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Irina-Kira [14]
2 years ago
10

Attached geometry question. Please help. 100 points, will mark branliest.

Mathematics
1 answer:
-BARSIC- [3]2 years ago
6 0

\large{\rm{\underline{\underline{Answer:}}}}

Since ∆AFB is similar to ∆ABC.

  • < F = < B (corresponding angle)
  • < G = < C (corresponding angle)
  • < A = common.

<u>In </u><u>∆</u><u>A</u><u>BC,</u>

⇛< A + < B + < C = 180°

⇛37° + 65° + < C = 180°

⇛102° + < C = 180°

⇛< C = 78°

We know that, < AGF / < G = < C

So, Measure of angle < AGF = <u>7</u><u>8</u><u>°</u><u> </u><u>(</u><u>ans)</u>

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Answer: \frac{20}{3}\ minutes or 6 \frac{2}{3}\ minutes

Step-by-step explanation:

For this exercise you can convert the mixed number to an improper fraction:

1. Multiply the whole number part by the denominator of the fraction.

2. Add the product obtained and the numerator of the fraction (This will be the new numerator).

3. The denominator does not change.

Then:

23\frac{1}{3}= \frac{(23*3)+1}{3}= \frac{70}{3}\ minutes

You know that he had 30 minutes in time-out, he counted spots on the ceiling for \frac{70}{3} minutes and the rest of the time he made faces at his stuffed tiger.

Then, in order to calculate the time Calvin spent making faces at his stuffed tiger, you need to subract 30 minutes and \frac{70}{3} minutes:

30\ min-\frac{70}{3}=(\frac{3(30)-70}{3})=\frac{20}{3}\ minutes or 6 \frac{2}{3}\ minutes

7 0
3 years ago
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How to multiply how far would a car go if it travels at 50 km an hour for 3 hours?
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A ball is launched from a 682.276 meter tall platform. the equation for the ball's height h at time t seconds after launch is h(
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The maximum height the ball achieves before landing is 682.276 meters at t = 0.

<h3>What are maxima and minima?</h3>

Maxima and minima of a function are the extreme within the range, in other words, the maximum value of a function at a certain point is called maxima and the minimum value of a function at a certain point is called minima.

We have a function:

h(t) = -4.9t² + 682.276

Which represents the ball's height h at time t seconds.

To find the maximum height first find the first derivative of the function and equate it to zero

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Find second derivative:

h''(t) = -9.8

At t = 0; h''(0) < 0 which means at t = 0 the function will be maximum.

Maximum height at t = 0:

h(0) = 682.276 meters

Thus, the maximum height the ball achieves before landing is 682.276 meters at t = 0.

Learn more about the maxima and minima here:

brainly.com/question/6422517

#SPJ1

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Answer:

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