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3 years ago

6

Greatest common factor of 8 and 14?

2 answers:

Leto [7]3 years ago

8 0

The Greatest common factors is 2 because it is The common factors of 8 and 14.☺

maxonik [38]3 years ago

3 0

The greatest common factor (GCF) of 8 and 14 is 2

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Given a polynomial f(x), if (x + 7) is a factor, what else must be true? Help plzzz im running out of time

solong [7]

Answer: f(-7) = 0
For the factored form of f(x), if x = -7, then one of the factors will be equal to 0, since (-7 + 7) = 0. This will cause the entire y-value to be 0, because anything multiplied by 0 = 0.

4 0

3 years ago

A rectangular box with a square base is open at the top and has a volume of 12 cubic feet. Each side of the base has a length of

Stolb23 [73]

We know that the side lengths of the square base are: x * x. The volume is 12, so for now, let's say that y is the other side length. Then, x * x * y = 12. We can solve for y: y = 12/x^2. Now, we find the surface area of the 5 sides.

Four of the sides have the same area: x * (12/x^2) = 12/x, so we multiply this by 4: 48/x.

The last side is the base: x * x = x^2.

We add 48/x to x^2:

x^2 + 48/x

So, the answer is the fourth choice, (d).

4 0

2 years ago

WILL GIVE BRAINLEST plz help meeeeeeeee

mixer [17]

Answer:

4

Step-by-step explanation:

3 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 <u>63.7 ft.</u>

The braking distance at 55 mph, is approximately <u>298.35 ft.</u>

The braking distance at 85 mph, is approximately <u>708.92 ft.</u>

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 <u>63.7 ft.</u>

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 <u>298.35 ft.</u>

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 <u>708.92 ft.</u>

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 ≈ <u>98.211 ft./s</u>

Learn more here:

brainly.com/question/18591940

8 0

1 year ago

What are the domain and range of the function f(x)=(√x-7)+9?

Yanka [14]

Answer:

the domain of the function f(x) is $x\ge 7$

the range of the function f(x) is $f(x)\ge 9$

Step-by-step explanation:

Consider the parent function $y=\sqrt{x}.$

The domain og this function is $x\ge 0,$ the range of this function is $y\ge 0.$

The function $f(x)=\sqrt{x-7}+9$ is translated function $y=\sqrt{x}$ 7 units to the right and 9 units up, so

the domain of the function f(x) is $x\ge 7$

the range of the function f(x) is $f(x)\ge 9$

3 0

3 years ago