All categories

1 year ago

−11 ≤ −5; Add 16 to both sides The resulting inequality is:

Mathematics

1 answer:

Lemur [1.5K]1 year ago

3 0

Answer:

True, (-∞, ∞) 5 ≤ 11

Step-by-step explanation:

−11 ≤ −5 Add 16 to both sides

−11 + 16 ≤ −5 + 16

5 ≤ 11

You might be interested in

Please help me solve this problem with step by step explanation.

Kisachek [45]

Answer:

Im not rally sure but maybe 5.71 , here is my explanation :

Step-by-step explanation:

Tan 55 =x/4

tan 55 = 1/42814

1/42814 = x /4

×=5.712592027 .

3 0

3 years ago

I need help on this and the first person that answer this correctly gets a BRANLIST(answer if you know)

irakobra [83]

Answer:

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Type 11/5 in the simplest form

aleksklad [387]

Answer:

$2\frac{1}{5}$

Step-by-step explanation:

11 ÷ 5 = 2 R 1 → $2\frac{1}{5}$

Hope this helps! :)

7 0

2 years ago

I don't know how to simplify

In-s [12.5K]

(5-2/x)/4-3/x^2

after simplifying these
(5x-2)/x(4x^2-3)/x^2
x(5x-2)/(4x^2-3)
5x^2-2x/4x^2-3

now u can solve it

6 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