Find the slope of the line that passes through the points (2 -5) and (7, 1)

Calculate the ROI, given the following: Andrew invested $20,000 in mutual funds and received a sum of $35,000 at the end of the

Dennis_Churaev [7]

We are given with an initial deposit of $20,000 and a future worth of $35,000. In this case, we are asked for the return of income (ROI) of the investment. in this case, we assume the number of years equal to 1. hence,

$35,000 = $20,000* (1+i) ^1
i or ROI then is equal to 0.75

3 0

3 years ago

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Which statement is true?

iVinArrow [24]

Answer:

the answer is " the slope of the function A is less then the slop of the function B

hope it helps

5 0

3 years ago

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NO LINKS OR FILES!

Archy [21]

(a) If the particle's position (measured with some unit) at time t is given by s(t), where

$s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}$

then the velocity at time t, v(t), is given by the derivative of s(t),

$v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}$

(b) The velocity after 3 seconds is

$v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}$

(c) The particle is at rest when its velocity is zero:

$\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}$

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

$\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|$

In interval notation, this happens for t in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt$

By definition of absolute value, we have

$|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)$

In part (d), we've shown that v(t) > 0 when -√11 < t < √11, so we split up the integral at t = √11 as

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt$

and by the fundamental theorem of calculus, since we know v(t) is the derivative of s(t), this reduces to

$s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}$

7 0

2 years ago

Which statement about the simplified binomial expansion of (a + b)", where n is a positive integer, is true?

Alja [10]

Answer:

$(a+b)^n ={n \choose 0}a^{(n)}b^{(0)} + {n \choose 1}a^{(n-1)}b^{(1)} + {n \choose 2}a^{(n-2)}b^{(2)} + ..... +{n \choose n}a^{(0)}b^{(n)}$

Step-by-step explanation:

The Given question is INCOMPLETE as the statements are not provided.

Now, let us try and solve the given expression here:

The given expression is: $(a +b)^n, n > 0$

Now, the BINOMIAL EXPANSION is the expansion which describes the algebraic expansion of powers of a binomial.

Here, $(a+b)^n = \sum_{k=0}^{n}{n \choose k}a^{(n-k)}b^{(k)}$

or, on simplification, the terms of the expansion are:

$(a+b)^n ={n \choose 0}a^{(n)}b^{(0)} + {n \choose 1}a^{(n-1)}b^{(1)} + {n \choose 2}a^{(n-2)}b^{(2)} + ..... +{n \choose n}a^{(0)}b^{(n)}$

The above statement holds for each n > 0

Hence, the complete expansion for the given expression is given as above.

4 0

3 years ago

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Use the place-valué charro to quite a Numbers in the table that is 1/10 of given Number?

Lana71 [14]

This is the answer for the Use the place-valué charro to quite a Numbers in the table that is 1/10 of given Number?

6 0

3 years ago