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3 years ago

5

What is the value of StartFraction 1 Over 6 Superscript 0 EndFraction? One-sixth 0 1 undefined

2 answers:

Allisa [31]3 years ago

8 0

Answer:

1

Step-by-step explanation:

just took the test :)

marusya05 [52]3 years ago

5 0

Answer:

1

Step-by-step explanation:

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Find the midpoint of the line segment with the given endpoints (-4,4) (5,-1)

seraphim [82]

Answer:

(1/2), (3/2)

Step-by-step explanation:

midpoint fomula

(x1+x2)/ 2, (y1+y2)/2

(5-4)= 1 1/2 = (1/2)

(-1+4)=3 3/2=(3/2)

3 0

3 years ago

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If the length of the major axis of an ellipse is 28 units and the foci are located 3.5 units away from the center, what is the e

Mandarinka [93]

Eccentricity = foci / half the length of the major axis
= 3.5 / 14 = 1/4

3 0

3 years ago

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two mountain bikers leave from the same parking lot and head in opposite directions on two different trails. the first rider goe

sveta [45]

Applying the required <em>rule </em>or <em>theorem</em>, it can be concluded that the second biker is <u>farther</u> from the <em>parking lot</em>. The distance of the bikers to the <em>parking lot</em> are:

i. First biker = 17.0 km

ii. Second biker = 20.22 km

The <u>path</u> of travel of both bikers would form a triangle. Applying the <u>Pythagoras</u> theorem to the path of the <em>first</em> biker would give his <u>distance</u> from the starting point. While applying the <u>cosine</u> rule to the path of <em>second</em> rider would gives his <u>distance</u> to the starting point.

Thus,

a. <u>To determine the distance of the first biker from the parking lot.</u>

Let the required <em>distance </em>be represented by x. Applying the Pythagoras theorem, we have:

$hyp^{2}$ = $adj 1^{2}$ + $adj 2^{2}$

$x^{2}$ = $8^{2}$ + $15^{2}$

= 64 + 225

= 289

x = $\sqrt{289}$

= 17

x = 17 km

Thus, the <u>first</u> biker is 17.0 km from the <em>starting</em> point.

b. <u>To determine the distance of the second biker from the parking lot.</u>

Let the required <em>distance</em> be represented by x. So that applying the cosine rule, we have:

$c^{2}$ = $a^{2}$ + $b^{2}$ - 2ab Cos θ

$x^{2}$ = $8^{2}$ + $15^{2}$ - 2(15*8) Cos (180 - 20)

= 64 + 225 - 240 Cos 160

= 289 - 240 * -0.5

$x^{2}$ = 289 + 120

= 409

x = $\sqrt{409}$

= 20.2237

x = 20.22 km

Thus, the <u>second</u> biker is 20.22 km from the <em>starting</em> point.

Therefore, the second biker is <u>farther</u> from the <em>parking lot</em>.

A sketch of the path of travel for the two bikers is attached for more clarifications.

Visit: brainly.com/question/22699651

7 0

2 years ago

While walking between gates at an airport, you notice a child running along a moving walkway. Estimating that the child runs at

tekilochka [14]

The speed of the moving walkway relative to the airport terminal exists at 1.84 m/s.

<h3>How to estimate the speed of the moving walkway relative to the airport terminal?</h3>

Let x be the speed of the walkway.

(2.8 + x) = speed of child moving in direction of the walkway

(2.8 - x) = speed of child moving against the direction of the walkway

Travel time = distance/speed

Travel time of child moving in direction of walkway = 23/(2.8+x)

Total elapsed time given = 29s

23/(2.8 + x)+ 23 / (2.8-x) = 29

LCD = (2.8 + x)(2.8 - x)

$23(2.8 - x) + 23(2.8 + x) = 29(2.8 + x)(2.8 -x)$

simplifying the equation, we get

$23*2.8-23x+23*2.8+23x=29(2.8^2-x^2)$

$23(2.8+2.8)/29=2.8^2-x^2$

$x^2=(2.8)^2-(23*5.6)/29)=3.4$

$x=\sqrt{3.4}=1.84m/s$

Speed of walkway = 1.84 m/s

The speed of the moving walkway relative to the airport terminal exists at 1.84 m/s.

To learn more about Speed refer to:

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4 0

2 years ago

The Fruit Juice Company sold 5,909,999,222 cans of juice last year. This

elixir [45]

A) 2000000000

5,909,999,222-3,999,002,135= 1,910,997,087 which is closest to answer A

5 0

2 years ago