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3 years ago

Simplify the following expression(-5)(-3)(4)

Mathematics

2 answers:

rusak2 [61]3 years ago

8 0

(-5)(-3)(4)
(15)(4)
(60)

60 is the answer

marissa [1.9K]3 years ago

4 0

Answer:

$\displaystyle 60$

Step-by-step explanation:

All you do is multiply the terms straight across to get the above answer.

I am joyous to assist you anytime.

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Please help I will give brainliest please show work and find the volume of this odd looking pyramid

JulsSmile [24]

Answer:

206 2/3 ft³

Step-by-step explanation:

The volume of this solid is given by V = (1/3)(base area)(height).

Here, V = (1/3) (62 ft²) (10 ft ) = 206 2/3 ft³

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3 years ago

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lakkis [162]

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Step-by-step explanation:

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3 years ago

What's -9 x 7 plssssssssss answerrrrr quicklyyyy

NNADVOKAT [17]

It’s -63 all you would be multiple like normal 63 which then you put a negative symbol on front

6 0

3 years ago

Read 2 more answers

The science fair judges will be science and volunteers each judge will only have time to view 5 projects there are 133 science t

antiseptic1488 [7]

5 times amount of people has to be bigger than 2112

5(133+v)≥2112
distribute
665+5v≥2112
minus 665 from both sides
5v≥1447
divide both sides by 5
v≥289.4
yo can't have 0.4 volunterer
v≥290

has to have at least 290 volunteers

8 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

3 years ago