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Alexxandr [17]
3 years ago
10

The coordinates of the point I are (6,2) and the coordinates of point J are (6,-4).

Mathematics
1 answer:
jarptica [38.1K]3 years ago
5 0

Answer:

6 units

(Use the distance formula and plug in everything)

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A student is given that point P(a, b) lies on the terminal ray of angle Theta, which is between StartFraction 3 pi Over 2 EndFra
Harman [31]

Answer:

<em>A.</em>

<em>The student made an error in step 3 because a is positive in Quadrant IV; therefore, </em>

<em />cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}

Step-by-step explanation:

Given

P\ (a,b)

r = \± \sqrt{(a)^2 + (b)^2}

cos\theta = \frac{-a}{\sqrt{a^2 + b^2}} = -\frac{\sqrt{a^2 + b^2}}{a^2 + b^2}

Required

Where and which error did the student make

Given that the angle is in the 4th quadrant;

The value of r is positive, a is positive but b is negative;

Hence;

r = \sqrt{(a)^2 + (b)^2}

Since a belongs to the x axis and b belongs to the y axis;

cos\theta is calculated as thus

cos\theta = \frac{a}{r}

Substitute r = \sqrt{(a)^2 + (b)^2}

cos\theta = \frac{a}{\sqrt{(a)^2 + (b)^2}}

cos\theta = \frac{a}{\sqrt{a^2 + b^2}}

Rationalize the denominator

cos\theta = \frac{a}{\sqrt{a^2 + b^2}} * \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 + b^2}}

cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}

So, from the list of given options;

<em>The student's mistake is that a is positive in quadrant iv and his error is in step 3</em>

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GIVING BRAINLIEST FOR BEST ANSWER!
ra1l [238]

Answer:

its B

Step-by-step explanation:

Hope it helps

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Any value as long as P = Q

For the equation to have infinitely many solutions, we require both sides of the equation to have exactly the same terms.

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3 years ago
What are two fractions between 3/5 and 4/5
Elza [17]
6/10 and 13/20, sorry i would explain this but its late, cheers.
Hope this Helps.
4 0
4 years ago
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