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4 years ago

Can u plz answer questions 5 through8 plz thank u so much

Mathematics

1 answer:

Andrew [12]4 years ago

6 0

Question 5: 4) 2006-2012

Question 6: 8x^2+8x+1

Question 7: 2) 2

Question 8: a is picture on top, b is picture on bottom

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A person stands 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 meters

In-s [12.5K]

Answer:

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

Step-by-step explanation:

Given that,

A person stand 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 m/s.

From Pythagorean Theorem,

(The distance between car and person)²= (The distance of the car from intersection)²+ (The distance of the person from intersection)²+

Assume that the distance of the car from the intersection and from the person be x and y at any time t respectively.

∴y²= x²+10²

$\Rightarrow y=\sqrt{x^2+100}$

Differentiating with respect to t

$\frac{dy}{dt}=\frac{1}{2\sqrt{x^2+100}}. 2x\frac{dx}{dt}$

$\Rightarrow \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}. \frac{dx}{dt}$

Since the car driving towards the intersection at 13 m/s.

so, $\frac{dx}{dt}=-13$

$\therefore \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}.(-13)$

Now

$\therefore \frac{dy}{dt}|_{x=24}=\frac{24}{\sqrt{24^2+100}}.(-13)$

$=\frac{24\times (-13)}{\sqrt{676}}$

$=\frac{24\times (-13)}{26}$

= -12 m/s

Negative sign denotes the distance between the car and the person decrease.

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

8 0

3 years ago

An urn contains 6 red balls, 8 green balls, and 10 white balls. If balls are selected one by one (without replacing them after t

PIT_PIT [208]

Answer:

0.43478

Step-by-step explanation:

From the information given:

$The \ required \ probability = P(1st ball = red) \times P(2nd ball = red) + P(1st ball = red) \times P(2nd ball = white) + P(2nd ball = red) \times P(1st ball = white) + P(1st ball = white) \times P(2nd ball = white)$

$= \dfrac{6}{6+8+10}\times \dfrac{5}{5+8+10} + \dfrac{6}{6+8+10} \times \dfrac{10}{5+8+10} + \dfrac{10}{6+8+10} \times \dfrac{6}{5+8+10} \times \dfrac{10}{6+8+10} \times \dfrac{9}{5+8+9}$

$=(\dfrac{6}{24}\times \dfrac{5}{23})+ (\dfrac{6}{24}\times \dfrac{10}{23}) +( \dfrac{10}{24}\times \dfrac{6}{23} ) + (\dfrac{10}{24}\times \dfrac{9}{23})$