Find the area of the figure. A. 50 ft^2 B. 41 ft^2 C. 42 ft^2 D. 38 ft^2

The present age of the father is twice the age of his daughter. Ten years ago, the

Elenna [48]

Answer:

<h3> The father is 40 and the daughter is 20.</h3>

Step-by-step explanation:

x - the present age of the daughter

2x - the present age of the fathter

x - 10 - the age of the daughter ten year ago

2x - 10 - the age of the fathter ten year ago

Father is older than his dauther, so:

2x - 10 = (x - 10) + 20

2x - 10 = x - 10 + 20

2x - 10 = x + 10 {subtract x from both sides}

x - 10 = 10 {add 10 to both sides}

x = 20

2x = 2·20 = 40

6 0

2 years ago

£440 is divided between David, Mark & Henry so that David gets twice as much as Mark, and Mark gets three times as much as H

Lesechka [4]

Mark received £ 132

Solution:

Given that £440 is divided between David, Mark & Henry

Let "d" be the share of david

Let "m" be the share of mark

Let "h" be the share of henry

Total amount is 440

Therefore,

share of david + share of mark + share of henry = 440

d + m + h = 440 ------- eqn 1

David gets twice as much as Mark

d = 2m ----- eqn 2

Mark gets three times as much as Henry

m = 3h

$h = \frac{m}{3}$ ------ eqn 3

Substitute eqn 2 and eqn 3 in eqn 1

$2m + m + \frac{m}{3} = 440\\\\\frac{6m + 3m + m}{3} = 440\\\\10m = 1320\\\\m = 132$

Thus Mark received £ 132

8 0

3 years ago

A number line contains points Q, R, S, and T. Point Q is on the coordinate 24, R is on the coordinate 28, S is on the coordinate

const2013 [10]

$|\Omega|=QT=42-24=18\\ |A|=ST=42-29=13\\\\ P(A)=\dfrac{13}{18}\approx72.2\%\Rightarrow \text{D}$

7 0

3 years ago

Read 2 more answers

The graph of a polynomial function approaches - as x approaches -co, and approaches + op as x approaches + Which could be the de

wolverine [178]

Answer:

degree 5, leading coefficient 1

Step-by-step explanation:

When the sign of the end behavior matches that of x, the leading coefficient is positive, and the degree is odd.

One possibility is ...

degree 5, leading coefficient 1

6 0

3 years ago

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