Answer:
B
Step-by-step explanation:
We know that,
Julia can finish a 20-mile bike ride in 1.2 hours.
The distance Julia travels is 20 miles and the time she takes is 1.2 hours.
So, Julia's speed =
= 16.67 mph
Katie can finish the same bike ride in 1.6 hours.
The distance Katie travels is 20 miles and the time she takes is 1.6 hours.
So, Katie's speed =
= 12.5 mph
Now, to find how much faster Julia rides than Katie we subtract Katie's speed from Julia's speed.
So, 16.67 mph - 12.5 mph = 4.17 mph = 4.2 mph (approximately)
Thus, Julia rides 4.2 mph faster than Katie.
Answer:
(arranged from top to bottom)
System #3, where x=6
System #1, where x=4
System #7, where x=3
System #5, where x=2
System #2, where x=1
Step-by-step explanation:
System #1: x=4

To solve, start by isolating your first equation for y.

Now, plug this value of y into your second equation.

System #2: x=1

Isolate your second equation for y.

Plug this value of y into your first equation.

System #3: x=6

Isolate your first equation for y.

Plug this value of y into your second equation.

System #4: all real numbers (not included in your diagram)

Plug your value of y into your second equation.

<em>all real numbers are solutions</em>
System #5: x=2

Isolate your second equation for y.

Plug in your value of y to your first equation.

System #6: no solution (not included in your diagram)

Isolate your first equation for y.

Plug your value of y into your second equation.

<em>no solution</em>
System #7: x=3

Plug your value of y into your second equation.

0.3% is the answer because the decimal is over one meaning the percent is over 100
These 2 equations has no solution and the equations are independent of each other.
<u>Step-by-step explanation:</u>
-10x² -10y² = -300 ----a
5x² + 5y² = 150 ---- b
While trying to solve this,
We can multiply the eq. b by 2 so we will get eq. c and then add to eq. a we will get 0 as the solution.
10x² + 10y² = 300 ----c
-10x² -10y² = -300 ---a
Everything cutoff, we will get 0, and there is no solution to these equations.