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3 years ago

9

Suppose a railroad is 5 Kilometers and it expands on a hot day by 16 centimeters in length. Approximately how many meters would

the center of the rail rise above the ground?

2 answers:

Butoxors [25]3 years ago

3 0

Answer:

17.32 meters above the ground.

Step-by-step explanation:

ASHA 777 [7]3 years ago

3 0

Answer:

20.0

Step-by-step explanation:

You might be interested in

Sales rise from £400 a week to £504 a week. Calculate the percentage increase

Andrew [12]

Answer:

26%

Step-by-step explanation:

percent change = (new number - old number)/(old number) * 100%

Here, the new number is £504, and the old number is £400.

percent change = (£504 - £400)/(£400) * 100% =

= £104/£400 * 100% = 0.26 * 100% = 26%

5 0

3 years ago

Write the equation of the line perpendicular to 2x+3y=9 that passes through (-2,5). Write your answer in slope-intercept form. S

Harman [31]

ANSWER

$y = \frac{3}{2} x + 8$

EXPLANATION

The line given to us has equation,

$2x + 3y = 9$

We need to write this equation in the slope intercept form to obtain,

$3y = - 2x + 9$

$\Rightarrow \: y = - \frac{2}{3}x + 3$

The slope of this line is

$m_1 = - \frac{2}{3}$
Let the slope of the perpendicular line be

$m_2$

Then
$m_1 \times m_2 = - 1$

$- \frac{2}{3} m_2= - 1$

This implies that,

$m_2 = - 1 \times - \frac{3}{2}$

$m_2 = \frac{3}{2}$

Let the equation of the perpendicular line be,

$y = mx + b$

We substitute the slope to get,

$y = \frac{3}{2} x + b$

Since this line passes through
$(-2,5)$
it must satisfy its equation.

This means that,

$5= \frac{3}{2} ( - 2)+ b$

$5 = - 3 + b$

$5 + 3 = b$

$b = 8$

Wherefore the slope-intercept form is

$y = \frac{3}{2} x + 8$

6 0

3 years ago

!!!!!!!!PLEASE HELP ASAP!!!!!!!!!

Slav-nsk [51]

Welp someone already answered and I just gotta answer questions so goodluck :)

4 0

3 years ago

Read 2 more answers

Write an algebraic expression to find the AREA of a rectangle with a length of 4x-5 and a width of 7.

blagie [28]

Answer:

28x-35

Step-by-step explanation:

4x*7=28x

5*7=35

8 0

3 years ago

Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured across a line (x + 2)/3 = (2y + 3)/4

dimaraw [331]

Answer:

Distance of the point from its image = 8.56 units

Step-by-step explanation:

Given,

Co-ordinates of point is (-2, 3,-4)

Let's say

$x_1\ =\ -2$

$y_1\ =\ 3$

$z_1\ =\ -4$

Distance is measure across the line

$\dfrac{x+2}{3}\ =\ \dfrac{2y+3}{4}\ =\ \dfrac{3z+4}{5}$

So, we can write

$\dfrac{x-x_1+2}{3}\ =\ \dfrac{2(y-y_1)+3}{4}\ =\ \dfrac{3(z-z_1)+4}{5}\ =\ k$

$=>\ \dfrac{x-(-2)+2}{3}\ =\ \dfrac{2(y-3)+3}{4}\ =\ \dfrac{3(z-(-4))+4}{5}\ =\ k$

$=>\ \dfrac{x+4}{3}\ =\ \dfrac{2y-3}{4}\ =\ \dfrac{3z+16}{5}\ =\ k$

$=>\ x\ =\ 3k-4,\ y\ =\ \dfrac{4k+3}{2},\ z\ =\ \dfrac{5k-16}{3}$

Since, the equation of plane is given by

x+y+z=3

The point which intersect the point will satisfy the equation of plane.

So, we can write

$3k-4+\dfrac{4k+3}{2}+\dfrac{5k-16}{3}\ =\ 3$

$=>6(3k-4)+3(4k+3)+2(5k-16)\ =\ 18$

$=>18k-24+12k+9+10k-32\ =\ 18$

$=>\ k\ =\dfrac{13}{8}$

So,

$x\ =\ 3k-4$

$=\ 3\times \dfrac{13}{8}-4$

$=\ \dfrac{7}{4}$

$y\ =\ \dfrac{4k+3}{2}$

$=\ \dfrac{4\times \dfrac{13}{8}+3}{2}$

$=\ \dfrac{19}{4}$

$z\ =\ \dfrac{5k-16}{3}$

$=\ \dfrac{5\times \dfrac{13}{8}-16}{3}$

$=\ \dfrac{-21}{8}$

Now, the distance of point from the plane is given by,

$d\ =\ \sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}$

$=\ \sqrt{(-2-\dfrac{7}{4})^2+(3-\dfrac{19}{4})^2+(-4+\dfrac{21}{8})^2}$

$=\ \sqrt{(\dfrac{-15}{4})^2+(\dfrac{-7}{4})^2+(\dfrac{9}{8})^2}$

$=\ \sqrt{\dfrac{225}{16}+\dfrac{49}{16}+\dfrac{81}{64}}$

$=\ \sqrt{\dfrac{1177}{64}}$

$=\ 4.28$

So, the distance of the point from its image can be given by,

D = 2d = 2 x 4.28

= 8.56 unit

So, the distance of a point from it's image is 8.56 units.

4 0

3 years ago