Plz help me thank you. 4 cm of the ice tray

Mathematics

2 answers:

disa [49]3 years ago

6 0

I had to do this and it’s fake, my whole class got it wrong even though we all put the right answer so ya

Readme [11.4K]3 years ago

4 0

I don’t know bud I need point so yea

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Over which interval is the graph of f(x) = –x2 + 3x + 8 increasing? (–∞, 1.5) (–∞, 10.25) (1.5, ∞) (10.25, ∞)

Vadim26 [7]

F(x) = -x^2 + 3x + 8
f'(x) = -2x + 3

when f(x) is increasing f'(x) > 0

-2x + 3 > 0

-2x > -3

x < 3/2 or 1.5

So the required interval is (-infinity,1.5).The first choice.

6 0

3 years ago

Read 2 more answers

A projectile is fired with muzzle speed 250 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe

qaws [65]

The projectile's horizontal and vertical positions at time $t$ are given by

$x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ\,t$

$y=30\,\mathrm m+\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ\,t-\dfrac g2t^2$

where $g=9.8\dfrac{\rm m}{\mathrm s^2}$ . Solve $y=0$ for the time $t$ it takes for the projectile to reach the ground:

$30+\dfrac{250}{\sqrt2}t-4.9t^2=0\implies t\approx36.2458\,\mathrm s$

In this time, the projectile will have traveled horizontally a distance of

$x=\dfrac{250\frac{\rm m}{\rm s}}{\sqrt2}(36.2458\,\mathrm s)\approx6400\,\mathrm m$

The projectile's horizontal and vertical velocities are given by

$v_x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ$

$v_y=\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ-gt$

At the time the projectile hits the ground, its velocity vector has horizontal component approx. 176.77 m/s and vertical component approx. -178.43 m/s, which corresponds to a speed of about $\sqrt{176.77^2+(-178.43)^2}\dfrac{\rm m}{\rm s}\approx250\dfrac{\rm m}{\rm s}$ .