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3 years ago

Compare decimals 0. 7 and 0. 6 9 9 9 is it A< be >. c

Mathematics

1 answer:

il63 [147K]3 years ago

5 0

Answer:

0.7 is larger than 0.6999 by 0.01

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Please help click on the image below

damaskus [11]

Yeah the answer would be B. 80 minutes

3 0

3 years ago

4. The revenue of a company for a given month is represented as ????(x) = 1,500x − x^2 and its costs as ????(x) = 1,500 + 1,000x

mrs_skeptik [129]

Answer:

The profit will be maximum on x = 250.

Step-by-step explanation:

From the given information:

Revenue = 1500x - x²

Cost = 1500 + 1000x

As we know that

Profit = Revenue - Cost ; Let say it equation 1

Then after putting the values of revenue and cost in equation 1 we have:

Profit = (1500x - x²) - (1500 + 1000x)

Profit = 1500x - x² - 1500 - 1000x

Profit = -x² + 500x - 1500

We know that at the max or min the slope of the graph formed by the profit function will be zero, therefore we find the slope of profit function by taking the first derrivative w.r.t. x as under:

d(Profit)/dx = d/dx(-x² + 500x - 1500)

d(Profit)/dx = -2x + 500

By putting the above slope equal to zero we get:

d(Profit)/dx = -2x + 500 = 0

-2x + 500 = 0

-2x = -500

x = 250

Therefore it is concluded that the profit will be maximum when x will be equal to 250.

4 0

3 years ago

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 $\sqrt{176.77^2+(-178.43)^2}\dfrac{\rm m}{\rm s}\approx250\dfrac{\rm m}{\rm s}$ .