P(x)=x^6-11x^5+30x^4 Number 15

33. Suppose you were on a planet where the

tatyana61 [14]

Answer:

a) 3.675 m

b) 3.67m

Explanation:

We are given acceleration due to gravity on earth = $9.8ms^-2$

And on planet given = $2.0ms^-2$

A) Since the maximum jump height is given by the formula

$\mathrm{H}=\frac{\left(\mathrm{v} 0^{2} \times \sin 2 \emptyset\right)}{2 \mathrm{g}}$

Where H = max jump height,

v0 = velocity of jump,

Ø = angle of jump and

g = acceleration due to gravity

Considering velocity and angle in both cases

$\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 2}{\mathrm{g} 1}$

Where H1 = jump height on given planet,

H2 = jump height on earth = 0.75m (given)

g1 = 2.0 $ms^-2$ and

g2 = 9.8 $ms^-2$

Substituting these values we get H1 = 3.675m which is the required answer

B) Formula to find height of ball thrown is given by

$\mathrm{h}=(\mathrm{v} 0 * \mathrm{t})+\frac{\mathrm{a} *\left(t^{2}\right)}{2}$

which is due to projectile motion of ball

Now h = max height,

v0 = initial velocity = 0,

t = time of motion,

a = acceleration = g = acceleration due to gravity

Considering t = same on both places we can write

$\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 1}{\mathrm{g} 2}$

where h1 and h2 are max heights ball reaches on planet and earth respectively and g1 and g2 are respective accelerations

substituting h2 = 18m, g1 = 2.0 $ms^-2$ and g2 = 9.8 $ms^-2$

We get h1 = 3.67m which is the required height

6 0

3 years ago

5 watermelons 6 cantaloupes cost $54. 4 watermelons and 8 cantaloupes $56. how much is a cantaloupe

GREYUIT [131]

Answer:

A cantaloupe costs $4

Step-by-step explanation:

Watermelons = W

Cantaloupes = C

Write given statements as equations

5W + 6C = 54

4W + 8C = 56

Simplify the second equation

4W + 8C = 56

Subtract 8C from both sides of the equation

4W = 56 - 8C

Divide both sides of the equation by 4

W = 14 - 2C

Substitute into second equation

5W + 6C = 54

5(14 - 2C) + 6C = 54

Multiply 5(14 - 2C)

70 - 10C + 6C = 54

Add -10C + 6C

70 - 4C = 54

Add 4C to both sides of the equation

70 = 54 + 4C

Subtract 54 from both sides of the equation

16 = 4C

Divide both sides of the equation by 4

C = 4

A cantaloupe costs $4

5W + 6(4) = 54

Multiply 6(4)

5W + 24 = 54

Subtract 24 from both sides of the equation

5W = 30

Divide both sides of the equation by 5

W = 6

A cantaloupe costs $4

A watermelon costs $6

Hope this helps :)

8 0

3 years ago

The ratio of the width to the length of a flower is 10:19. How long is the flower bed if it’s width is 6 feet?

Lubov Fominskaja [6]

Just set up an equation comparing the two ratios and solve for the length, in this case x.

(10/19)=(6/x)
19/10=x/6
114/10=x

Answer= 11.4 feet for the length of the flower

6 0

3 years ago

If the measure of angle JKP is 3r + 12 and the measure of angle PKN is 4r - 2, what is the measure of angle JKN

il63 [147K]

Answer should be 108 would you mind sending a visual so I can be sure

8 0

2 years ago

Find the distance from the origin to the graph of 7x+9y+11=0

Cerrena [4.2K]

One way to do it is with calculus. The distance between any point $(x,y)=\left(x,-\dfrac{7x+11}9\right)$ on the line to the origin is given by

$d(x)=\sqrt{x^2+\left(-\dfrac{7x+11}9\right)^2}=\dfrac{\sqrt{130x^2+154x+121}}9$

Now, both $d(x)$ and $d(x)^2$ attain their respective extrema at the same critical points, so we can work with the latter and apply the derivative test to that.

$d(x)^2=\dfrac{130x^2+154x+121}{81}\implies\dfrac{\mathrm dd(x)^2}{\mathrm dx}=\dfrac{260}{81}x+\dfrac{154}{81}$

Solving for $(d(x)^2)'=0$ , you find a critical point of $x=-\dfrac{77}{130}$ .

Next, check the concavity of the squared distance to verify that a minimum occurs at this value. If the second derivative is positive, then the critical point is the site of a minimum.

You have

$\dfrac{\mathrm d^2d(x)^2}{\mathrm dx^2}=\dfrac{260}{81}>0$

so indeed, a minimum occurs at $x=-\dfrac{77}{130}$ .

The minimum distance is then

$d\left(-\dfrac{77}{130}\right)=\dfrac{11}{\sqrt{130}}$

4 0

3 years ago