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3 years ago

Plz help with this I don’t really get it

Mathematics

1 answer:

Greeley [361]3 years ago

6 0

Answer:

Step-by-step explanation:

Associative Property of Multiplication is a(bc) = (ab)c

d is the only one that follows the rule

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Pls help 5[9+3(3.2-3)]

SIZIF [17.4K]

Answer:

Step-by-step explanation:

3.2 - 3 = .2

3(.2) = .6

9 + .6 = 9.6

5 * 9.6 = 48

4 0

3 years ago

Read 2 more answers

P=2=2/+2w help solve for w

mr Goodwill [35]

Answer:

P = 2l + 2w

solve for w

re-write

2l + 2w = P

subtract 2l from both sides

2w = P - 2l

divide both sides by 2

w = (P - 2l) / 2

Answer is B

w = (P - 2l) / 2

4 0

2 years ago

123556 cube root pls

Korvikt [17]

Answer: ❄︎ ✭ ❄︎

The cube root of 123556 is: ✩ ✶ ✲

49.807.... ★ ✮

It is irrational, which means it goes on forever. ✧ ☆

Step-by-step explanation: ✺ ✹

$*VirtuosoTeen*$ ✰ ✴︎

✴︎ ❃ ✴︎ ❈ ✴︎ ✩ ✦ ✣

4 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,

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