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3 years ago

HELP ASAP Find the 15th term or the geometric sequence 7, -21, 63,

Mathematics

1 answer:

mihalych1998 [28]3 years ago

5 0

<h3>Answer:</h3>

33480783

<h2>Step-by-step explanation:</h2>

Finding the Common Ratio:

We know that the formula for the common ratio of a GP is:

r = aₙ / (aₙ₋₁) (Where n is any integer)

replacing the values, taking n=2

r = -21 / 7

r = -3

Solving for the 15th Term:

We know that the Formula for the nth term of a GP:

G(n) = a * r ⁿ⁻¹ (where a is the first term and r is the common ratio)

replacing the values (taking n = 15 since we need the 15th term)

G(15) = 7 * (-3)¹⁴

G(15) = 33480783

Hence, the 15th term of the given GP is 33480783

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3 years ago

walnut grower estimates from past records that if 20 trees are planted per acre, then each tree will average 60 pounds of nuts p

laila [671]

Answer:

5 trees should be planted to maximize the yield per acre,

The maximum yield would be 1250

Step-by-step explanation:

Given,

The original number of trees per acre = 20,

Average pounds of nuts by a tree = 60,

Let x be the times of increment in number of trees,

So, the new number of trees planted per acre = 20 + x

∵ for each additional tree planted per acre, the average yield per tree drops 2 pounds,

So, the new number of pounds of nut = (60 - 2x)

Thus, the total yield per acre,

$Y(x) = (20+x)(60-2x)$

Differentiating with respect to t ( time ),

$Y'(x) = (20+x)(-2) + 60 - 2x = -40 - 2x + 60 - 2x = 20 - 4x$

Again differentiating with respect to t,

$Y''(x) = -4$

For maxima or minima,

$Y'(x) = 0$

⇒ 20 - 4x = 0

⇒ 20 = 4x

⇒ x = 5,

For x = 5, Y''(x) = negative,

Hence, Y(x) is maximum for x = 5,

And, maximum value of Y(x) = (20+5)(60 - 10) = 25(50) = 1250,

i.e. 5 trees should be planted to maximize the yield per acre,

and the maximum yield would be 1250 pounds

4 0

3 years ago

WILL UPVOTE

amm1812

X=6 you need to multiply by 2

7 0

4 years ago

15. If x=a Sin2t (I+Cos2t) and y=b Cos 2t (1-Cos2t) then find dy/dx at =22/7*4

Sav [38]

By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}$

It looks like we're given

$\begin{cases}x=a\sin(2t)(1+\cos(2t))\\y=b\cos(2t)(1-\cos(2t))\end{cases}$

where a and b are presumably constant.

Recall that

$\cos^2t=\dfrac{1+\cos(2t)}2$

$\sin^2t=\dfrac{1-\cos(2t)}2$

so that

$\begin{cases}x=2a\sin(2t)\cos^2t\\y=2b\cos(2t)\sin^2t\end{cases}$

Then we have

$\dfrac{\mathrm dx}{\mathrm dt}=4a\cos(2t)\cos^2t-4a\sin(2t)\cos t\sin t$

$\dfrac{\mathrm dy}{\mathrm dt}=-4b\sin(2t)\sin^2t+4b\cos(2t)\sin t\cos t$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{4b\cos(2t)\sin t\cos t-4b\sin(2t)\sin^2}{4a\cos(2t)\cos^2t-4a\sin(2t)\cos t\sin t}$

$\implies\boxed{\dfrac{\mathrm dy}{\mathrm dx}=\dfrac ba\tan t}$

where the last reduction follows from dividing through everything by $\cos(2t)\cos^2t$ and simplifying.

I'm not sure at which point you're supposed to evaluate the derivative (22/7*4, as in 88/7? or something else?), so I'll leave that to you.

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3 years ago