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3 years ago

HELP ASAP!!! WILL GIVE BRAINLEIST!!!

Mathematics

1 answer:

netineya [11]3 years ago

8 0

Answer:

The first question is (n*8) - 4

The second question is no.

Step-by-step explanation:

The first question, you add 8. The first is 1. 1 * 8 = 8 - 4 = 4.

Therefore, the equation is (n*8) - 4

The second question, you can plug in numbers. You can solve for a term. -100 + 17 = -83. -83 is not a multiple of -4. -100 is not a term.

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After two consecutive years of 5% rates of return, what rate of return in the third year will produce a cumulative gain of 16%?

murzikaleks [220]

Using decimal multipliers, it is found that a rate of return of 5.2% in the third year will produce a cumulative gain of 16%.

The <u>decimal multiplier</u> for a increase of a% is given by:

$M = \frac{100 + a}{100}$

In this problem, two increases of 5%, thus:

$M = \frac{100 + 5}{100} = 1.05$

Another of x, that we want to find, and the result is a increase of 16%, thus:

$M = \frac{100 + 16}{100} = 1.16$

The three increases(5%, 5% and x%) result in a increase of 16%, thus:

$1.05(1.05)(x) = 1.16$

$x = \frac{1.16}{(1.05)^2}$

$x = 1.052$

1.052 - 1 = 0.052

A rate of return of 5.2% in the third year will produce a cumulative gain of 16%.

A similar problem is given at brainly.com/question/21806362

5 0

2 years ago

Joshua gave away 1/3 of his marbles to Kale.If Kale got 25 marbles,how many marbles did Joshua have at first?

OLEGan [10]

Answer:

Step-by-step explanation:

joshua - x

kale - 1/3x

1/3x = 25

x = 25(3)

x = 75

6 0

2 years ago

The radius of a cone is increasing at a constant rate of 7 meters per minute, and the volume is decreasing at a rate of 236 cubi

storchak [24]

Answer:

The rate of change of the height is 0.021 meters per minute

Step-by-step explanation:

From the formula

$V = \frac{1}{3}\pi r^{2}h$

Differentiate the equation with respect to time t, such that

$\frac{d}{dt} (V) = \frac{d}{dt} (\frac{1}{3}\pi r^{2}h)$

$\frac{dV}{dt} = \frac{1}{3}\pi \frac{d}{dt} (r^{2}h)$

To differentiate the product,

Let r² = u, so that

$\frac{dV}{dt} = \frac{1}{3}\pi \frac{d}{dt} (uh)$

Then, using product rule

$\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h\frac{du}{dt}]$

Since $u = r^{2}$

Then, $\frac{du}{dr} = 2r$

Using the Chain's rule

$\frac{du}{dt} = \frac{du}{dr} \times \frac{dr}{dt}$

∴ $\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h(\frac{du}{dr} \times \frac{dr}{dt})]$

Then,

$\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}]$

Now,

From the question

$\frac{dr}{dt} = 7 m/min$

$\frac{dV}{dt} = 236 m^{3}/min$

At the instant when $r = 99 m$

and $V = 180 m^{3}$

We will determine the value of h, using

$V = \frac{1}{3}\pi r^{2}h$

$180 = \frac{1}{3}\pi (99)^{2}h$

$180 \times 3 = 9801\pi h$

$h =\frac{540}{9801\pi }$

$h =\frac{20}{363\pi }$

Now, Putting the parameters into the equation

$\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}]$

$236 = \frac{1}{3}\pi [(99)^{2} \frac{dh}{dt} + (\frac{20}{363\pi }) (2(99)) (7)]$

$236 \times 3 = \pi [9801 \frac{dh}{dt} + (\frac{20}{363\pi }) 1386]$

$708 = 9801\pi \frac{dh}{dt} + \frac{27720}{363}$

$708 = 30790.75 \frac{dh}{dt} + 76.36$

$708 - 76.36 = 30790.75\frac{dh}{dt}$

$631.64 = 30790.75\frac{dh}{dt}$

$\frac{dh}{dt}= \frac{631.64}{30790.75}$

$\frac{dh}{dt} = 0.021 m/min$

Hence, the rate of change of the height is 0.021 meters per minute.

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