So
y=ax^2+bx+c
(x,y)
sub the points and solve
(4.28,6.48)
6.48=a(4.28)^2+b(4.28)+c
(12.61,15.04)
15.04=a(12.61)^2+b(12.61)+c
well, for 3 variables, we need equations and therefor 3 points
maybe we are supposed to assume it starts at (0,0)
so then
0=a(0)^2+b(0)+c
0=c
so then
6.48=a(4.28)^2+b(4.28)
15.04=a(12.61)^2+b(12.61)
solve for a by subsitution
first equation, minut a(4.28)^2 from both sides
6.48-a(4.28)^2=b(4.28)
divide both sides by 4.28
(6.48/4.28)-4.28a=b
sub that for b in other equation
15.04=a(12.61)^2+b(12.61)
15.04=a(12.61)^2+((6.48/4.28)-4.28a)(12.61)
expand
15.04 =a(12.61)^2+(81.7128/4.28)-53.9708a
minus (81.7128/4.28) both sides
15.04-(81.7128/4.28)=a(12.61)^2-53.9708a
15.04-(81.7128/4.28)=a((12.61)^2-53.9708)
(15.04-(81.7128/4.28))/(((12.61)^2-53.9708))=a
that's the exact value of a
to find b, subsitute to get
(6.48/4.28)-4.28((15.04-(81.7128/4.28))/(((12.61)^2-53.9708)))=b
if we aprox
a≈-0.038573167896199
b≈1.6791118501845
so then the equation is
y=-0.038573167896199x²+1.6791118501845x
The length of the brace required is 4.3m
What is sine rule?
In a ΔABC a, b and c are the sides and A, B and C are angles then,

We can find the length, l as shown below:
Let AB=3m, BC=2m and AC=l
Let ∠A=25°
So, in ΔABC




∠C=39.34°
∠A+∠B+∠C=180°
∠B=180°-25°-39.34°
∠B=115.66°



l=4.2659
Rounding to nearest tenth of meter.
l=4.3m
Hence, the length of the brace required is 4.3m
Learn more about Sine Rule here:
brainly.com/question/25852087
#SPJ1
Parallelogram have 4 sides.
Use the distributive property. 2 multiply the last and negative 4 in the parentheses.
2l + 2(l - 4) = 72 m
Combine Like terms
2l + 2l - 8 = 72 m
4l - 8 = 72 m
Get l all by itself by adding 8 to both sides
4l - 8 = 72 m
+8. +8
Divide both sides by 4
4l = 80
4. 4
l= 20
The width of the parallelogram is four meters less than its length.
4 sides
2 length = 20
2 width = 16
I got 16 by subtracting 4 from 20
Comment if you need any other help :)
Answer:
True. That is a function.
Step-by-step explanation:
A function just means that every input has only one output. The simpler way to do these is the vertical line test. Draw a vertical line anywhere on the graph and see if the vertical line is intersected in two places. If the vertical line is crossed twice, the graph isn't a function.