Answer:
27 and 29.
Step-by-step explanation:
Let the numbers be n and n + 2
2(n + 2) + n = 85
3n + 4 = 85
3n = 81
n = 27.
Answer:
43 truffles
Step-by-step explanation:
First add all of the weights together
5.4+6.1+5.7=17.2
Divide the total chocolate by the weight of a truffle
17.2/0.4=43
Answer:
The vertex: ![(\frac{3}{4},-\frac{41}{4} )](https://tex.z-dn.net/?f=%28%5Cfrac%7B3%7D%7B4%7D%2C-%5Cfrac%7B41%7D%7B4%7D%20%29)
The vertical intercept is: ![y=-8](https://tex.z-dn.net/?f=y%3D-8)
The coordinates of the two intercepts of the parabola are
and ![(\frac{3-\sqrt{41} }{4} , 0)](https://tex.z-dn.net/?f=%28%5Cfrac%7B3-%5Csqrt%7B41%7D%20%7D%7B4%7D%20%2C%200%29)
Step-by-step explanation:
To find the vertex of the parabola
you need to:
1. Find the coefficients <em>a</em>, <em>b</em>, and <em>c </em>of the parabola equation
<em>
</em>
2. You can apply this formula to find x-coordinate of the vertex
, so
![x=-\frac{-6}{2\cdot 4}\\x=\frac{3}{4}](https://tex.z-dn.net/?f=x%3D-%5Cfrac%7B-6%7D%7B2%5Ccdot%204%7D%5C%5Cx%3D%5Cfrac%7B3%7D%7B4%7D)
3. To find the y-coordinate of the vertex you use the parabola equation and x-coordinate of the vertex (
)
![f(-\frac{b}{2a})=a(-\frac{b}{2a})^2+b(-\frac{b}{2a})+c\\f(\frac{3}{4})=4\cdot (\frac{3}{4})^2-6\cdot (\frac{3}{4})-8\\y=\frac{-41}{4}](https://tex.z-dn.net/?f=f%28-%5Cfrac%7Bb%7D%7B2a%7D%29%3Da%28-%5Cfrac%7Bb%7D%7B2a%7D%29%5E2%2Bb%28-%5Cfrac%7Bb%7D%7B2a%7D%29%2Bc%5C%5Cf%28%5Cfrac%7B3%7D%7B4%7D%29%3D4%5Ccdot%20%28%5Cfrac%7B3%7D%7B4%7D%29%5E2-6%5Ccdot%20%28%5Cfrac%7B3%7D%7B4%7D%29-8%5C%5Cy%3D%5Cfrac%7B-41%7D%7B4%7D)
To find the vertical intercept you need to evaluate x = 0 into the parabola equation
![f(x)=4x^2-6x-8\\f(0)=4(0)^2-6\cdot 0-0\\f(0)=-8](https://tex.z-dn.net/?f=f%28x%29%3D4x%5E2-6x-8%5C%5Cf%280%29%3D4%280%29%5E2-6%5Ccdot%200-0%5C%5Cf%280%29%3D-8)
To find the coordinates of the two intercepts of the parabola you need to solve the parabola by completing the square
![\mathrm{Add\:}8\mathrm{\:to\:both\:sides}](https://tex.z-dn.net/?f=%5Cmathrm%7BAdd%5C%3A%7D8%5Cmathrm%7B%5C%3Ato%5C%3Aboth%5C%3Asides%7D)
![x^2-6x-8+8=0+8](https://tex.z-dn.net/?f=x%5E2-6x-8%2B8%3D0%2B8)
![\mathrm{Simplify}](https://tex.z-dn.net/?f=%5Cmathrm%7BSimplify%7D)
![4x^2-6x=8](https://tex.z-dn.net/?f=4x%5E2-6x%3D8)
![\mathrm{Divide\:both\:sides\:by\:}4](https://tex.z-dn.net/?f=%5Cmathrm%7BDivide%5C%3Aboth%5C%3Asides%5C%3Aby%5C%3A%7D4)
![\frac{4x^2-6x}{4}=\frac{8}{4}\\x^2-\frac{3x}{2}=2](https://tex.z-dn.net/?f=%5Cfrac%7B4x%5E2-6x%7D%7B4%7D%3D%5Cfrac%7B8%7D%7B4%7D%5C%5Cx%5E2-%5Cfrac%7B3x%7D%7B2%7D%3D2)
![\mathrm{Write\:equation\:in\:the\:form:\:\:}x^2+2ax+a^2=\left(x+a\right)^2](https://tex.z-dn.net/?f=%5Cmathrm%7BWrite%5C%3Aequation%5C%3Ain%5C%3Athe%5C%3Aform%3A%5C%3A%5C%3A%7Dx%5E2%2B2ax%2Ba%5E2%3D%5Cleft%28x%2Ba%5Cright%29%5E2)
