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2 years ago

What do exponential functions model in the real world? How does the

1 answer:

3 0

They model various situations where there is an increasing curve. An example is if you throw a baseball and it curves back down as gravity has more and more of a affect on it. This would be a different equation than if you threw a basketball as the basketball wouldn't go as high but would have about the same downwards acceleration due to gravity.

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Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 90.5 N , Jill pul

Radda [10]

Answer:

Fn= 174.9 N : Magnitude of the net force the people exert on the donkey.

Step-by-step explanation:

We find the components of the forces in x-y-z

Force of Jack in z =F₁z=90.5 N in direction (+z)

Force of Jill in x = F₂x= -82.3*cos45°= - 58.19 N (-x)

Force of Jill in y =F₂y=-82.3*sin45°= + 58.19 N (+y)

Force of Jane in x =F₃x=125*cos45°= + 88.4 N (+x)

Force of Jane in y =F₃y=125*sin45°= + 88.4 N (+y)

Calculating of the components of the net force the people exert on the donkey.

Fnx= F₂x+F₃x=( - 58.19+ 88.4 )N=30.2N (+x)

Fny= F₂y+F₃y=( 58.19+88.4 ) = 146.59 N (+y)

Fnz =F₁z=90.5 N (+z)

Calculating of the magnitude of the net force the people exert on the donkey.

$F_{n} =\sqrt{(F_{nx})^{2}+(F_{ny}) ^{2} +(F_{nz}) ^{2} }$

$F_{n} =\sqrt{(30.2)^{2}+( 146.59) ^{2} +(90.5) ^{2} }$

$F_{n} = 174.9 N$

7 0

3 years ago

NEED HELP NEED HELP please if you know show me how you got the answer

OLEGan [10]

Answer:

It says to Graph y = 1/2x - 4

So equation y = 1/2x - 4

this corresponds to slope intercept form which is y = mx + b

M = slope = rise / run

B = y - intercept

First, start at the origin, (0,0)

Then go down 4, you should now be on the coordinate (0,4)

Now we need to use the slope.

The slope is 1 / 2, this means rise 1, run 2 (in this case 2 to the right because it is positive)

So, start at (0 , 4) and go up 1 unit then 2 to the right, then place a dot, keep repeating until you meet the end of the paper.

Then do the opposite so that you can graph the other side.

Down 1 and 2 to the left.

4 0

3 years ago

Read 2 more answers

The image of the polygon ABCDE reflected across the x-axis. A(-1,-1),B(0,1),C(4,2),D(6,0),E(3,-3) The vertices of the image are.

4vir4ik [10]

A reflection across the x-axis has the rule:

(x,y)→(x,-y).

Then:

A(-1,-1)→A'(-1,1),
B(0,1)→B'(0,-1),
C(4,2)→C'(4,-2),
D(6,0)→D'(6,0),
E(3,-3)→E'(3,3).

Answer: the vertices of the image are A'(-1,1), B'(0,-1), C'(4,-2), D'(6,0) and E'(3,3).

6 0

3 years ago

Please help asap, The area of the living room is:<br> A. 20 m²<br> B. 25 m²<br> C. 15 m²<br> D. 9 m²

Inga [223]

Answer:

All these rooms are rectangles so to find the

area of the rooms multiply length by width.

Living Room: 5x4 = 20 m^2

7 0

2 years ago

Read 2 more answers

A box designer has been charged with the task of determining the surface area of various open boxes (no lid) that can be constru

Viktor [21]

Answer:

1) $S = 2\cdot w\cdot l - 8\cdot x^{2}$ , 2) The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$ , 3) $S = 176\,in^{2}$ , 4) $x \approx 4.528\,in$ , 5) $S = 164.830\,in^{2}$

Step-by-step explanation:

1) The function of the box is:

$S = 2\cdot (w - 2\cdot x)\cdot x + 2\cdot (l-2\cdot x)\cdot x +(w-2\cdot x)\cdot (l-2\cdot x)$

$S = 2\cdot w\cdot x - 4\cdot x^{2} + 2\cdot l\cdot x - 4\cdot x^{2} + w\cdot l -2\cdot (l + w)\cdot x + l\cdot w$

$S = 2\cdot (w+l)\cdot x - 8\cdpt x^{2} + 2\cdot w \cdot l - 2\cdot (l+w)\cdot x$

$S = 2\cdot w\cdot l - 8\cdot x^{2}$

2) The maximum cutout is:

$2\cdot w \cdot l - 8\cdot x^{2} = 0$

$w\cdot l - 4\cdot x^{2} = 0$

$4\cdot x^{2} = w\cdot l$

$x = \frac{\sqrt{w\cdot l}}{2}$

The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$

3) The surface area when a 1'' x 1'' square is cut out is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1\,in)^{2}$

$S = 176\,in^{2}$

4) The size is found by solving the following second-order polynomial:

$20\,in^{2} = 2 \cdot (8\,in)\cdot (11.5\,in)-8\cdot x^{2}$

$20\,in^{2} = 184\,in^{2} - 8\cdot x^{2}$

$8\cdot x^{2} - 164\,in^{2} = 0$

$x \approx 4.528\,in$

5) The equation of the box volume is:

$V = (w-2\cdot x)\cdot (l-2\cdot x) \cdot x$

$V = [w\cdot l -2\cdot (w+l)\cdot x + 4\cdot x^{2}]\cdot x$

$V = w\cdot l \cdot x - 2\cdot (w+l)\cdot x^{2} + 4\cdot x^{3}$

$V = (8\,in)\cdot (11.5\,in)\cdot x - 2\cdot (19.5\,in)\cdot x^{2} + 4\cdot x^{3}$

$V = (92\,in^{2})\cdot x - (39\,in)\cdot x^{2} + 4\cdot x^{3}$

The first derivative of the function is:

$V' = 92\,in^{2} - (78\,in)\cdot x + 12\cdot x^{2}$

The critical points are determined by equalizing the derivative to zero:

$12\cdot x^{2}-(78\,in)\cdot x + 92\,in^{2} = 0$

$x_{1} \approx 4.952\,in$

$x_{2}\approx 1.548\,in$

The second derivative is found afterwards:

$V'' = 24\cdot x - 78\,in$

After evaluating each critical point, it follows that $x_{1}$ is an absolute minimum and $x_{2}$ is an absolute maximum. Hence, the value of the cutoff so that volume is maximized is:

$x \approx 1.548\,in$

The surface area of the box is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1.548\,in)^{2}$

$S = 164.830\,in^{2}$

4 0

2 years ago