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astraxan [27]
3 years ago
14

A man flies a small airplane from Fargo to Bismarck, North Dakota --- a distance of 180 miles. Because he is flying into a head

wind, the trip takes him 2 hours. On the way back, the wind is still blowing at the same speed, so the return trip takes only .9 hours. What is the plane's speed in still air, and how fast is the wind blowing?
Mathematics
1 answer:
Mice21 [21]3 years ago
7 0

Answer:

90 mi/h

Step-by-step explanation:

Calculate speed, distance, or time using the formula d = st, distance equals speed times time. The Speed Distance Time Calculator can solve for the unknown sdt value given two known values.

Time can be entered or solved for in units of seconds (s), minutes (min), hours (hr), or hours and minutes and seconds (hh:mm:ss).  

To solve for distance use the formula for distance d = st, or distance equals speed times time.

distance = speed x time

Rate and speed are similar since they both represent some distance per unit time like miles per hour or kilometers per hour. If rate r is the same as speed s, r = s = d/t. You can use the equivalent formula d = rt which means distance equals rate times time.

distance = rate x time

To solve for speed or rate use the formula for speed, s = d/t which means speed equals distance divided by time.

speed = distance/time

To solve for time use the formula for time, t = d/s which means time equals distance divided by speed.

time = distance/speed

Plane speed in still air: 120mph

Wind speed: 30pmh

Speed of plane to Bismarck: 90mph

Speed of plane to Fargo: 150mph

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Answer:
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6 0
3 years ago
Triangle P R Q is shown. The length of side P R is 5 n. The length of side R Q is 32 + n. Angles R P Q and P Q R are congruent.
snow_tiger [21]

Answer:

<h3>20</h3>

Step-by-step explanation:

According to Triangle P R Q, if  angles R P Q and P Q R are congruent, this means that two of the sides of the triangle a re also congruent (Isosceles triangle).

We can say then that length of side PR is equal to that of RQ i.e PR = RQ

Given

PR = 5n

RQ = 32+n

Required

Length of PR

Since the two sides are equal i.e PR = RQ

5n = 32+n

5n - n = 32

4n = 32

n = 32/8

n = 4

Get PR;

Since PR = 5n

PR = 5(4)

PR = 20

Hence the length of PR is 20.

6 0
2 years ago
Read 2 more answers
Jessie draws triangle ABC on a coordinate grid. The slope of line segment AB is Jessie then transforms triangle
balu736 [363]

Answer:

1) Supports Jessie's Claim

2) Does Not Support Jessi's Claim

3) Supports Jessie's Claim

4) Does Not Support Jessi's Claim

Step-by-step explanation:

The given transformations are;

1) Rotation of 180° around the origin

For a rotation of 180° around the origin, either clockwise or anti clockwise, for a given coordinate of the preimage (x, y), the coordinate of the image is (-x, -y)

Therefore, whereby the slope of the preimage, given two points (0, 0) and (2, 2), = (2 - 0)/(2 - 0) = 1

For the image with the points (0, 0) and (-2, -2), we have;

(-2 - 0)/(-2 - 0) = 1

Therefore, the slope of the preimage and the image are equal

Therefore, supports Jessie's Claim

2) For a reflection across the line y = 2, we have

We note that the line y = 2 is parallel to the x-axis

For a reflection across the x-axis, for a preimage (x, y), we have the coordinates of the image (x, -y)

However for the reflection across the line y = 2, we have;

For a preimage, (x, y), the coordinate of the image is (x, -y+4)

Given two points, of the preimage (0, 0) and (2, 2), we have the image given as (0, 4) and (2, -2 + 4) = (2, 2);

The slope of the preimage is (2 - 0)/(2 - 0) = 1

The slope of the image is (2 - 4)/(2 - 0) = -1

The slope of the line of the preimage and the image are different

Therefore, does Not Support Jessi's Claim

3) For a translation up 1.25 units, we note that the difference in the y and x values of the coordinates of the preimage and the image will be equal when finding the slope, and therefore, the slope of the figure of the preimage and the slope of the figure of the image will be equal

Therefore, supports Jessie's Claim

4) For a reflection across the x-axis, a point on the preimage, with coordinates (x, y) will form a point on the image with coordinates (x, - y)

For a preimage with points (0, 0) and (2, 2), we have the image as (0, 0) and (2, -2)

The slope of the preimage is (2 - 0)/(2 - 0) = 1

The slope of the image is (-2 - 0)/(2 - 0) = -1

The slope of the line of the preimage and the image are different

Therefore, does Not Support Jessi's Claim

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yaroslaw [1]

\huge \tt \red {√Solution}

\tt \pink {a= 180°-47°}

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\tt \orange {c=133°}

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