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2 years ago

For how many values of x: (4.9 cm < x < infinity) is it true that ex = 0?

Mathematics

1 answer:

RUDIKE [14]2 years ago

5 0

E^x is not zero for any finite value of x.

The number of values for which e^x is zero is zero.

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If we measured the heights of different plants and got 50”, 55”, 48”, 50” and 52”. What

mr Goodwill [35]

Answer:

Step-by-step explanation:

50+55+48+50+52=255

255/5=51

Anwser 51

5 0

2 years ago

What is the perimeter of the garage if d=3<br><br> I’m marking branliest

mart [117]

Answer:

perimeter= 46

Step-by-step explanation:

(3(3)-2) + (3(3)-2) + (5(3)+1) + (5(3)+1)

7 + 7 + 16 + 16

= 46

5 0

3 years ago

1 substitute p=5, q=7, r = -2

lora16 [44]

a.) 2q + 5r

2(7) + 5(-2)

14 - 10 = 4

b.) 3(p + 6) + q + r Plug in the numbers

3(5 + 6) + 7 - 2 Solve inside the parentheses first

3(11) + 7 - 2

33 + 5 = 38

a.) m(3m + 4n)

2(3(2) + 4(3))

2(6 + 12)

2(18) = 36

b.) n²(m + p²)

(3)²(2 + (-5)²)

9(2 + 25)

9(27) = 243

c.) 3m(8 + n) + n²

3(2) (8 + 3) + 3²

6(11) + 9

66 + 9 = 75

3 0

2 years ago

Point A represents which fraction? A)1/3 B)4/8 C)5/6 D)5/8

ELEN [110]

I believe the answer would b ( B) 1/3

Hope this helps!

4 0

2 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