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kakasveta [241]
3 years ago
13

What is the solution to the system that is created by the equation y=-x+6 and the graph shown below?

Mathematics
2 answers:
34kurt3 years ago
6 0
1st let/s find the equation of the graph shown. It has te form of y₁ =mx since it oases by the origin. Moreover we notice that the slope m = rise/run = 1/2. hence the equation IS y₁ = (1/2)x.
To find the solution of y₁ = 1/2 x and y= -x + 6, let's y=y₁

-x + 6 = 1/2 x ↔ -2x +12 = x ↔ -3x= -12  and x = 4. For x = 4 , y =2.

And the solution is the intersection point between y and y₁ which is nothing but the solution or te coordinate of this point that is (4,2)
mina [271]3 years ago
3 0
The graph graphed is y=1/2x

so
y=1/2x and y=-x+6
y=y so
1/2x=-x+6
times 2 both sides
x=-2x+12
add 2x both sides
3x=12
x=4

sub back
y=1/2x
y=1/2(4)
y=2

(4,2) is the solution
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3 years ago
A rectangular solid has width w, a length of 7 more than the width, and a height that is equivalent to 15 decreased by 3 times t
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Answer:

For a rectangular solid with:

width = w

length = l

height = h

The volume is equal to:

V = w*l*h

in this case we know that:

width = w.

"length of 7 more than the width"

l = w + 7.

"a height that is equivalent to 15 decreased by 3 times the width."

h = 15 - 3*w

Then the volume will be:

V = w*l*h = w*(w + 7)*(15 - 3w) = (w^2 + 7*w)*(15 - 3*w)

V =  ( -3*w^3 + 15*w^2 + 105*w - 21*w^2)

V = (-3*w^3 - 6*w^2 + 105*w)

Now, the maximum volume will be for the value of w such that:

V'(w) = 0.

and:

V''(W) < 0

Where:

dV/dw = V'(w).

dV'/dw = V''(w)

Then first we need to differentiate the equation for the volume.

V'(w) = dV/dw = ( 3*(-3*w^2) + 2*(-6*w) + 105)

V'(w) = -9*w^2 - 12*w + 105.

Then we need to find the solution for:

-9*w^2 - 12*w + 105 = 0.

We can use the Bhaskara formula, and we will get:

w = \frac{+12 +-\sqrt{(-12)^2 - 4*(-9)*105} }{2*-9}  = \frac{+12 +- 62.6}{-18}

Then the two solutions are:

w₁ = (+12 - 62.6)/(-18) = 2.81

w₂ = (+12 + 62.6)/(-18) = -15.5

But we can not have a negative width, so we can just discard the second solution.

Now let's check the second condition for the maximum, we must have:

V''(2.81) < 0.

V'' = dV'/dw = 2*(-9*w) - 12 = -18*w - 12

V''(2.81) = -18*2.81 - 12 = -62.58 < 0 .

Then the volume is maximized when w = 2.81, and the maximum volume will be:

V(2.81) =  (-3*(2.81)^3 - 6*(2.81)^2 + 105*2.81) = 180.1

 

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