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beks73 [17]
3 years ago
10

2r-7=-1. Help please thanks

Mathematics
1 answer:
UkoKoshka [18]3 years ago
6 0

Answer:

-3

Step-by-step explanation:

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A 1200 gram sample of radioactive matter will completely decay (be undetectable) in 10 hours. There exists a set of ordered pair
julia-pushkina [17]

Using it's concept, the range of the function is given as follows:

0 ≤ m ≤ 1200.

<h3>What is the range of a function?</h3>

The range of a function is the set that contains all possible output values for the function.

For this problem, we have to consider these two bullet points next, considering the mass is the output value of the function.

  • The smallest possible mass for the substance is of 0 grams, as after the substance decays to 0 grams, it will not assume a negative value, it will just disappear.
  • The greatest possible mass for the substance is the initial mass of 1200 grams, as the substance does not adquire mass with time, it just loses it.

Considering these masses, the range of the function is given as follows:

0 ≤ m ≤ 1200.

More can be learned about the range of a function at brainly.com/question/10197594

#SPJ1

3 0
1 year ago
if the domain of a coordinate transformation that is reflected over the y-axis is (-1,3),(2,1),(5,-1)and(-1,-2) what is the rang
AURORKA [14]

Answer:

Range remains the same.

Step-by-step explanation:

(1,3),(-2,1),(-5,-1) and (1,-2)

7 0
4 years ago
Read 2 more answers
A donkey weighs 570 pounds and an elephant weighs 5 tons how much more does the elephant weigh in pounds and ounces?
goldfiish [28.3K]
The elephant weighs 9,430 pounds more than the donkey.
6 0
3 years ago
Does anyone understand this
melomori [17]
I hope this helps you

7 0
3 years ago
HELP PRECALC DO NOT UNDERSTAND WILL GIVE BRAINLIEST
Olegator [25]

Answer:

  the lower right matrix is the third correct choice

Step-by-step explanation:

Your problem statement shows that you have correctly selected the matrices representing the initial problem setup (middle left) and the problem solution (middle right).

Of the remaining matrices, the upper left is an incorrect setup, and the lower left is an incorrect solution matrix.

__

We notice that in the remaining matrices on the right that the (2,3) term is 0, and the (3,2) and (3,3) terms are both 1.

The easiest way to get a 0 in the 3rd column of row 2 is to add the first row to the second. When you do that, you get ...

  \left[\begin{array}{ccc|c}1&1&1&29000\\1+2&1-3&1-1&1000(29+1)\\0&0.15&0.15&2100\end{array}\right] =\left[\begin{array}{ccc|c}1&1&1&29000\\3&-2&0&30000\\0&0.15&0.15&2100\end{array}\right]

Already, we see that the second row matches that in the lower right matrix.

The easiest way to get 1's in the last row is to divide that row by 0.15. When we do that, the (3,4) entry becomes 2100/0.15 = 14000, matching exactly the lower right matrix.

The correct choices here are the two you have selected, and <em>the lower right matrix</em>.

3 0
4 years ago
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