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7nadin3 [17]
3 years ago
14

PLSSSS HURRY I HAVE A TIME LIMIT MUST SHOW WORK

Mathematics
1 answer:
Goryan [66]3 years ago
7 0

Answer: XVR: 125 ; RVS: 55 ; WVS: 125 ; RST: 110 ; RSV: 70

Step-by-step explanation:

XVR: XVR is equal to WVS (alternate angles), and WVS plus XVW equals 180° (definition of a straight line)

So, 180° - 55° = 125° (this is the measure of SVW, but remember, SVW is equal to XVR)

RVS: RVS is equal to XVW since they're alternate angles, so we know that RVS is equal to 55°

WVS: We already solved this in the beginning

RST: First, we need to find the measure of RSV. To find the measure of RSV, use the fact that a triangle adds up to 180°. We know that the angle RVS equals 55°, and that angle VRS is also equal to 55°. So, we can use this equation:

RSV = 180° - (55° + 55°)

RSV = 70°

Now that we know RSV = 70°, we can find RST

180 - (RSV + RST)

180 - (70° + RST)

RST = 110°

RSV: We already found this

Sorry, that was a lot. Hope it wasn't too confusing.

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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
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