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juin [17]
3 years ago
5

It’s the same distance from the second base to the first base from the second base to third-base the angle formed by the first b

ase second base and Homeplate has the same measure as the angle formed by the base second base home plate what can you conclude about the distance from first base to home plate from home plate to third-base explaining using your knowledge of congruent angles
Mathematics
1 answer:
Margaret [11]3 years ago
3 0

Answer:

Distance of home plate to 1st base is equal to home plate to 3rd base

Step-by-step explanation:

Let's first assign each position:

A: home base

B: 1st base

C: 2nd base

D: 3rd base

The information given:

'It’s the same distance from the second base to the first base from the second base to third-base' tells us BC = CD

'the angle formed by the first base second base and Homeplate has the same measure as the angle formed by the base second base home plate' tells us angle(BCA) = angle(DCA)

Check for congruency between triangle ABC and ADC

BC = CD - side congruency (S)

angle(BCA) = angle(DCA) - angle congruency (A)

CA = CA - side congruency (S)

So triangle ABC and ADC fullfill the criteria of congruence property as SAS.

Since both triangle are congruent, another side of the triangles AB and AD must be the same too.

AB = home plate to 1st base

AD = home plate to 3rd base

Therefore we can conclude that the distance of home plate to 1st base is equal to home plate to 3rd base

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Answer:

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Step-by-step explanation:

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3 years ago
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Graph the solution to this inequality on the number line.<br><br> −3x&lt;4.2
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Answer:

3x < 4 .2

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3 years ago
Jason jumped off a cliff into the ocean in Acapulco while vacationing with some friends. His height as a function of time could
WINSTONCH [101]

Answer:

484feet

Step-by-step explanation:

If the height as a function of time could be modeled by the function h(t)=−16t^2 +16t+480, where t is the time in seconds and h is the height in feet.

The maximum height of Jason occurs when dh/dt = 0 (velocity is zero)

dh/dt = -32t + 16

0 = -32t + 16

32t = 16

t = 16/32

t = 1/2

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Get the maximum height

Recall that h(t)=−16t^2 +16t+480

Substitute t = 0.5

h(0.5)=−16(0.5)^2 +16(0.5)+480

h(0.5) = -4 + 8 + 480

h(0.5) = 4 + 480

h(0.5) = 484

Hence the maximum height reached by Jason is 484feet

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3 years ago
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