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2 years ago

what is the perimeter measured in centimeters of rectangle picture below do not include units in your answer

Mathematics

1 answer:

scZoUnD [109]2 years ago

8 0

Answer:

Step-by-step explanation:

Formula for Perimeter of Rectangle

P = 2(l+w)

P = 2(30*18)

P = 96

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What is the answer to 7(1 + 4k)

Sophie [7]

How to solve it:
•1. multiply 7 x 1= 7
•2. multiply 7 x 4k= 28k

so the answer is: 7+28k

i hope i have been helpful :)

8 0

2 years ago

Read 2 more answers

7. The largest of the Great Pyramids in Egypt was built with a square base,

strojnjashka [21]

Answer:

Difference between the original and current volumes = 4013140.30 cubic feet

Step-by-step explanation:

Largest pyramid has the square base with one edge = 754 feet

Area of the square base = (Side)²

= (754)²

= 568516 square feet

Volume of the pyramid = $\frac{1}{3}(\text{Area of the base})(\text{Height})$

= $\frac{1}{3}\times 568516\times 481$

= 91152065.3 cubic feet

After erosion,

Area of the base = (745)² = 555025 square feet

Volume of the pyramid = $\frac{1}{3}(555025)(471)$

= 87138925 cubic feet

Difference between the original and current volumes of the pyramid

= 91152065.30 - 87138925

= 4013140.30 cubic feet

5 0

2 years ago

Ill mark brainliest I just don't understand

Misha Larkins [42]

Answer:

a triangle always equals 180 degrees so

63+ 2x+8 + x+7 =180

solve for x

and then plug it back into original equation

let me know if this makes sense

glad to help you more if needed

Step-by-step explanation:

3 0

2 years ago

suppose a population of 200 crickets double in size every 4 months. How many crickets will there be after 2 years?

tangare [24]

Well 2 years is 24 months. 24 divided by 4 is 6. 200 multiplied by 6 would equal 1200.

The total population of crickets after two years would be 1200.

4 0

2 years ago

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$