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BlackZzzverrR [31]
3 years ago
10

Complete the table. Then

Mathematics
1 answer:
elena-14-01-66 [18.8K]3 years ago
3 0

Answer:

y               x

-2           1/4

-1           1/2

0             1

1               2

2               4

Step-by-step explanation:

its easy algbra

doing it rn to

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Which inequality matches the given situation if w represents weight in pounds:Baby tigers can be no larger than 4 pounds.
ozzi

Given:

w be the weight in pounds of a baby tigers.

To find:

The inequality if baby tigers can be no larger than 4 pounds.

Solution:

Baby tigers can be no larger than 4 pounds. It mean weight of baby tigers cannot be greater than 4. In other words, the weight of the baby tigers must be less than or equal to 4.

Let w represents weight in pounds and the weight of the baby tigers must be less than or equal to 4. So,

w\leq 4

Therefore, the required inequality is w\leq 4.

4 0
3 years ago
Please help me. Por favorrrrrr
irakobra [83]
The statements that are correct are A and E. A because the y-intercept for y=5x+12 is 12 and the table of values it is 5. E is correct because the rate of change for the table of values is 24 and y=5x+12, it is 5.
4 0
3 years ago
Which number is an integer?<br> -34 -0.4 0.4 3.4 PLS HELP I'M TIMED EDGE 2020
Stella [2.4K]

Answer:

Step-by-step explanation:         The four numbers:  - 34, -0.4, 0.4, 3.4

so if you look at it it  The answer is -34

7 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
Is it - or + 60 plz help me
Artemon [7]
+60 two negatives equal a positive
6 0
3 years ago
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