All categories

3 years ago

Express in numerals two thousand one hundred fiteen

Mathematics

1 answer:

Irina18 [472]3 years ago

4 0

The numerals of two thousand one hundred fifteen is 2115

<span />

You might be interested in

What's -0.25 as a fraction

sattari [20]

Answer:

0.25 is a quarter so 1/4

then -1/4

5 0

3 years ago

A pre-algebra book is 8 1 half inches by 9 1 fourth inches. Determine the perimeter of the textbook.

Arisa [49]

Answer:

Rectangular book--

two 8 1/2 sides

two 9 1/4 sides

8 1/2*2=17

9 1/4*2=18 1/2

17+18 1/2= 35 1/2

The perimeter of the textbook is 35 1/2 inches or 35.5 inches.

-this is not mine btw creds to sunflowerem :))

8 0

3 years ago

Order the following from greatest to least: -12,12,-19,1 and 1/2,5.....

den301095 [7]

-19, -12, 1/2, 1, 5, 12 ----> least to greatest
12, 5, 1, 1/2, -12, -19 ----> greatest to least

Do the same as the last one, starting with the largest negative number, and ending with the largest positive number. The only difference in this case is that it is greatest to least. I tend to go from least to greatest first and then flip it around to get greatest to least (:

3 0

3 years ago

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$