Answer:
Original position: base is 1.5 meters away from the wall and the vertical distance from the top end to the ground let it be y and length of the ladder be L.
Step-by-step explanation:
By pythagorean theorem, L^2=y^2+(1.5)^2=y^2+2.25 Eq1.
Final position: base is 2 meters away, and the vertical distance from top end to the ground is y - 0.25 because it falls down the wall 0.25 meters and length of the ladder is also L.
By pythagorean theorem, L^2=(y -0.25)^2+(2)^2=y^2–0.5y+ 0.0625+4=y^2–0.5y+4.0625 Eq 2.
Equating both Eq 1 and Eq 2: y^2+2.25=y^2–0.5y+4.0625
y^2-y^2+0.5y+2.25–4.0625=0
0.5y- 1.8125=0
0.5y=1.8125
y=1.8125/0.5= 3.625
Using Eq 1: L^2=(3.625)^2+2.25=15.390625, L=(15.390625)^1/2= 3.92 meters length of ladder
Using Eq 2: L^2=(3.625)^2–0.5(3.625)+4.0625
L^2=13.140625–0.90625+4.0615=15.390625
L= (15.390625)^1/2= 3.92 meters length of ladder
<em>hope it helps...</em>
<em>correct me if I'm wrong...</em>
Find the common denominator
It is 6
Multiply everything by 6
2/3*(6)= 4g
1/2 *(6)= 3g
14*(6)= 84
4g +3g = 84
7g= 84
g= 12
Answer:
Step-by-step explanation:
In order to write the equation of the line perpendicular to the given line, we first have to know what the slope of the given line is, and there's no way to tell by looking at it in its current form, which is standard. We need to solve that equation for y to determine the slope of that line. Solving for y:
and
3y = 4x - 5 (just change all the signs so our y term isn't negative anymore...yes, you're "allowed" to do that!) and
So we can see now that the slope of this line is 4/3. That means that the perpendicular slope is -3/4. Passing through the given point (3, 5):
* and
and
so
** and, in standard form:
4y = -3x + 29 and
3x + 4y = 29***
* : point-slope form
** : slope-intercept form
*** : standard form
Answer:
282.7miles
Step-by-step explanation:
334+514=848inch
12inch=4miles
848inch=x
![x = \frac{818 \times 4}{12} = 282.7](https://tex.z-dn.net/?f=x%20%3D%20%20%5Cfrac%7B818%20%5Ctimes%204%7D%7B12%7D%20%20%3D%20282.7)
An inch or 2 so yeah if in shapes then around a rectangle