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densk [106]
2 years ago
15

Question 4

Mathematics
1 answer:
Anna [14]2 years ago
5 0

Answer:

f(1)= 12-31+7

Step-by-step explanation:

f(1) = -12

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Ther are 9 thirds in 3 ones a multiply and divide equashon
Ivanshal [37]

Answer:

That is a division equation.

Step-by-step explanation:

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3 years ago
The bill in a restaurant is $102.25. A tip of 20% is added to the bill. What id tyhe final amount paid?
motikmotik

Answer: $117.59

Step-by-step explanation:

8 0
3 years ago
Please help me with this question. Algebra 2 is hard!!
trapecia [35]

Answer:

  • domain: {x ∈ ℝ : x ≤ 5}
  • range: {y ∈ ℝ : y ≤ -1}

Step-by-step explanation:

<u>Domain</u>

The domain of a function is the set of x values for which the function is defined. Here, the domain is limited by the values of x that make the square root defined. That is, the expression under the radical cannot be negative:

  -3x +15 ≥ 0

  15 ≥ 3x . . . . . . add 3x

  5 ≥ x . . . . . . . . divide by 3

  x ≤ 5 . . . . . . . . put x on the left (swap sides)

The rest of the notation in the domain expression simply says x is a real number.

  domain: {x ∈ ℝ : x ≤ 5} . . . . . . matches the first choice

__

<u>Range</u>

The range of a function is the set of values that f(x) can have. We know the square root can be zero or any positive number. When it is zero, f(x) = -1.

When it is a positive number, that value is multiplied by -4 and added to -1, so f(x) is a number more negative than -1. Then the range of the function is all numbers -1 and below:

  range: {y ∈ ℝ : y ≤ -1} . . . . . . matches the last choice

_____

<em>Comment on domain/range problems</em>

When working domain and range problems, it works well to have a good understanding of the domain and range limitations of the functions we usually work with: polynomials, square root, logarithm, trig functions, exponential functions. Domain and range problems generally involve combinations of these or ratios of combinations of these.

3 0
3 years ago
.08 is 10 times as much as
Mazyrski [523]
It is 10 times as much as .008

Do you see the pattern?
5 0
3 years ago
Dubnium-262 has a half-life of 34 s. How long will it take for 500.0 grams to
dmitriy555 [2]

Answer:

the time taken for the radioactive element to decay to 1 g is 304.8 s.

Step-by-step explanation:

Given;

half-life of the given Dubnium = 34 s

initial mass of the given Dubnium, m₀ = 500 grams

final mass of the element, mf = 1 g

The time taken for the radioactive element to decay to its final mass is calculated as follows;

1 = 500 (0.5)^{\frac{t}{34}} \\\\\frac{1}{500} =  (0.5)^{\frac{t}{34}}\\\\log(\frac{1}{500}) = log [(0.5)^{\frac{t}{34}}]\\\\log(\frac{1}{500})  = \frac{t}{34} log(0.5)\\\\-2.699 = \frac{t}{34} (-0.301)\\\\t = \frac{2.699 \times 34}{0.301} \\\\t = 304.8 \ s

Therefore, the time taken for the radioactive element to decay to 1 g is 304.8 s.

4 0
2 years ago
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