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koban [17]
3 years ago
5

On a particular day, the wind added 4 miles per hour to Alfonso’s rate when he was cycling with the wind and subtracted 4 miles

per hour from his rate on his return trip. Alfonso found that in the same amount of time he could cycle 68 miles with the wind, he could go only 30 miles against the wind. What is his normal bicycling speed with no wind?
Mathematics
1 answer:
Greeley [361]3 years ago
8 0

d = r · t ⇒ t = d/r

going

t = 54/(r + 4)

coming

t = 30/(r - 4)

54/(r + 4) = 30/(r - 4) times are equal

54(r - 4) = 30(r + 4) product means/extremes

54r - 216 = 30r + 120 distribute

24r - 216 = 120 subtract 30r from both sides

24r = 336 add 216 to both sides

r = 14 divide both sides by 24

Alfonso bikes at 24 mph.

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Find the minimum distance between the point (-3,2) and the line y = -x + 1
Natasha2012 [34]

The minimum distance  between the point(-3,2) and the line y = -x + 1  is \sqrt{2} units.

<h3>What is a line segment?</h3>

The line that joins  two points on a cartesian plane is a line segment.

Analysis:

The formula for calculating distance between a line and a point is

d = \frac{Ax1 + By1 + c}{\sqrt{A^{2} + B^{2}  } }

where A = coefficient of x term of the line, B = coefficient of y in the line and C is the constant term of the line.

x1 and y1 are coordinates of the point.

for the line y = -x + 1 which is  = y + x -1 = 0, A = 1, B = 1, C = -1, x1 = -3, y = 2
d = \frac{1(-3) + 1(2) + (-1)}{\sqrt{(-3)^{2}  + 2^{2} } }= \sqrt{2} units

In conclusion, the distance between the point and the line is  \sqrt{2} units

Learn more about distance between line segments: brainly.com/question/2437195

#SPJ1

4 0
2 years ago
HELP ASAP (Geometry)
Andrei [34K]

1) Parallel line: y=-2x-3

2) Rectangle

3) Perpendicular line: y = 0.5x + 2.5

4) x-coordinate: 2.7

5) Distance: d=\sqrt{(4-3)^2+(7-1)^2}

6) 3/8

7) Perimeter: 12.4 units

8) Area: 8 square units

9) Two slopes of triangle ABC are opposite reciprocals

10) Perpendicular line: y-5=-4(x-(-1))

Step-by-step explanation:

1)

The equation of a line is in the form

y=mx+q

where m is the slope and q is the y-intercept.

Two lines are parallel to each other if they have same slope m.

The line given in this problem is

y=-2x+7

So its slope is m=-2. Therefore, the only line parallel to this one is the line which have the same slope, which is:

y=-2x-3

Since it also has m=-2

2)

We can verify that this is a rectangle by checking that the two diagonals are congruent. We have:

- First diagonal: d_1 = \sqrt{(-3-(-1))^2+(4-(-2))^2}=\sqrt{(-2)^2+(6)^2}=6.32

- Second diagonal: d_2 = \sqrt{(1-(-5))^2+(0-2)^2}=\sqrt{6^2+(-2)^2}=6.32

The diagonals are congruent, so this is a rectangle.

3)

Given points A (0,1) and B (-2,5), the slope of the line is:

m=\frac{5-1}{-2-0}=-2

The slope of a line perpendicular to AB is equal to the inverse reciprocal of the slope of AB, so:

m'=\frac{1}{2}

And using the slope-intercept for,

y-y_0 = m(x-x_0)

Using the point (x_0,y_0)=(7,1) we find:

y-1=\frac{1}{2}(x-7)

And re-arranging,

y-1 = \frac{1}{2}x-\frac{7}{2}\\y=\frac{1}{2}x-\frac{5}{2}\\y=0.5x-2.5

4)

The endpoints of the segment are X(1,2) and Y(6,7).

