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4 years ago

If 16 counters are 4/9 then what is One

1 answer:

3 0

36...

divide your 16 by 4, so that you have 1/9. You then multiply the resulting number(4) by your denominator, 9. 4 times 9 = 36

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Sam's age is 125% of mary's age. mary's age is p% of sam's age. what is the value of p?

ddd [48]

Let Sam be (S+8) years while his sister is S years.

S(S+8) = 105

S^2+8S-105 = 0

(S+15)(S-7) = 0

Sam’s sister is 7 years and Sam is 15 years old.

7 0

2 years ago

your local radio station is having their yearly mp3 and concert ticket giveaway for one minute, every 5th caller will win an mp3

a_sh-v [17]

You find the lcm of both
5 10 15 20 25 30 35
7 14 21 28 35
You were the 35 caller

5 0

4 years ago

Read 2 more answers

Helppppp i need help ASAP

algol13

Answer:

a) 10.10 a.m. , (b) 10.20 a.m. which is 10 minutes after sprints start , (c) The students work at 10.50 a.m. , (d) For 60 minutes from 10 a.m. to 11 a.m. which equals to 1 hour

4 0

3 years ago

Differentiate with respect to x and simplify your answer. Show all the appropriate steps? 1.e^-2xlog(ln x)^3 2.e^-2x(log(ln x))^

serious [3.7K]

(1) I assume "log" on its own refers to the base-10 logarithm.

$\left(e^{-2x}\log(\ln x)^3\right)'=\left(e^{-2x}\right)'\log(\ln x)^3+e^{-2x}\left(\log(\ln x)^3\right)'$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{e^{-2x}}{\ln10(\ln x)^3}\left((\ln x)^3\right)'$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}(\ln x)^2}{\ln10(\ln x)^3}\left(\ln x\right)'$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}(\ln x)^2}{\ln10\,x(\ln x)^3}$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}}{\ln10\,x\ln x}$

Note that writing $\log(\ln x)^3=3\log(\ln x)$ is one way to avoid using the power rule.

(2)

$\left(e^{-2x}(\log(\ln x))^3\right)'=(e^{-2x})'(\log(\ln x))^3+e^{-2x}\left(\log(\ln x))^3\right)'$

$=-2e^{-2x}(\log(\ln x))^3+3e^{-2x}(\log(\ln x))^2(\log(\ln x))'$

$=-2e^{-2x}(\log(\ln x))^3+3e^{-2x}(\log(\ln x))^2\dfrac{(\ln x)'}{\ln10\,\ln x}$

$=-2e^{-2x}(\log(\ln x))^3+\dfrac{3e^{-2x}(\log(\ln x))^2}{\ln10\,x\ln x}$

(3)

$\left(\sin(xe^x)^3\right)'=\left(\sin(x^3e^{3x})\right)'=\cos(x^3e^{3x}(x^3e^{3x})'$

$=\cos(x^3e^{3x})((x^3)'e^{3x}+x^3(e^{3x})')$

$=\cos(x^3e^{3x})(3x^2e^{3x}+3x^3e^{3x})$

$=3x^2e^{3x}(1+x)\cos(x^3e^{3x})$

(4)

$\left(\sin^3(xe^x)\right)'=3\sin^2(xe^x)\left(\sin(xe^x)\right)'$

$=3\sin^2(xe^x)\cos(xe^x)(xe^x)'$

$=3\sin^2(xe^x)\cos(xe^x)(x'e^x+x(e^x)')$

$=3\sin^2(xe^x)\cos(xe^x)(e^x+xe^x)$

$=3e^x(1+x)\sin^2(xe^x)\cos(xe^x)$

(5) Use implicit differentiation here.

$(\ln(xy))'=(e^{2y})'$

$\dfrac{(xy)'}{xy}=2e^{2y}y'$

$\dfrac{x'y+xy'}{xy}=2e^{2y}y'$

$y+xy'=2xye^{2y}y'$

$y=(2xye^{2y}-x)y'$

$y'=\dfrac y{2xye^{2y}-x}$

8 0

3 years ago

Help me please help me

Vlad [161]

999999999999999999999999999999Answer:9999999999999999999 poppplygojhyijjuhhnnjnnnjjndjbjkghdhhghbenhdjgetpodjfhdj

Step-by-step explanation:

4 0

3 years ago