Solve graphically: 2x - y = 7 and x - 2y = 5.
1 answer:
Answer:
(3, - 1)
Step-by-step explanation:
We want to solve the system of equations:
2*x - y = 7
x - 2*y = 5
graphically.
To do it, we need to graph both lines, it will be easier if we rewrite both lines in a more general way, like in the slope-intercept form:
y = -7 + 2*x
y = -(5/2) + x/2
Now, a solution of the system is a point (x, y) such that the point is a solution to both equations, then, when we graph the lines, as that point belongs to both equations, we should see that both lines pass through that point.
Then to find the solution of the system, we just need to graph the lines and find the point where the lines intersect.
Below, you can see a graph of the lines, and is easy to see that the lines do intersect at the point (3, - 1)
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Square means L=W
V=62.5=LWH
L=W so
V=62.5=HL^2
SA=2(L^2+2LH)
we have
V=62.5=HL^2
solve for H
divide both sides by L^2
62.5/L^2=H
sub that for H in other equation
SA=2(L^2+2L(62.5/L^2))
SA=2(L^2+125/L)
SA=2L^2+250/L
find minimum of 2L^2+250L^-1
take the derivitive
4L-250L^-2, or 2(2L^3-125)/L^2
find where it equals zero
it equals zero at L=2.5∛4
L=W
if we evaluate 2L^2+250/L at L=2.5∛4, the value is 75∛2
H=62.5/L^2
H=![\frac{25 \sqrt[3]{2} }{4}](https://tex.z-dn.net/?f=%20%5Cfrac%7B25%20%5Csqrt%5B3%5D%7B2%7D%20%7D%7B4%7D%20)
dimentions are
L=2.5∛4
W=2.5∛4
H=
minimum surface area is 75∛2 in^2 or aprox 94.4941 in^2
V=62.5=LWH
L=W so
V=62.5=HL^2
SA=2(L^2+2LH)
we have
V=62.5=HL^2
solve for H
divide both sides by L^2
62.5/L^2=H
sub that for H in other equation
SA=2(L^2+2L(62.5/L^2))
SA=2(L^2+125/L)
SA=2L^2+250/L
find minimum of 2L^2+250L^-1
take the derivitive
4L-250L^-2, or 2(2L^3-125)/L^2
find where it equals zero
it equals zero at L=2.5∛4
L=W
if we evaluate 2L^2+250/L at L=2.5∛4, the value is 75∛2
H=62.5/L^2
H=
dimentions are
L=2.5∛4
W=2.5∛4
H=
minimum surface area is 75∛2 in^2 or aprox 94.4941 in^2
Answer:
Step-by-step explanation:
Given :
In the given quadrilateral ABCD,
BN and DM are the perpendiculars drawn to AC such that,
BN = DM
To prove:
Point O is the midpoint of segment BD.
Or
OD = OB
Solution:
In ΔOMD and ΔONB,
∠MOD ≅ NOB [Vertical angles]
∠M ≅ ∠N ≅ 90° [Given]
Therefore, by AA property of similarity,
ΔOMD ~ Δ ONB
Therefore, their corresponding sides will be proportional,
Since BN = DM,
OD = OB
Hence O is the midpoint of BD.