The standard form of a parabola is y=ax²+bx+c
use the three given points to find the three unknown constants a, b, and c:
-2=a+b+c............1
-2=4a+2b+c......... 2
-4=9a+3b+c...........3
equation 2 minus equation 1: 3a+b=0..........4
equation 3 minus equation 2: 5a+b=-2.........5
equation 5 minus equation 4: 2a=-2, so a=-1
plug a=-1 in equation 4: -3+b=0, so b=3
Plug a=-1, b=3 in equation 1: -2=-1+3+c, so c=-4
the parabola is y=-x²+3x-4
double check: when x=1, y=-1+3-4=-2
when x=2, y=-4+6-4=-2
when x=3, y=-9+9-4=-4
Yes.
The solution to this system set is: "x = 4" , "y = 0" ; or write as: [4, 0] .
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Given:
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y = - 4x + 16 ;
4y − x + 4 = 0 ;
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"Solve the system using substitution" .
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First, let us simplify the second equation given, to get rid of the "0" ;
→ 4y − x + 4 = 0 ;
Subtract "4" from each side of the equation ;
→ 4y − x + 4 − 4 = 0 − 4 ;
→ 4y − x = -4 ;
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So, we can now rewrite the two (2) equations in the given system:
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y = - 4x + 16 ; ===> Refer to this as "Equation 1" ;
4y − x = -4 ; ===> Refer to this as "Equation 2" ;
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Solve for "x" and "y" ; using "substitution" :
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We are given, as "Equation 1" ;
→ " y = - 4x + 16 " ;
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→ Plug in this value for [all of] the value[s] for "y" into {"Equation 2"} ;
to solve for "x" ; as follows:
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Note: "Equation 2" :
→ " 4y − x = - 4 " ;
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Substitute the value for "y" {i.e., the value provided for "y"; in "Equation 1}" ;
for into the this [rewritten version of] "Equation 2" ;
→ and "rewrite the equation" ;
→ as follows:
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→ " 4 (-4x + 16) − x = -4 " ;
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Note the "distributive property" of multiplication :
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a(b + c) = ab + ac ; AND:
a(b − c) = ab <span>− ac .
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As such:
We have:
</span>
→ " 4 (-4x + 16) − x = - 4 " ;
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AND:
→ "4 (-4x + 16) " = (4* -4x) + (4 *16) = " -16x + 64 " ;
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Now, we can write the entire equation:
→ " -16x + 64 − x = - 4 " ;
Note: " - 16x − x = -16x − 1x = -17x " ;
→ " -17x + 64 = - 4 " ; Solve for "x" ;
Subtract "64" from EACH SIDE of the equation:
→ " -17x + 64 − 64 = - 4 − 64 " ;
to get:
→ " -17x = -68 " ;
Divide EACH side of the equation by "-17" ;
to isolate "x" on one side of the equation; & to solve for "x" ;
→ -17x / -17 = -68/ -17 ;
to get:
→ x = 4 ;
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Now, Plug this value for "x" ; into "{Equation 1"} ;
which is: " y = -4x + 16" ; to solve for "y".
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→ y = -4(4) + 16 ;
= -16 + 16 ;
→ y = 0 .
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The solution to this system set is: "x = 4" , "y = 0" ; or write as: [4, 0] .
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Now, let us check our answers—as directed in this very question itself ;
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→ Given the TWO (2) originally given equations in the system of equation; as they were originally rewitten;
→ Let us check;
→ For EACH of these 2 (TWO) equations; do these two equations hold true {i.e. do EACH SIDE of these equations have equal values on each side} ; when we "plug in" our obtained values of "4" (for "x") ; and "0" for "y" ??? ;
→ Consider the first equation given in our problem, as originally written in the system of equations:
→ " y = - 4x + 16 " ;
→ Substitute: "4" for "x" and "0" for "y" ; When done, are both sides equal?
→ "0 = ? -4(4) + 16 " ?? ; → "0 = ? -16 + 16 ?? " ; → Yes! ;
{Actually, that is how we obtained our value for "y" initially.}.
→ Now, let us check the other equation given—as originally written in this very question:
→ " 4y − x + 4 = ?? 0 ??? " ;
→ Let us "plug in" our obtained values into the equation;
{that is: "4" for the "x-value" ; & "0" for the "y-value" ;
→ to see if the "other side of the equation" {i.e., the "right-hand side"} holds true {i.e., in the case of this very equation—is equal to "0".}.
→ " 4(0) − 4 + 4 = ? 0 ?? " ;
→ " 0 − 4 + 4 = ? 0 ?? " ;
→ " - 4 + 4 = ? 0 ?? " ; Yes!
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→ As such, from "checking [our] answer (obtained values)" , we can be reasonably certain that our answer [obtained values] :
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→ "x = 4" and "y = 0" ; or; write as: [0, 4] ; are correct.
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Hope this lenghty explanation is of help! Best wishes!
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I don't exactly know what a net is, but I can figure out the surface area (maybe). When you put this pattern together, you have to make a couple of assumptions.
First of all the 2 triangles are connected to a base that's 8 by 13.
Base
The area of the base = 8 * 13 = 104 square feet.
Two Triangles.
The area of any triangle is
Area = 1/2 b * h
Area = 1/2 *12 * 5
Area = 30 square feet. There are 2 triangles so ...
The total area of both is 60 square feet.
Overlap
Now we come to the tough part. how big is it from the first dotted line to the second one?
The answer is 5. But why? Well the 5 comes from the the length of the side that has to be covered by this pattern. The 5 is part of the triangle.
How far across is the dotted line? The answer is 8. Why again? That's the width of the whole prism or pattern.
So the area between the two dotted lines is 5 * 8 = 40
Finally the area over the 12.
The question is how far is it from the second dotted line to the last boundary? The answer is 12. And it's width is 8
So the area = 12 * 8 = 96
Total area
96 + 60 + 40 + 104 = 300 square feet.
Answer:
38.5
Step-by-step explanation:
First do 77/2 then you get 38.5, if anything is wrong please tell me!!
