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3 years ago

Log m + log n = 2 log (m+n/4)

1 answer:

3 0

Hello! are we solving for m or n? i can help you get the answer in the comments if u can clarify which we are solving for

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Which could be the area of a face of the rectangular prism?

Alex17521 [72]

Answer:

6m squared

Step-by-step explanation:

cause one of the faces is 2 times 3 and that is 6m squared

3 0

3 years ago

Read 2 more answers

Their sum is −7 and their difference is 14 what are my numbers

Mashcka [7]

Answer:

Answer: 3 1/2 and -10 1/2 are the two numbers.

Step-by-step explanation:

Let x and y be the two unknown numbers.

x+y=-7 [Given]

x-y=14 [Given]

x=y+14 [Add y to both sides]

x+y=-7 [Given]

(y+14)+y=-7 [Subtitution]

2y+14=-7 [Combine like terms]

2y=-21 [Subtract 14 from both sides]

y=-21/2 [Divide both sides by 2]

y=-10 1/2 [Division]

x-y=14 [Given]

x=y+14 [Add y to both sides]

x=-10 1/2 + 14 [Substitution]

x= 3 1/2 [Addition]

Check:

x+y=-7 [Given]

3 1/2 + -10 1/2?=-7 [Substition]

-7=-7 [Addition]

QED

x-y=14 [Given]

3 1/2 - -10 1/2?=14 [Substitution]

3 1/2 + 10 1/2?=14 [Change the sign of the subtrahend and add]

14=14 [Addition]

QED

Answer: 3 1/2 and -10 1/2 are the two numbers.

8 0

3 years ago

What is -d=|300-48t|

spayn [35]

I believe it would be b= -264

(I’m not 100% sure though so you might wanna check)

3 0

3 years ago

Find thd <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D" id="TexFormula1" title="\frac{dy}{dx}" alt="\frac{dy}{dx}" a

NARA [144]

$x^3y^2+\sin(x\ln y)+e^{xy}=0$

Differentiate both sides, treating $y$ as a function of $x$ . Let's take it one term at a time.

Power, product and chain rules:

$\dfrac{\mathrm d(x^3y^2)}{\mathrm dx}=\dfrac{\mathrm d(x^3)}{\mathrm dx}y^2+x^3\dfrac{\mathrm d(y^2)}{\mathrm dx}$

$=3x^2y^2+x^3(2y)\dfrac{\mathrm dy}{\mathrm dx}$

$=3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(\sin(x\ln y)}{\mathrm dx}=\cos(x\ln y)\dfrac{\mathrm d(x\ln y)}{\mathrm dx}$

$=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)$

$=\cos(x\ln y)\left(\ln y+\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}$

$=e^{xy}\left(\dfrac{\mathrm d(x)}{\mathrm dx}y+x\dfrac{\mathrm d(y)}{\mathrm dx}\right)$

$=e^{xy}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}$

The derivative of 0 is, of course, 0. So we have, upon differentiating everything,

$3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0$

Isolate the derivative, and solve for it:

$\left(6x^3y+\dfrac{\cos(x\ln y)}y+xe^{xy}\right)\dfrac{\mathrm dy}{\mathrm dx}=-\left(3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}\right)$

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}}{6x^3y+\frac{\cos(x\ln y)}y+xe^{xy}}$

(See comment below; all the 6s should be 2s)

We can simplify this a bit by multiplying the numerator and denominator by $y$ to get rid of that fraction in the denominator.

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^3+y\cos(x\ln y)\ln y-y^2e^{xy}}{6x^3y^2+\cos(x\ln y)+xye^{xy}}$

3 0

3 years ago

A football team was training for four hours. During the first hour, they practiced for 5/8 of an hour. During the second hour, t

Leviafan [203]

Answer: 2 19/24 hours was spent in practising.

Step-by-step explanation:

During the first hour, they practiced for 5/8 of an hour. During the second hour, they practiced for 2/3 of an hour. This means that the total time for which they practiced in the first 2 hours would be

5/8 + 2/3 = 31/24 hours

During the last two hours, they first practiced for 3/5 of an hour, took a 1/2 hour break and then practiced the rest of the time. This means that the rest of the time for which they practiced is

2 - (3/5 + 1/2) = 2 - 11/10 = 9/10

Therefore, the time they spent practicing in total would be

31/24 + 3/5 + 9/10 = 67/24 =

2 19/24 hours

7 0

3 years ago

Log m + log n = 2 log (m+n/4)​

Log m + log n = 2 log (m+n/4)