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3 years ago

NEED HELP!!

1 answer:

6 0

Wait
omg i just did this at school ughhhhhhh
one sec ill remember like at some point
getting ma book
nothinnnnnnn
really?
2x-35=somethin
and
somethin-9=88+.........
like that?/??????

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A man starts walking north at 2 ft/s from a point P. Five minutes later a woman starts walking south at 3 ft/s from a point 500

Vedmedyk [2.9K]

Answer:

The rate at which both of them are moving apart is 4.9761 ft/sec.

Step-by-step explanation:

Given:

Rate at which the woman is walking, $\frac{d(w)}{dt}$ = 3 ft/sec

Rate at which the man is walking, $\frac{d(m)}{dt}$ = 2 ft/sec

Collective rate of both, $\frac{d(m+w)}{dt}$ = 5 ft/sec

Woman starts walking after 5 mins so we have to consider total time traveled by man as (5+15) min = 20 min

Now,

Distance traveled by man and woman are $m$ and $w$ ft respectively.

⇒ $m=2\ ft/sec=2\times \frac{60}{min} \times 20\ min =2400\ ft$

⇒ $w=3\ ft/sec = 3\times \frac{60}{min} \times 15\ min =2700\ ft$

As we see in the diagram (attachment) that it forms a right angled triangle and we have to calculate $\frac{dh}{dt}$ .

Lets calculate h.

Applying Pythagoras formula.

⇒ $h^2=(m+w)^2+500^2$

⇒ $h=\sqrt{(2400+2700)^2+500^2} = 5124.45$

Now differentiating the Pythagoras formula we can calculate the rate at which both of them are moving apart.

Differentiating with respect to time.

⇒ $h^2=(m+w)^2+500^2$

⇒ $2h\frac{d(h)}{dt}=2(m+w)\frac{d(m+w)}{dt} + \frac{d(500)}{dt}$

⇒ $\frac{d(h)}{dt} =\frac{2(m+w)\frac{d(m+w)}{dt} }{2h}$ ...as $\frac{d(500)}{dt}= 0$

⇒ Plugging the values.

⇒ $\frac{d(h)}{dt} =\frac{2(2400+2700)(5)}{2\times 5124.45}$ ...as $\frac{d(m+w)}{dt} = 5$ ft/sec

⇒ $\frac{d(h)}{dt} =4.9761$ ft/sec

So the rate from which man and woman moving apart is 4.9761 ft/sec.

3 0

3 years ago

A local eat-in pizza restaurant wants to investigate the possibility of starting to deliver pizzas. The owner of the store has

gavmur [86]

Answer:

Step-by-step explanation:

The owner of the store has determined that home delivery will be successful if the average time spent on the deliveries does not exceed 34 minutes. This is the null hypothesis. It is written as

H0 : µ ≤ 34

The alternative hypothesis would be

Ha : µ > 34

This is a right tailed test because of the greater then symbol in the alternative hypothesis. Since the p value for the test was found to be 0.0281281, if we use a significant level of 0.05, then the conclusion would be

Reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the average time spent on the deliveries does exceed 34 minutes.

6 0

3 years ago

....Help Please......

Bogdan [553]

Answer:

9984000

Step-by-step explanation:

subtract the total by the other number (coverage)

3 0

2 years ago

Read 2 more answers

Which function is graphed below?

maxonik [38]

I’m not entirely sure, but try Desmos or Symbolab calculators :)

4 0

3 years ago

Read 2 more answers

Write an equation in slope-intercept form for the line that passes through the point ( -1 , -2 ) and is perpendicular to the l

mamaluj [8]

The equation in slope-intercept form for the line that passes through the point ( -1 , -2 ) and is perpendicular to the line − 4 x − 3 y = − 5 is $y = \frac{3}{4}x - \frac{5}{4}$

Solution:

The slope intercept form is given as:

y = mx + c ----- eqn 1

Where "m" is the slope of line and "c" is the y - intercept

Given that the line that passes through the point ( -1 , -2 ) and is perpendicular to the line − 4 x − 3 y = − 5

Given line is perpendicular to − 4 x − 3 y = − 5

− 4 x − 3 y = − 5

-3y = 4x - 5

3y = -4x + 5

$y = \frac{-4x}{3} + \frac{5}{3}$

On comparing the above equation with eqn 1, we get,

$m = \frac{-4}{3}$

We know that product of slope of a line and slope of line perpendicular to it is -1

$\frac{-4}{3} \times \text{ slope of line perpendicular to it}= -1\\\\\text{ slope of line perpendicular to it} = \frac{3}{4}$

Given point is (-1, -2)

Now we have to find the equation of line passing through (-1, -2) with slope $m = \frac{3}{4}$

Substitute (x, y) = (-1, -2) and m = 3/4 in eqn 1

$-2 = \frac{3}{4}(-1) + c\\\\-2 = \frac{-3}{4} + c\\\\c = - 2 + \frac{3}{4}\\\\c = \frac{-5}{4}$

$\text{ substitute } c = \frac{-5}{4} \text{ and } m = \frac{3}{4} \text{ in eqn 1}$

$y = \frac{3}{4} \times x + \frac{-5}{4}\\\\y = \frac{3}{4}x - \frac{5}{4}$

Thus the required equation of line is found

8 0

3 years ago