<span>$y=ax^3+bx^2+cx+d$ y = a x 3+ b x 2+ c x + d .</span>
Based on your problem, you only have one given. So, you can't make an equation for this because there are not limits to the equation. The only thing that you know is that the three numbers are consecutive even integers. My way of solution for this is trial-and-error. However, it's really quite easy.
For example: 42 + 46 = 88. I have to increase the numbers more to reach 136. Suppose: 82 + 86 = 168. That exceeded 136. So, it must be between 46 and 82. Suppose again: 66 + 70 = 136. Therefore, the sequence of the consecutive even integers are 66, 68, and 70.
Answer:
A≈78.54
Step-by-step explanation:
A¹=7.7×6.6=50.82(area of rectangle inside)
16.1-7.7=8.4/2=4.2
A²=6.6×4.2=27.72(area of triangles)
A=50.82+27.72=78.54
Answer:
A.) gf(x) = 3x^2 + 12x + 9
B.) g'(x) = 2
Step-by-step explanation:
A.) The two given functions are:
f(x) = (x + 2)^2 and g(x) = 3(x - 1)
Open the bracket of the two functions
f(x) = (x + 2)^2
f(x) = x^2 + 2x + 2x + 4
f(x) = x^2 + 4x + 4
and
g(x) = 3(x - 1)
g(x) = 3x - 3
To find gf(x), substitute f(x) for x in g(x)
gf(x) = 3( x^2 + 4x + 4 ) - 3
gf(x) = 3x^2 + 12x + 12 - 3
gf(x) = 3x^2 + 12x + 9
Where
a = 3, b = 12, c = 9
B.) To find g '(12), you must first find the inverse function of g(x) that is g'(x)
To find g'(x), let g(x) be equal to y. Then, interchange y and x for each other and make y the subject of formula
Y = 3x + 3
X = 3y + 3
Make y the subject of formula
3y = x - 3
Y = x/3 - 3/3
Y = x/3 - 1
Therefore, g'(x) = x/3 - 1
For g'(12), substitute 12 for x in g' (x)
g'(x) = 12/4 - 1
g'(x) = 3 - 1
g'(x) = 2.
The circumference will be 18.683
