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3 years ago

Please help I will brainliest!! Really urgent!!

Mathematics

1 answer:

madreJ [45]3 years ago

8 0

Answer:

$\frac{1}{6}$ x-19

Step-by-step explanation:

$\frac{2}{3}$ x-9- $\frac{1}{2}$ x-10 = $\frac{2}{3}$ x- $\frac{1}{2}$ x-19

If you minus 9 and then minus 10, you are deducting 19 from the equation in total

$\frac{2}{3}$ x- $\frac{1}{2}$ x-19 = $\frac{4}{6}$ x- $\frac{3}{6}$ x-19

You would want to make the denominator of the fractions the same by multiplying the first fraction by 2, and the second fraction by 3 so that both denominators are 6.

$\frac{4}{6}$ x- $\frac{3}{6}$ x-19 = $\frac{1}{6}$ x-19

Now that their denominators are the same, you can directly do the working for the numerators (4-3=1) and keep the denominator the same. Tada!

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How many numbers bewteen 50 and 150 have at least one digit that is 2

mr Goodwill [35]

Answer:

52 62 72 82 92 102 112 120 121 122 123 124 125 126 127 128 129 132 142

there are 19.

Step-by-step explanation:

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Siddique is making a wedding cake. He needs 2.5 kg of flour. Each bag of flour weighs 500g. How many bags will he need to buy?

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3 years ago

Discuss the continuity of the function on the closed interval.Function Intervalf(x) = 9 − x, x ≤ 09 + 12x, x > 0 [−4, 5]The f

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Answer:

It is continuous since $\lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)$

Step-by-step explanation:

We are given that the function is defined as follows $f(x) = 9-x, x\leq 0$ and $f(x) = 9+12x, x>0$ and we want to check the continuity in the interval [-4,5]. Note that this a piecewise function whose only critical point (that might be a candidate of a discontinuity) x=0 since at this point is where the function "changes" of definition. Note that 9-x and 9+12x are polynomials that are continous over all $\mathbb{R}$ . So F is continous in the intervals [-4,0) and (0,5]. To check if f(x) is continuous at 0, we must check that

$\lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)$ (this is the definition of continuity at x=0)

Note that if x=0, then f(x) = 9-x. So, f(0)=9. On the same time, note that

$\lim_{x\to 0^{-}} f(x) = \lim_{x\to 0^{-}} 9-x = 9$ . This result is because the function 9-x is continous at x=0, so the left-hand limit is equal to the value of the function at 0.

Note that when x>0, we have that f(x) = 9+12x. In this case, we have that

$\lim_{x\to 0^{+}} f(x) = \lim_{x\to 0^{+}} 9+12x = 9$ . As before, this result is because the function 9+12x is continous at x=0, so the right-hand limit is equal to the value of the function at 0.

Thus, $\lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)=9$ , so by definition, f is continuous at x=0, hence continuous over the interval [-4,5].

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Answer:

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