17.5
Divide 7 by 8 to get .875 and multiply that by 20 to get your answer.
Answer:
no solutions
Step-by-step explanation:
Hi there!
We're given this system of equations:
x+y=3
4y=-4x-4
and we need to solve it (find the point where the lines intersect, as these are linear equations)
let's solve this system by substitution, where we will set one variable equal to an expression containing the other variable, substitute that expression to solve for the variable the expression contains, and then use the value of the solved variable to find the value of the first variable
we'll use the second equation (4y=-4x-4), as there is already only one variable on one side of the equation. Every number is multiplied by 4, so we'll divide both sides by 4
y=-x-1
now we have y set as an expression containing x
substitute -x-1 as y in x+y=3 to solve for x
x+-x-1=3
combine like terms
-1=3
This statement is untrue, meaning that the lines x+y=3 and 4y=-4x-4 won't intersect.
Therefore the answer is no solutions
Hope this helps! :)
The graph below shows the two equations graphed; they are parallel, which means they will never intersect. If they don't intersect, there's no common solution
To find the rate in miles per hour, divide the miles by the hours.
(1 1/2 miles)/(3/5 hour) =
= 3/2 miles * (5/3 hour)
= 5/2 mph
= 2 1/2 mph
Her walking rate is 2 1/2 mph.
(4 1/2 miles) / ( 2 1/2 mph) =
= 9/2 miles / (5/2 mph)
= 9/2 * 2/5 hours
= 9/5 hours
= 1 4/5 hours
From 9:00 a.m. to 11 a.m., she has 2 hours, but she only needs 1 4/5 hours to walk, so she will make it to work on time.
![\text{Let the product of two natural numbers p and q is 590, and their HCF is 59}\\ \\ \text{we know that the product of LCM and HCF of any two numbers is equal}\\ \text{to the product of the numbers. that is}\\ \\ \text{HCF}\times \text{ LCM}=p\times q\\ \\ \Rightarrow 59 \times \text{LCM}=590\\ \\ \Rightarrow \text{LCM}=\frac{590}{59}\\ \\ \Rightarrow \text{LCM}=10\\ \\ \text{for any two natural numbers, their Least Common Multiple (LCM) is always}](https://tex.z-dn.net/?f=%20%5Ctext%7BLet%20the%20product%20of%20two%20natural%20numbers%20p%20and%20q%20is%20590%2C%20and%20their%20HCF%20is%2059%7D%5C%5C%0A%5C%5C%0A%5Ctext%7Bwe%20know%20that%20the%20product%20of%20LCM%20and%20HCF%20of%20any%20two%20numbers%20is%20equal%7D%5C%5C%0A%5Ctext%7Bto%20the%20product%20of%20the%20numbers.%20that%20is%7D%5C%5C%0A%5C%5C%0A%5Ctext%7BHCF%7D%5Ctimes%20%5Ctext%7B%20LCM%7D%3Dp%5Ctimes%20q%5C%5C%0A%5C%5C%0A%5CRightarrow%2059%20%5Ctimes%20%5Ctext%7BLCM%7D%3D590%5C%5C%0A%5C%5C%0A%5CRightarrow%20%5Ctext%7BLCM%7D%3D%5Cfrac%7B590%7D%7B59%7D%5C%5C%0A%5C%5C%0A%5CRightarrow%20%5Ctext%7BLCM%7D%3D10%5C%5C%0A%5C%5C%0A%5Ctext%7Bfor%20any%20two%20natural%20numbers%2C%20their%20Least%20Common%20Multiple%20%28LCM%29%20is%20always%7D%20)
![\text{greater than their HCF.}\\ \\ \text{but here we can see that }LCM](https://tex.z-dn.net/?f=%20%5Ctext%7Bgreater%20than%20their%20HCF.%7D%5C%5C%0A%5C%5C%0A%5Ctext%7Bbut%20here%20we%20can%20see%20that%20%7DLCM%20%3CHCF%20)
Hence there is no such natural numbers exist.