We have to divide the sgment into 1/3 and 2/3 parts from X to Y, so for the x-coordinate we get:

x' = x_0 + \frac{1}{3}(x_1 - x_0) = 1+\frac{1}{3}(6-1)=2.7

5)

The distance between two points A(x_A,y_A) and B(x_B,y_B) is given by

d=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}

In this problem, the two points are

E(3,1)

F(4,7)

So the distance is given by

d=\sqrt{(4-3)^2+(7-1)^2}

6)

We have:

A(3,4)

B(11,3)

Point C divides the segment into two parts with 3:5 ratio.

The distance between the x-coordinates of A and B is 8 units: this means that the x-coordinate of C falls 3 units to the right of the x-coordinate of A and 5 units to the left of the x-coordinate of B, so overall, the x-coordinate of C falls at

\frac{3}{3+5}=\frac{3}{8}

of the  distance between A and B.

7)

To find the perimeter, we have to calculate the length of each side:

d_{EF}=\sqrt{(x_E-x_F)^2+(y_E-y_F)^2}=\sqrt{(-1-2)^2+(6-4)^2}=3.6

d_{FG}=\sqrt{(x_G-x_F)^2+(y_G-y_F)^2}=\sqrt{(-1-2)^2+(3-4)^2}=3.2

d_{GH}=\sqrt{(x_G-x_H)^2+(y_G-y_H)^2}=\sqrt{(-1-(-3))^2+(3-3)^2}=2

d_{EH}=\sqrt{(x_E-x_H)^2+(y_E-y_H)^2}=\sqrt{(-1-(-3))^2+(6-3)^2}=3.6

So the perimeter is

p = 3.6 + 3.2 + 2 + 3.6 = 12.4

8)

The area of a triangle is

A=\frac{1}{2}(base)(height)

For this triangle,

Base = XW

Height = YZ

We calculate the length of the base and of the height:

Base =XW=\sqrt{(x_X-x_W)^2+(y_X-y_W)^2}=\sqrt{(6-2)^2+(3-(-1))^2}=5.7

Height =YZ=\sqrt{(x_Y-x_Z)^2+(y_Y-y_Z)^2}=\sqrt{(7-5)^2+(0-2)^2}=2.8

So the area is

A=\frac{1}{2}(XW)(YZ)=\frac{1}{2}(5.7)(2.8)=8

9)

A triangle is a right triangle when there is one right angle. This means that two sides of the triangle are perpendicular to each other: however, two lines are perpendicular when their slopes are opposite reciprocals. Therefore, this means that the true statement is

"Two slopes of triangle ABC are opposite reciprocals"

10)

The initial line is

y=\frac{1}{4}x-6

A line perpendicular to this one must have a slope which is the opposite reciprocal, so

m'=-4

Using the slope-intercept form,

y-y_0 = m'(x-x_0)

And using the point

(x_0,y_0)=(-1,5)

we find:

y-5=-4(x-(-1))

Learn more about parallel and perpendicular lines:

brainly.com/question/3414323

brainly.com/question/3569195

#LearnwithBrainly

8 0
3 years ago
E = MV^2 ÷ 2 + mgh, Make V Subject of the formula​
Stels [109]

Answer:

\red{ \bold{ V = \sqrt{\frac{2(E - mgh)}{M}}}}

Step-by-step explanation:

E = MV^2 ÷ 2 + mgh \\  \\ E - mgh = MV^2 ÷ 2  \\  \\  2(E - mgh)= MV^2   \\  \\  \frac{2(E - mgh)}{M}=V^2  \\  \\  \sqrt{\frac{2(E - mgh)}{M}}  = V \\  \\ \red{ \bold{ V = \sqrt{\frac{2(E - mgh)}{M}}}}

4 0
3 years ago
List all pairs of matching angles
anyanavicka [17]

Answer:

Listed

Please mark me as the brainliest and thanks

3 0
3 years ago
Write the equation of the line that passes through the given points. (-1,16.5) And (0,6.5)
11111nata11111 [884]

Answer: -10x+6.5

Step-by-step explanation:

8 0
3 years ago
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