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3 years ago

Find the indicated values, where g(t)= t^2 - t and f(x)= 1+x

Mathematics

2 answers:

SVEN [57.7K]3 years ago

5 0

Answer:

Option b is correct

g(f(3)-2f(1)) =0

Step-by-step explanation:

Given the functions:

$g(t) = t^2-t$

and

$f(x) = 1+x$

We have to find g(f(3)-2f(1)) .

At x = 3

$f(3) = 1+3 = 4$

at x = 1

$f(1) = 1+1= 2$

then;

g(f(3)-2f(1)) = $g(4-2(2)) = g(4-4) = g(0)$

then;

$g(f(3)-2f(1)) = 0^2-0 = 0$

Therefore, the indicated value of $g(f(3)-2f(1))$ is , 0

ki77a [65]3 years ago

4 0

Your answer would be B

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3 years ago

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AURORKA [14]

Answer:

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Step-by-step explanation:

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3 years ago

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0, -2/5, 0,67, 7/9, -0.5 in order from least to greatest

inessss [21]

Answer:

-0.5,-2/5,0,0.67,7/9

Step-by-step explanation:

Well since the bigger the negative the lesser the number it is itself so -0.5 is the least. The fraction -2/5 turn It into a decimal and you’ll get -0.4 which is greater then -0.5. Then the next number is 0 which is greater then any negative number. Then its 0.67 becuase the other positive fraction Is 0.78 simplified.

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3 years ago

Express the integral as a limit of Riemann sums. Do not evaluate the limit. (Use the right endpoints of each subinterval as your

Darina [25.2K]

Answer:

Given definite integral as a limit of Riemann sums is:

$\lim_{n \to \infty} \sum^{n} _{i=1}3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]$

Step-by-step explanation:

Given definite integral is:

$\int\limits^7_4 {\frac{x}{2}+x^{3}} \, dx \\f(x)=\frac{x}{2}+x^{3}---(1)\\\Delta x=\frac{b-a}{n}\\\\\Delta x=\frac{7-4}{n}=\frac{3}{n}\\\\x_{i}=a+\Delta xi\\a= Lower Limit=4\\\implies x_{i}=4+\frac{3}{n}i---(2)\\\\then\\f(x_{i})=\frac{x_{i}}{2}+x_{i}^{3}$

Substituting (2) in above

$f(x_{i})=\frac{1}{2}(4+\frac{3}{n}i)+(4+\frac{3}{n}i)^{3}\\\\f(x_{i})=(2+\frac{3}{2n}i)+(64+\frac{27}{n^{3}}i^{3}+3(16)\frac{3}{n}i+3(4)\frac{9}{n^{2}}i^{2})\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{3}{2n}i+\frac{144}{n}i+66\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{291}{2n}i+66\\\\f(x_{i})=3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]$

Riemann sum is:

$= \lim_{n \to \infty} \sum^{n} _{i=1}3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]$

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3 years ago

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Svetlanka [38]

75% of the gallon of milk is left

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3 years ago