We have to see how many times 130 fits into 1494.23:
![1494.23/130=11.4940....](https://tex.z-dn.net/?f=1494.23%2F130%3D11.4940....)
So now we have to multiply this by the amount of time needed for 130 meters:
11.4940*1=11.4940.
So the answer is 11.4940 minutes.
Answer:
y = -
(x - 5)² + 7
Step-by-step explanation:
The equation of a quadratic in vertex form is
y = a(x - h)² + k
where (h, k ) are the coordinates of the vertex and a is a multiplier
Here (h, k ) = (5, 7 ) , then
y = a(x - 5)² + 7
To find a substitute (10, - 3 ) into the equation
- 3 = a(10 - 5)² + 7 ( subtract 7 from both sides )
- 10 = 5²a = 25a ( divide both sides by 25 )
= a , that is
a = - ![\frac{2}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B5%7D)
y = -
(x - 5)² + 7 ← in vertex form
Answer:
Colin has <em>8 sheets </em>left for his third class.
Step-by-step explanation:
Given that:
Total Number of pieces of papers = ![x](https://tex.z-dn.net/?f=x)
Number of pieces of papers used for 1st class = 5 fewer than half of the pieces in the pad
Writing the equation:
![\text{Number of pieces of papers used for 1st class =} \dfrac{x}{2} -5 ...... (1)](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20pieces%20of%20papers%20used%20for%201st%20class%20%3D%7D%20%5Cdfrac%7Bx%7D%7B2%7D%20-5%20......%20%281%29)
Also, Given that number of pieces of papers used for the 2nd class are 2 more than that of papers used in the 1st class.
![\text{Number of pieces of papers used for 2nd class =} \dfrac{x}{2} -5+2 = \dfrac{x}2 -3 ...... (2)](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20pieces%20of%20papers%20used%20for%202nd%20class%20%3D%7D%20%5Cdfrac%7Bx%7D%7B2%7D%20-5%2B2%20%3D%20%5Cdfrac%7Bx%7D2%20-3%20......%20%282%29)
Now, number of pieces of papers left for the third class = Total number of pieces of papers in the pad - Number of pieces of papers used in the first class - Number of pieces of papers used in the first class
![\text{number of pieces of papers left for the third class = }x-(\dfrac{x}{2}-5)-(\dfrac{x}{2}-3)\\\Rightarrow x-\dfrac{x}2-\dfrac{x}2+5+3\\\Rightarrow x-x+5+3\\\Rightarrow 8](https://tex.z-dn.net/?f=%5Ctext%7Bnumber%20of%20pieces%20of%20papers%20left%20for%20the%20third%20class%20%3D%20%7Dx-%28%5Cdfrac%7Bx%7D%7B2%7D-5%29-%28%5Cdfrac%7Bx%7D%7B2%7D-3%29%5C%5C%5CRightarrow%20x-%5Cdfrac%7Bx%7D2-%5Cdfrac%7Bx%7D2%2B5%2B3%5C%5C%5CRightarrow%20x-x%2B5%2B3%5C%5C%5CRightarrow%208)
So, the answer is:
Colin has <em>8</em> <em>sheets </em>left for his third class.
Question 1: 5/8=0.625=62.5% 13/22=about 0.59(repeats 09)=9%
22/30=about 0.73(repeats 3)=73% 12/25=0.45=45%
Store 3 had the greatest percentage of smart phones.
Question 2: 0.775/7.75%=0.775/0.0775=10
For mobile phone 3, it uses o.775 hour of battery life if 7.75% of battery is used.
Answer:
x < -11/6.
Step-by-step explanation:
−12x + 13 > 35
-12x > 35 - 13
-12x > 22
Divide both sides by -12 and invert the inequality sign:
x < -22/12
x < -11/6.