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3 years ago

Draw a shape with 4 congruent angles, but only opposite sides are congruent

Mathematics

1 answer:

ANEK [815]3 years ago

4 0

A trapezoid falls under the definition.
Hope it helped!

~JustastudentXD

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Sadie computes the perimeter of a rectangle by adding the length, l , and width, w , and doubling this sum. Eric computes the pe

larisa [96]

a) P = 2(l+w) Sadie

P =2l+2w Eric

b) Sadie : P = 2(30+75) = 2(105) = 210

Eric: P = 2(30) +2(75) = 60+150 =210

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2 years ago

What part of 35 is 56? *not a percent* help pls

lesantik [10]

Expressed as a fraction 56 is 56/35 of 35. That fraction can be reduced, and expressed several ways.

56/35 = 8/5 = 1 3/5 = 1.6

56 is 1 3/5 of 35

56 is 1.6 times 35

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3 years ago

Find each sum (-4y+3) + (11y-5)

Airida [17]

Answer:

7y−2

Step-by-step explanation:

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3 years ago

. The time of a clock as seen in a mirror is 3:15. The correct time is

Oksana_A [137]

Answer:

3:15 cause a mirror cant change time

Step-by-step explanation:

go in ur mirror and look at the time did it somehow get darker or lighter???

8 0

3 years ago

Read 2 more answers

Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$

Therefore the half-life is between 600 and 700 years

5 0

3 years ago