multiply 4 by 7 which gives you 28

then double 28

you would multiply 28 by 2

which then gives you an answer of 56

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34.15 round to the nearest hundred

I am Lyosha [343]

It’s zero, I think. Because
3 doesn’t round up. It’s stays the same. But since there is nothing in the hundred place. It is 0.

4 0

3 years ago

Please help with these

kherson [118]

The answers are as follows.

a. 9x

In order to get this, you need to reevaluate 64 as a base of 4. Since 64 is equal to 4^3, we can rewrite the right side as

64^3x = (4^3)^3x = 4^9x

Which gives you the first answer.

b. 10

Similar to the first problem, we need to express 16 as a base of 2. 2^4 is equal to 16, so we use that in it's place and simplify.

16^5/2 = (2^4)^5/2 = 2^20/2 = 2^10

c. Sqrt(7.31)

This one is more simple. Raising something to a 1/2 power (.5) is the same as taking the square root.

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4 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

Austin's truck has a mass of 2000 kg when traveling at 22.0 m/s, it brakes to a stop in 4.0 s. show that the magnitude of the br

olga2289 [7]

Since F=m•a, you want to show that a = -5.5

5 0

3 years ago

Read 2 more answers

If the weight of a class 4 truck is increased by two tons will it still be classified as a class 4 truck

Assoli18 [71]

If the truck is 14,000 pounds and increases by 2 tons yes. However if it is above 14,000 then no it will not be a class 4 truck.

8 0

4 years ago