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g100num [7]
3 years ago
8

If there is 15 liters of milk in the refrigerator, after how many days will more milk need to be purchased?

Mathematics
1 answer:
adoni [48]3 years ago
7 0

Answer:

Depends on Howe much you can drink. It varies.

Step-by-step explanation:

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Find the value of in the triangle shown below<br><br>9, 7, and x around the triangle <br>​
borishaifa [10]

Answer:

Step-by-step explanation: 56

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2 years ago
What are the first 3 common multiples of 3 and 5
marshall27 [118]
<span>The least common multiple is the smallest number that is divisible by each of the numbers being considered. Example: While 15, 30, 45, 60, 75, ... are all commonmultiplies of 3 and 5, 15 is the smallest number divisible by both 3 and 5. It is the leastcommon multiple.</span>
5 0
3 years ago
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Which graph represents the solution set of the system of inequalities? {x+y&lt;12 y≥x−4
RSB [31]

Answer: The graph in the bottom right-hand corner

(see figure 4 in the attached images below)

===========================================

Explanation:

Let's start off by graphing x+y < 1. The boundary equation is x+y = 1 since we simply change the inequality sign to an equal sign. Solve for y to get x+y = 1 turning into y = -x+1. This line goes through (0,1) and (1,0). The boundary line is a dashed line due to the fact that there is no "or equal to" in the original inequality sign. So x+y < 1 turns into y < -x+1 and we shade below the dashed line. The "less than" means "shade below" when y is fully isolated like this. See figure 1 in the attached images below.

Let's graph 2y >= x-4. Start off by dividing everything by 2 to get y >= (1/2)x-2. The boundary line is y = (1/2)x-2 which goes through the two points (0,-2) and (4,0). The boundary line is solid. We shade above the boundary line. Check out figure 2 in the attached images below.

After we graph each individual inequality, we then combine the two regions on one graph. See figure 3 below. The red and blue shaded areas in figure 3 overlap to get the purple shaded area you see in figure 4, which is the final answer. Any point in this purple region will satisfy both inequalities at the same time. The solution point cannot be on the dashed line but it can be on the solid line as long as the solid line is bordering the shaded purple region. Figure 4 matches up perfectly with the bottom right corner in your answer choices.

5 0
3 years ago
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Lis 10 múltiples of 8
andrey2020 [161]
10 times 8 would = 80
6 0
3 years ago
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Please explain how they plugged the removable discontinuity in this question.
kondor19780726 [428]

In the limit

\displaystyle \lim_{x\to c} f(x)

we're interested in the value that f(x) converges to as x gets closer to c. So in fact x\neq c.

In the given example, f(x) is factorized to reveal a common factor of x-1 in the numerator and denominator. We have x\neq1 if x\to1, so x-1\neq0 so we can simplify

\dfrac{x-1}{x-1} = 1

and *remove* the discontinuity.

Then

\displaystyle \lim_{x\to1} \frac{2(x-1)}{(x+1)(x-1)} = \lim_{x\to1} \frac2{x+1} = \frac2{1+1} = 1

8 0
1 year ago
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