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3 years ago

Help me in Geometry!!! I don’t know nothing about this help. PHOTO ABOVE

Mathematics

1 answer:

SOVA2 [1]3 years ago

6 0

Answer:

m<1 = 39

m<2 = 51

Step-by-step explanation:

For this problem, you need to understand that a little square in the bottom of two connecting lines represents a right-angle (an angle this 90 degrees). This problem, gives you two relationships for angle 1 and angle 2 within a right-angle. Using this information, we can solve for the measures of the two angles.

Let's write the two relations:

m< 1 = 3x

m< 2 = x + 38

And now let's right an equation that represents the two angles to the picture:

m<1 + m<2 = 90

Using this information, let's substitute the expressions we have for the two angles and solve for x. Once we have the value of x, we can find the measure of the two angles.

m< 1 + m< 2 = 90

(3x) + (x + 38) = 90

3x + x + 38 = 90

x ( 3 + 1 ) + 38 = 90

x ( 4 ) + 38 = 90

4x + 38 = 90

4x + 38 - 38 = 90 - 38

4x = 90 - 38

4x = 52

4x * (1/4) = 52 * (1/4)

x = 52 * (1/4)

x = 13

Now that we have the value of x, we simply plug it back into our expressions for the m<1 and m<2.

m<1 = 3x = 3(13) = 39

m<2 = x + 38 = 13 + 38 = 51

And we can verify this is correct with the relational equation:

m<1 + m<2 = 90

39 + 51 ?= 90

90 == 90

Hence, we have found the values of m<1 and m<2.

Cheers.

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4 years ago

A yo-yo is moving up and down a string so that its velocity at time t is given by v(t) = 3cos(t) for time t ≥ 0. The initial pos

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Part A - The average value of v(t) over the interval (0, π/2) is 6/π

Part B - The displacement of the yo-yo from time t = 0 to time t = π is 0 m

Part C - The total distance the yo-yo travels from time t = 0 to time t = π is 6 m.

<h3>Part A: Find the average value of v(t) on the interval (0, π/2)</h3>

The average value of a function f(t) over the interval (a,b) is

$f(t)_{avg} = \frac{1}{b - a} \int\limits^b_a {f(t)} \, dx$

So, since velocity at time t is given by v(t) = 3cos(t) for time t ≥ 0. Its average value over the interval (0, π/2) is given by

$v(t)_{avg} = \frac{1}{\frac{\pi }{2} - 0} \int\limits^{\frac{\pi }{2} }_0 {v(t)} \, dt$

Since v(t) = 3cost, we have

$v(t)_{avg} = \frac{1}{\frac{\pi }{2} - 0} \int\limits^{\frac{\pi }{2} }_0 {3cos(t)} \, dt\\= \frac{3}{\frac{\pi }{2}} \int\limits^{\frac{\pi }{2} }_0 {cos(t)} \, dt\\= \frac{6}{{\pi}} [{sin(t)}]^{\frac{\pi }{2} }_{0} \\= \frac{6}{{\pi}} [{sin(\frac{\pi }{2})} - sin0]\\ = \frac{6}{{\pi}} [1 - 0]\\ = \frac{6}{{\pi}} [1]\\ = \frac{6}{{\pi}}$

So, the average value of v(t) over the interval (0, π/2) is 6/π

<h3>Part B: What is the displacement of the yo-yo from time t = 0 to time t = π?</h3>

To find the displacement of the yo-yo, we need to find its position.

So, its position x = ∫v(t)dt

= ∫3cos(t)dt

= 3∫cos(t)dt

= 3sint + C

Given that at t = 0, x = 3. so

x = 3sint + C

3 = 3sin0 + C

3 = 0 + C

C = 3

So, x(t) = 3sint + 3

So, its displacement from time t = 0 to time t = π is

Δx = x(π) - x(0)

= 3sinπ + 3 - (3sin0 + 3)

= 3 × 0 + 3 - 0 - 3

= 0 + 3 - 3

= 0 + 0

= 0 m

So, the displacement of the yo-yo from time t = 0 to time t = π is 0 m

<h3>Part C: Find the total distance the yo-yo travels from time t = 0 to time t = π. (10 points)</h3>

The total distance the yo-yo travels from time t = 0 to time t = π is given by

$x(t) = \int\limits^{\pi}_0 {v(t)} \, dt\\= \int\limits^{\pi }_0 {3cos(t)} \, dt\\= 3 \int\limits^{\pi }_0 {cos(t)} \, dt\\ = 3 \int\limits^{\frac{\pi }{2} }_0 {cos(t)} \, dt + 3\int\limits^{\pi }_{\frac{\pi }{2}} {cos(t)} \, dt\\= 3 \times 2\int\limits^{\frac{\pi }{2} }_0 {cos(t)} \, dt\\= 6 [{sin(t)}]^{\frac{\pi }{2} }_{0} \\= 6[{sin\frac{\pi }{2} - sin0]\\\\= 6[1 - 0]\\= 6(1)\\= 6$

So, the total distance the yo-yo travels from time t = 0 to time t = π is 6 m.

Learn more about average value of a function here:

brainly.com/question/15870615

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1 year ago