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zvonat [6]
2 years ago
12

3. What is the GCF of 12, 24, and 36? A. 12 C. 24 B. 18 D. 48​

Mathematics
2 answers:
agasfer [191]2 years ago
7 0

Answer:

A

Step-by-step explanation:

GCF is greatest common factor, which means that it is the greatest number all the numbers can be divisible by

for these numbers it is 12

Ludmilka [50]2 years ago
6 0

Hey there!

Finding the GCF (Greatest Common Factor) of 12, 24, & 48

List of the factors

• 12: 1, 2, 3, 4, 6, & 12

• 24: 1, 2, 3, 4, 6, 8, 12, & 24

• 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, & 48

Common factors

• 1, 2, 3, 4, 6, & 12

Biggest numbers out of the factors

• 6 & 12

GREATEST common factor

• 12

Therefore, your answer is: GCF (Greatest Common Factor) → 12

• Option A.

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

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I NEED HELP NOW <br> ur really gonna need to solve this <br> 10t-5=60
faltersainse [42]

Answer:

\huge \: t =  \frac{13}{2}

Step-by-step explanation:

10t - 5 = 60

Using the addition property add 5 to both sides of the equation

That's

10t - 5 + 5 = 60 + 5 \\ 10t = 65

<u>Divide both sides by 10</u>

\frac{10t}{10}  =  \frac{65}{10}  \\  t =  \frac{65}{10}

<u>Reduce the fraction with 5</u>

We have the final answer as

t =  \frac{13}{2}  \\

Hope this helps you

3 0
2 years ago
Read 2 more answers
7 yards and 2feet equals how many feet?
Vladimir [108]

Answer:

23 ft

Step-by-step explanation:

a yard is 3 feet multiply it by 7 and you get 21 add 2 and you get 23. 23ft is your answer

4 0
2 years ago
If the circle below is cut from the square of plywood below, how many square inches of plywood would be left over? Use π = 3.14,
Elis [28]

Answer:

86in²

Step-by-step explanation:

Find the complete question attached

Area of the circle = πr²

Area of the square = L²

Length of the side of the square

r is the radius of the circle

r = L/2 = 20/2 =10in

Given

Length of the side of the circle = 20in

Area of the square = 20²

Area of the square = 400in²

Area of the circle =π(10)²

Area of the circle = 3.14(100)

Area of the circle= 314in²

Amount of plywood left over = 400in²-314in² = 86in²

8 0
3 years ago
1. )The function f(x)=12,500(0.87)^x models the value of a car x years after it is purchased.
sweet-ann [11.9K]
1] Given that the value of x has been modeled by f(x)=12500(0.87)^x, then:
the rate of change between years 1 and 5 will be:
rate of change is given by:
[f(b)-f(a)]/(b-a)
thus:
f(1)=12500(0.87)^1=10875
f(5)=12500(0.87)^5=6230.3
rate of change will be:
(6230.3-10875)/(5-1)
=-1161.2
 rate of change in years 11 to 15 will be:
f(11)=12500(0.87)^11=2701.6
f(15)=12500(0.87)^15=1,547.74
thus the rate of change will be:
(1547.74-2701.6)/(15-11)
=-288
dividing the two rates of change we get:
-288/-1161.2
-=1/4
comparing the two rate of change we conclude that:
The average rate of change between years 11 and 15  is about 1/4  the rate between years 1 and 5.
The answer is D]
2] Given that the population of beavers decreases exponentially at the rate of 7.5% per year, the monthly rate will be:
monthly rate=(n/12)
where n is the number of months
=7.5/12
=0.625
This is approximately equal to 0.65%. The correct answer is A. 0.65%

7 0
3 years ago
Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th
trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

    t = 20: 999 fishes

    t = 30: 1168 fishes

(c)

P(\infty) = 1200

Step-by-step explanation:

Given

P(t) =\frac{d}{1+ke^-{ct}}

d = 1200\\k = 11\\c=0.2

Solving (a): Fishes at t = 0

This gives:

P(0) =\frac{1200}{1+11*e^-{0.2*0}}

P(0) =\frac{1200}{1+11*e^-{0}}

P(0) =\frac{1200}{1+11*1}

P(0) =\frac{1200}{1+11}

P(0) =\frac{1200}{12}

P(0) = 100

Solving (a): Fishes at t = 10, 20, 30

t = 10

P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483

t = 20

P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999

t = 30

P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168

Solving (c): \lim_{t \to \infty} P(t)

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of e^{-ct} decreases

This implies that:

{t \to \infty} = {e^{-ct} \to 0}

So:

The value of P(t) for large values is:

P(\infty) = \frac{1200}{1 + 11 * 0}

P(\infty) = \frac{1200}{1 + 0}

P(\infty) = \frac{1200}{1}

P(\infty) = 1200

5 0
2 years ago
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