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seraphim [82]
3 years ago
12

Find the total number of jobs lost due to the pandemic in 2020. What proportion of

Mathematics
1 answer:
S_A_V [24]3 years ago
5 0

Answer:

.jsocnwnsjfownn1oqoajajajaosndbd xjxob2bed

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PLEASE HELP ME!!!
padilas [110]

Answer:

p(-5)=-11,987

p(3)=-227

Step-by-step explanation:

To find the value of the function p(-5) just substitute x=-5 into the function expression instead of x:

p(-5)=2\cdot (-5)^5-9\cdot (-5)^4-2\cdot (-5)^2+12\cdot (-5)-2=\\ \\=-6,250-5,625-50-60-2=-11,987

To find the value of the function p(3) just substitute x=3 into the function expression instead of x:

p(3)=2\cdot 3^5-9\cdot 3^4-2\cdot 3^2+12\cdot 3-2=\\ \\=486-729-18+36-2=-227

3 0
3 years ago
Read 2 more answers
Consider the following theorem. Theorem If f is integrable on [a, b], then b a f(x) dx = lim n→[infinity] n i = 1 f(xi)Δx where
mel-nik [20]

Split up the interval [1, 9] into <em>n</em> subintervals of equal length (9 - 1)/<em>n</em> = 8/<em>n</em> :

[1, 1 + 8/<em>n</em>], [1 + 8/<em>n</em>, 1 + 16/<em>n</em>], [1 + 16/<em>n</em>, 1 + 24/<em>n</em>], …, [1 + 8 (<em>n</em> - 1)/<em>n</em>, 9]

It should be clear that the left endpoint of each subinterval make up an arithmetic sequence, so that the <em>i</em>-th subinterval has left endpoint

1 + 8/<em>n</em> (<em>i</em> - 1)

Then we approximate the definite integral by the sum of the areas of <em>n</em> rectangles with length 8/<em>n</em> and height f(x_i) :

\displaystyle \int_1^9 (x^2-4x+6) \,\mathrm dx \approx \sum_{i=1}^n \frac8n\left(\left(1+\frac8n(i-1)\right)^2-4\left(1+\frac8n(i-1)\right)+6\right)

Take the limit as <em>n</em> approaches infinity and the approximation becomes exact. So we have

\displaystyle \int_1^9 (x^2-4x+6) \,\mathrm dx = \lim_{n\to\infty} \sum_{i=1}^n \frac8n\left(\left(1+\frac8n(i-1)\right)^2-4\left(1+\frac8n(i-1)\right)+6\right) \\\\ = \lim_{n\to\infty} \frac8n \sum_{i=1}^n \left(1+\frac{16}n(i-1)+\frac{64}{n^2}(i-1)^2-4-\frac{32}n(i-1)+6\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \sum_{i=1}^n \left(64(i-1)^2-16n(i-1)+3n^2\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \sum_{i=0}^{n-1} \left(64i^2-16ni+3n^2\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \left(64\sum_{i=0}^{n-1}i^2 - 16n\sum_{i=0}^{n-1}i + 3n^2\sum{i=0}^{n-1}1\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \left(\frac{64(2n-1)n(n-1)}{6} - \frac{16n^2(n-1)}{2} + 3n^3\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \left(\frac{49n^3}3-24n^2+\frac{32n}3\right) \\\\= \lim_{n\to\infty} \frac{8\left(49n^2-72n+32\right)}{3n^2} = \boxed{\frac{392}3}

3 0
3 years ago
Write 11/75 as a decimal
Maru [420]

Answer:

0.147

Step-by-step explanation:

This is rounded to the nearest thousandths place. 11 divided by 75 is 0.147.

3 0
3 years ago
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Links will get banned​
goblinko [34]

Answer:

22 cm long

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4 0
3 years ago
Fraction has a fraction in the<br> numerator, denominator, or both
lions [1.4K]
A fraction is numerator over denominator, most of the time whole numbers
7 0
2 years ago
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