We know that 5 days and 12 hours is as much as 5.5 days, so all we need to do is to multiply 1800 watt-hours by 5.5 days:
1800 * 5.5 = 9900
Answer: <u>A light bulb consumes 9900 watt-hours each 5 days and 12 hours</u>.
Answer:
Step-by-step explanation:
Slope= -3/1
plot from point (-2,4)
Answer:
30%
Step-by-step explanation:
<h2><u>Percentage change </u></h2><h3>formula :</h3>
= 50 - 35 = 15
= 
<h3>=
30 %</h3>
let's firstly change the 1.2 to a fraction
![1.\underline{2}\implies \cfrac{12}{1\underline{0}}\implies \cfrac{6}{5} \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{10}~,~\stackrel{y_1}{6})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{\frac{6}{5}}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{\frac{6}{5}}-\stackrel{y1}{6}}}{\underset{run} {\underset{x_2}{4}-\underset{x_1}{10}}}\implies \cfrac{~~ \frac{6-30}{5}~~}{-6}\implies \cfrac{~~ \frac{-24}{5}~~}{-6}\implies \cfrac{~~ -\frac{24}{5}~~}{-\frac{6}{1}}](https://tex.z-dn.net/?f=1.%5Cunderline%7B2%7D%5Cimplies%20%5Ccfrac%7B12%7D%7B1%5Cunderline%7B0%7D%7D%5Cimplies%20%5Ccfrac%7B6%7D%7B5%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B10%7D~%2C~%5Cstackrel%7By_1%7D%7B6%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B4%7D~%2C~%5Cstackrel%7By_2%7D%7B%5Cfrac%7B6%7D%7B5%7D%7D%29%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7Bslope%7D%7Bm%7D%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%20%7B%5Cstackrel%7By_2%7D%7B%5Cfrac%7B6%7D%7B5%7D%7D-%5Cstackrel%7By1%7D%7B6%7D%7D%7D%7B%5Cunderset%7Brun%7D%20%7B%5Cunderset%7Bx_2%7D%7B4%7D-%5Cunderset%7Bx_1%7D%7B10%7D%7D%7D%5Cimplies%20%5Ccfrac%7B~~%20%5Cfrac%7B6-30%7D%7B5%7D~~%7D%7B-6%7D%5Cimplies%20%5Ccfrac%7B~~%20%5Cfrac%7B-24%7D%7B5%7D~~%7D%7B-6%7D%5Cimplies%20%5Ccfrac%7B~~%20-%5Cfrac%7B24%7D%7B5%7D~~%7D%7B-%5Cfrac%7B6%7D%7B1%7D%7D)
![-\cfrac{\stackrel{4}{~~\begin{matrix} 24 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}{5}\cdot -\cfrac{1}{\underset{1}{~~\begin{matrix} 6 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}\implies \boxed{\cfrac{4}{5}}](https://tex.z-dn.net/?f=-%5Ccfrac%7B%5Cstackrel%7B4%7D%7B~~%5Cbegin%7Bmatrix%7D%2024%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%7D%7D%7B5%7D%5Ccdot%20-%5Ccfrac%7B1%7D%7B%5Cunderset%7B1%7D%7B~~%5Cbegin%7Bmatrix%7D%206%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%7D%7D%5Cimplies%20%5Cboxed%7B%5Ccfrac%7B4%7D%7B5%7D%7D)
Call (F) the age of the father and (J) the age of Julio
The F & J are related in this way: F=4J
Now you have a restriction in the form of inequality: The sum of both ages has to be greater or equal than 55.
Algebraically that is: F + J ≥ 55
You can substitute F with 4J to find the solution for J:
4J + J ≥ 55
5J ≥ 55
Now divide both sides by 5
5J/5 ≥ 55/5
J ≥ 11
That Imposes a lower boundary for the value of J of 11, meaning that the youngest age of Julio can be 11
