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KATRIN_1 [288]
2 years ago
5

In how many ways can a

Mathematics
1 answer:
stiks02 [169]2 years ago
3 0

Answer:

59,280

Step-by-step explanation:

the answer is base in my previous learning.. hope it helps..

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Find the limit as t approaches 8 for the function f(t)=7(t-1)(t-1)
Ratling [72]

Answer:

7³

Step-by-step explanation:

You could answer this after a quick inspection.  7 remains constant as t approaches 8.  Each of the (t -1) terms will approach 7.  So, multiplying these factors together, you'll get 7(7)(7) = 7³3 as your answer, the limit as t approaches 8 of the given expression.

5 0
3 years ago
Which name accurately describes the figure shown below and why?
mamaluj [8]

Answer: its A

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
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
2 years ago
Can someone help me with this?
LekaFEV [45]
The constant variation for the relationship being shown is 4
3 0
2 years ago
Read 2 more answers
Problem #3
AfilCa [17]

Answer:

b = 115.4 feet

Step-by-step explanation:

b^2 = a^2 + c^2 - 2 (a)(b) x cos B

b^2 = 230^2 + 360^2 - 2(230)(360) x cos(38 degrees)

b^2 = 52,900 + 129,600 - 2(230)(360) x cos(38 degrees)

b^2 = 182,500 - 165,600 x cos(38 degrees)

b = \sqrt{(16,900)(cos(38)) }

b = 115.4 feet

i hope this helps!! :)

8 0
2 years ago
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