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Between the time Iko woke up and lunchtime, the temperature rose by 11°. Then by the time he went to bed, the temperature droppe

Tanzania [10]

Answer: -3 degrees

Step-by-step explanation:

Lets say that the original temperature was "x."

The temperature rose by 11 degrees so it is now at 11+x.

Then the temperature fell by 14 degrees so you subtract. 11+x-14

11-14= -3 so you subtract 3 degrees from the original temperature and there fore the relative temperature is -3

I hope this helps!!!

4 0

2 years ago

Caleb draws two triangles

Arada [10]

Answer:

no no yes yes

Step-by-step explanation:

took the test

7 0

2 years ago

Calculus 2

FinnZ [79.3K]

Answer:

See Below.

Step-by-step explanation:

We want to estimate the definite integral:

$\displaystyle \int_1^47\sqrt{\ln(x)}\, dx$

Using the Trapezoidal Rule, Midpoint Rule, and Simpson's Rule with six equal subdivisions.

1)

The trapezoidal rule is given by:

$\displaystyle \int_{a}^bf(x)\, dx\approx\frac{\Delta x}{2}\Big(f(x_0)+2f(x_1)+...+2f(x_{n-1})+f(x_n)\Big)$

Our limits of integration are from x = 1 to x = 4. With six equal subdivisions, each subdivision will measure:

$\displaystyle \Delta x=\frac{4-1}{6}=\frac{1}{2}$

Therefore, the trapezoidal approximation is:

$\displaystyle =\frac{1/2}{2}\Big(f(1)+2f(1.5)+2f(2)+2f(2.5)+2f(3)+2f(3.5)+2f(4)\Big)$

Evaluate:

$\displaystyle =\frac{1}{4}(7)(\sqrt{\ln(1)}+2\sqrt{\ln(1.5)}+...+2\sqrt{\ln(3.5)}+\sqrt{\ln(4)})\\\\\approx18.139337$

2)

The midpoint rule is given by:

$\displaystyle \int_a^bf(x)\, dx\approx\sum_{i=1}^nf\Big(\frac{x_{i-1}+x_i}{2}\Big)\Delta x$

Thus:

$\displaystyle =\frac{1}{2}\Big(f\Big(\frac{1+1.5}{2}\Big)+f\Big(\frac{1.5+2}{2}\Big)+...+f\Big(\frac{3+3.5}{2}\Big)+f\Big(\frac{3.5+4}{2}\Big)\Big)$

Simplify:

$\displaystyle =\frac{1}{2}(7)\Big(f(1.25)+f(1.75)+...+f(3.25)+f(3.75)\Big)\\\\ =\frac{1}{2}(7) (\sqrt{\ln(1.25)}+\sqrt{\ln(1.75)}+...+\sqrt{\ln(3.25)}+\sqrt{\ln(3.75)})\\\\\approx 18.767319$

3)

Simpson's Rule is given by:

$\displaystyle \int_a^b f(x)\, dx\approx\frac{\Delta x}{3}\Big(f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+...+4f(x_{n-1})+f(x_n)\Big)$

So:

$\displaystyle =\frac{1/2}{3}\Big((f(1)+4f(1.5)+2f(2)+4f(2.5)+...+4f(3.5)+f(4)\Big)$

Simplify:

$\displaystyle =\frac{1}{6}(7)(\sqrt{\ln(1)}+4\sqrt{\ln(1.5)}+2\sqrt{\ln(2)}+4\sqrt{\ln(2.5)}+...+4\sqrt{\ln(3.5)}+\sqrt{\ln(4)})\\\\\approx 18.423834$