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3 years ago

Shari sent a total of 64 text messages to 4 friends. each friend received the same number of text messages. which number

Mathematics

1 answer:

Inga [223]3 years ago

4 0

Each freinds received 16 texts in total

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A rectangle has a length of x+7 and a width of x-3. write and expression that represents the AREA of the rectangle.

Lyrx [107]

Answer:

A = x^2 + 4x -21

Step-by-step explanation:

L = x+7

W = x-3

A = L*W

A = (x+7)(x-3)

A = x^2 + 4x - 21

6 0

3 years ago

Mr. Clifton was preparing for his upcoming birthday party. He ordered 51 white balloons and 47 red balloons. As he walked to his

Andrej [43]

Answer:

28 i think

Step-by-step explanation:

51+47=98

98-12 the ones that flew away that takes it to 86

86/3 the 3 balloons buntch than is 28

p.s. i mght of got some equations wrong but this is what i got

4 0

3 years ago

There are 37 cans of pineapple on each shelf at food mart if there are 7 shelves of pineapple estimate the number of cans pineap

Luba_88 [7]

Answer:

260

Step-by-step explanation:

7 shelves containing 37 cans of pineapples each would give you 259 cans of pineapple. Then, round to the nearest ten would give you 260.

7 0

3 years ago

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

2 years ago

Quadrilateral ABCD is inscribed in a circle. Find the measure of each of the angles of the quadrilateral. Show your work.

Xelga [282]

The opposite angles of a quadrilateral are supplementary so you can set them equal to 180.

If there was more information given then i could find the exact values.

Hope this helps :)

8 0

3 years ago