![x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=2+\left(-\frac{3}{4}\right)^2\\x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=\frac{41}{16}](https://tex.z-dn.net/?f=x%5E2-%5Cfrac%7B3x%7D%7B2%7D%2B%5Cleft%28-%5Cfrac%7B3%7D%7B4%7D%5Cright%29%5E2%3D2%2B%5Cleft%28-%5Cfrac%7B3%7D%7B4%7D%5Cright%29%5E2%5C%5Cx%5E2-%5Cfrac%7B3x%7D%7B2%7D%2B%5Cleft%28-%5Cfrac%7B3%7D%7B4%7D%5Cright%29%5E2%3D%5Cfrac%7B41%7D%7B16%7D)
![\left(x-\frac{3}{4}\right)^2=\frac{41}{16}](https://tex.z-dn.net/?f=%5Cleft%28x-%5Cfrac%7B3%7D%7B4%7D%5Cright%29%5E2%3D%5Cfrac%7B41%7D%7B16%7D)
![\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}](https://tex.z-dn.net/?f=%5Cmathrm%7BFor%5C%3A%7Df%5E2%5Cleft%28x%5Cright%29%3Da%5Cmathrm%7B%5C%3Athe%5C%3Asolutions%5C%3Aare%5C%3A%7Df%5Cleft%28x%5Cright%29%3D%5Csqrt%7Ba%7D%2C%5C%3A-%5Csqrt%7Ba%7D)
![x_1=\frac{\sqrt{41}+3}{4},\:x_2=\frac{-\sqrt{41}+3}{4}](https://tex.z-dn.net/?f=x_1%3D%5Cfrac%7B%5Csqrt%7B41%7D%2B3%7D%7B4%7D%2C%5C%3Ax_2%3D%5Cfrac%7B-%5Csqrt%7B41%7D%2B3%7D%7B4%7D)
Answer:
The limit of the function does not exists.
Step-by-step explanation:
From the graph it is noticed that the value of the function is 6 from all values of x which are less than 2. At x=2, the line y=6 has open circle. It means x=2 is not included.
For x<2
![f(x)=6](https://tex.z-dn.net/?f=f%28x%29%3D6)
The value of the function is -3 from all values of x which are greater than 2. At x=2, the line y=-3 has open circle. It means x=2 is not included.
For x>2
![f(x)=-3](https://tex.z-dn.net/?f=f%28x%29%3D-3)
The value of y is 1 at x=2, because of he close circles on (2,1).
For x=2
![f(x)=1](https://tex.z-dn.net/?f=f%28x%29%3D1)
Therefore the graph represents a piecewise function, which is defined as
![f(x)=\begin{cases}6& \text{ if } x2 \end{cases}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cbegin%7Bcases%7D6%26%20%5Ctext%7B%20if%20%7D%20x%3C2%5C%5C%201%26%20%5Ctext%7B%20if%20%7D%20x%3D2%20%5C%5C%20-3%26%20%5Ctext%7B%20if%20%7D%20x%3E2%20%5Cend%7Bcases%7D)
The limit of a function exist at a point a if the left hand limit and right hand limit are equal.
![lim_{x\rightarrow a^-}f(x)=lim_{x\rightarrow a^+}f(x)](https://tex.z-dn.net/?f=lim_%7Bx%5Crightarrow%20a%5E-%7Df%28x%29%3Dlim_%7Bx%5Crightarrow%20a%5E%2B%7Df%28x%29)
The function is broken at x=2, therefore we have to find the left and right hand limit at x=2.
![lim_{x\rightarrow 2^-}f(x)=6](https://tex.z-dn.net/?f=lim_%7Bx%5Crightarrow%202%5E-%7Df%28x%29%3D6)
![lim_{x\rightarrow 2^+}f(x)=-3](https://tex.z-dn.net/?f=lim_%7Bx%5Crightarrow%202%5E%2B%7Df%28x%29%3D-3)
![6\neq-3](https://tex.z-dn.net/?f=6%5Cneq-3)
Since the left hand limit and right hand limit are not equal therefore the limit of the function does not exists.