To find the correct function, just plug in 10 and 2 to see if the function equals zero with both numbers.
f(x)=x^2-12x+20
f(10)=100-120+20
f(10)=0 --> that's what we want, so let's check the other number
f(x)=x^2-12x+20
f(2)=4-24+20
f(2)=0 --> perfect. this function has zeros at both x=10 and x=2
The function that has zeros at both x=10 and x=2 is f(x)=x^2-12x+20.
Answer:
![\huge\boxed{\sqrt[4]{16a^{-12}}=2a^{-3}=\dfrac{2}{a^3}}](https://tex.z-dn.net/?f=%5Chuge%5Cboxed%7B%5Csqrt%5B4%5D%7B16a%5E%7B-12%7D%7D%3D2a%5E%7B-3%7D%3D%5Cdfrac%7B2%7D%7Ba%5E3%7D%7D)
Step-by-step explanation:
![16=2^4\\\\a^{-12}=a^{(-3)(4)}=\left(a^{-3}\right)^4\qquad\text{used}\ (a^n)^m=a^{nm}\\\\\sqrt[4]{16a^{-12}}=\bigg(16a^{-12}\bigg)^\frac{1}{4}\qquad\text{used}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\=\bigg(2^4(a^{-3})^4\bigg)^\frac{1}{4}\qquad\text{use}\ (ab)^n=a^nb^n\\\\=\bigg(2^4\bigg)^\frac{1}{4}\bigg[(a^{-3})^4\bigg]^\frac{1}{4}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=2^{(4)(\frac{1}{4})}(a^{-3})^{(4)(\frac{1}{4})}=2^1(a^{-3})^1=2a^{-3}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}](https://tex.z-dn.net/?f=16%3D2%5E4%5C%5C%5C%5Ca%5E%7B-12%7D%3Da%5E%7B%28-3%29%284%29%7D%3D%5Cleft%28a%5E%7B-3%7D%5Cright%29%5E4%5Cqquad%5Ctext%7Bused%7D%5C%20%28a%5En%29%5Em%3Da%5E%7Bnm%7D%5C%5C%5C%5C%5Csqrt%5B4%5D%7B16a%5E%7B-12%7D%7D%3D%5Cbigg%2816a%5E%7B-12%7D%5Cbigg%29%5E%5Cfrac%7B1%7D%7B4%7D%5Cqquad%5Ctext%7Bused%7D%5C%20a%5E%5Cfrac%7B1%7D%7Bn%7D%3D%5Csqrt%5Bn%5D%7Ba%7D%5C%5C%5C%5C%3D%5Cbigg%282%5E4%28a%5E%7B-3%7D%29%5E4%5Cbigg%29%5E%5Cfrac%7B1%7D%7B4%7D%5Cqquad%5Ctext%7Buse%7D%5C%20%28ab%29%5En%3Da%5Enb%5En%5C%5C%5C%5C%3D%5Cbigg%282%5E4%5Cbigg%29%5E%5Cfrac%7B1%7D%7B4%7D%5Cbigg%5B%28a%5E%7B-3%7D%29%5E4%5Cbigg%5D%5E%5Cfrac%7B1%7D%7B4%7D%5Cqquad%5Ctext%7Buse%7D%5C%20%28a%5En%29%5Em%3Da%5E%7Bnm%7D%5C%5C%5C%5C%3D2%5E%7B%284%29%28%5Cfrac%7B1%7D%7B4%7D%29%7D%28a%5E%7B-3%7D%29%5E%7B%284%29%28%5Cfrac%7B1%7D%7B4%7D%29%7D%3D2%5E1%28a%5E%7B-3%7D%29%5E1%3D2a%5E%7B-3%7D%5Cqquad%5Ctext%7Buse%7D%5C%20a%5E%7B-n%7D%3D%5Cdfrac%7B1%7D%7Ba%5En%7D)
Answer:
-7, -27
Step-by-step explanation:
J is at -17. K is 10 units from J.
Start at point J, -17:
a) If you go 10 units to the right, you end up at -17 + 10 = -7
b) If you go 10 units to the left, you end up at -17 - 10 = -27
J can be at -7 or at -27
Answer: -7, -27
Answer:
small: 6.25
large: 18.25
Step-by-step explanation:
Set the large boxes as x and the small boxes as y (The weight is the constant).
3x+5y=86
6x+2y=122
Now just solve the systems of equations using either substitution or elimination.
-6x-10y=-172 multiply the whole equation by -2 to eliminate
6x+2y=122, add the equations
-8y=-50
y= 6.25
Plug in y either equation:
3x+5(6.25)=86
3x=86-31.25
3x=54.75
x=18.25
Each small box weighs 6.25 kilograms, each large box weighs 18.25 kilograms, to check plug in the values to the other equation to see if they add up correctly.
