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2 years ago

What is the LCM of 8 and 10

Mathematics

1 answer:

iragen [17]2 years ago

3 0

40

8x5=40
10x4=40

Hope this helps

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How do i solve this.

nignag [31]

Let us set up some variables:

L: Length
W: Width

Now let us set up some equations based on the facts:

length of rectangle is 4 times its width --> L = 4W
perimeter of rectangle is 80 --> 2L + 2W = 80

Equations:

L = 4W -- equation 1

2L + 2W = 80 -- equation 2

Plug the value of L from equation 1 into equation 2

2(4W) + 2W = 80

8W + 2W = 80

10W = 80

W = 8 -- equation 3

Plug the value of W from equation 3 into equation 1:

L = 4W = 4 * 8 = 32

Therefore the length is 32 and the width is 8.

Hope that helps!

4 0

3 years ago

Pls help is easyyyyy geometry

Makovka662 [10]

Answer:

A = 30 unit^2

Step-by-step explanation:

To find the area of the triangle

A = 1/2 b*h where b is the length of the base and h is the height

b = 12 and h =5

A = 1/2 (12*5)

A = 1/2 *60

A = 30 unit^2

8 0

3 years ago

What is 1 3/4 as a percent

KiRa [710]

Answer:

267% and 400%

6 0

3 years ago

Read 2 more answers

In 1859, 24 rabbits were released into the wild in Australia, where they had no natural predators. Their population grew exponen

Mashcka [7]

Answer:

a) P' = P

$P(t) = 24e^{0.693t}$ where t is step of 6 months

b) 7.7 years

c)1064.67 rabbits/year

Step-by-step explanation:

The differential equation describing the population growth is

$\frac{dP}{dt} = P$

Where t is the range of 6 months, or half of a year.

P(t) would have the form of

$P(t) = P_0e^{kt}$

where $P_0 = 24$ is the initial population

After 6 month (t = 1), the population is doubled to 48

$P(1) = 24e^k = 48$

$e^k = 2$

$k = ln(2) = 0.693$

Therefore $P(t) = 24e^{0.693t}$

where t is step of 6 months

b. We can solve for t to get how long it takes to get to a population of 1,000,000:

$24e^{0.693t} = 1000000$

$e^{0.693t} = 1000000 / 24 = 41667$

$0.693t = ln(41667) = 10.64$

$t = 10.64 / 0.693 = 15.35$

So it would take 15.35 * 0.5 = 7.7 years to reach 1000000

c. $P' = P_0ke^{kt}$

We need to resolve for k if t is in the range of 1 year. In half of a year (t = 0.5), the population is 48

$24e^{0.5k) = 48$

$0.5k = ln2 = 0.693$

$k = 1.386$

Therefore, $P' = 1.386*24e^{1.386t}$

At the mid of the 3rd year, where t = 2.5, we can calculate P'

$P' = 1.386*24e^{1.386*2.5} = 1064.67$ rabbits/year

4 0

3 years ago

Someone help :) please

maxonik [38]

Answer:

The line between the points C and E is parallel to the line between the points F and B

The line between the points A and E is parallel to the line between the points B and D

The line between the points A and C is parallel to the line between the points F and D

3 0

3 years ago