<h3>
Answer: Perpendicular</h3>
The slopes -2/3 and 3/2 multiply to -1, which is sufficient info that the lines are perpendicular. Perpendicular slopes always multiply to -1 assuming neither line is vertical nor horizontal.
Note how -2/3 is the negative reciprocal of 3/2, and vice versa.
Answer:
It will cost $36.38 to travel 8.9 miles.
Step-by-step explanation:
In order to get how much it will cost to travel 8.9 miles, you need to first set up a linear equation (in slope-intercept form, which is
). You can do this with what the problem has given us. Since there's a flat fee of $3, that is the y-intercept or the
slope-intercept form. Then there is $3.75 added for every additional mile traveled, making it the slope or the
in slope-intercept form. That makes the equation look like:

The
in that equation represents how far you traveled. Therefore in order to find out 8.9 miles, plug in 8.9 for
.

Simply solve the equation and you will have how much it will cost to travel 8.9 miles.

Now add the like terms:

Since it needs to be rounded to the nearest cent, you will round to the nearest tenth:

Therefore, it will cost $36.38 to travel 8.9 miles.
Answer:
10.51
Always look to the right of the place you are rounding
so...
if you look at 10.507140429
the 0 after 5 is in the hundredths place
look to the right of 0 and you'll see a 7.
If the number to the right of the place you are supposed to round to is:
5 or greater - round up
less than 5 - round down
Hope this helped
<h3>
Answer:</h3>
- <u>20</u> kg of 20%
- <u>80</u> kg of 60%
<h3>
Step-by-step explanation:</h3>
I like to use a little X diagram to work mixture problems like this. The constituent concentrations are on the left; the desired mix is in the middle, and the right legs of the X show the differences along the diagonal. These are the ratio numbers for the constituents. Reducing the ratio 32:8 gives 4:1, which totals 5 "ratio units". We need a total of 100 kg of alloy, so each "ratio unit" stands for 100 kg/5 = 20 kg of constituent.
That is, we need 80 kg of 60% alloy and 20 kg of 20% alloy for the product.
_____
<em>Using an equation</em>
If you want to write an equation for the amount of contributing alloy, it works best to let a variable represent the quantity of the highest-concentration contributor, the 60% alloy. Using x for the quantity of that (in kg), the amount of copper in the final alloy is ...
... 0.60x + 0.20(100 -x) = 0.52·100
... 0.40x = 32 . . . . . . . . . . .collect terms, subtract 20
... x = 32/0.40 = 80 . . . . . kg of 60% alloy
... (100 -80) = 20 . . . . . . . .kg of 20% alloy