She would have 6 or 7 lessons 130 divided by 20 = 6.5
X < = 1 or x > 8
closed circle on 1, shaded to the left
open circle on 8, shaded to the right
If he wants to loose 30 points total, subtract what he lost already, 24, from 30. He needs to lose 6 more pounds.
Answer:
System A has 4 real solutions.
System B has 0 real solutions.
System C has 2 real solutions
Step-by-step explanation:
System A:
x^2 + y^2 = 17 eq(1)
y = -1/2x eq(2)
Putting value of y in eq(1)
x^2 +(-1/2x)^2 = 17
x^2 + 1/4x^2 = 17
5x^2/4 -17 =0
Using quadratic formula:

a = 5/4, b =0 and c = -17

Finding value of y:
y = -1/2x


System A has 4 real solutions.
System B
y = x^2 -7x + 10 eq(1)
y = -6x + 5 eq(2)
Putting value of y of eq(2) in eq(1)
-6x + 5 = x^2 -7x + 10
=> x^2 -7x +6x +10 -5 = 0
x^2 -x +5 = 0
Using quadratic formula:

a= 1, b =-1 and c =5

Finding value of y:
y = -6x + 5
y = -6(\frac{1\pm\sqrt{19}i}{2})+5
Since terms containing i are complex numbers, so System B has no real solutions.
System B has 0 real solutions.
System C
y = -2x^2 + 9 eq(1)
8x - y = -17 eq(2)
Putting value of y in eq(2)
8x - (-2x^2+9) = -17
8x +2x^2-9 +17 = 0
2x^2 + 8x + 8 = 0
2x^2 +4x + 4x + 8 = 0
2x (x+2) +4 (x+2) = 0
(x+2)(2x+4) =0
x+2 = 0 and 2x + 4 =0
x = -2 and 2x = -4
x =-2 and x = -2
So, x = -2
Now, finding value of y:
8x - y = -17
8(-2) - y = -17
-16 -y = -17
-y = -17 + 16
-y = -1
y = 1
So, x= -2 and y = 1
System C has 2 real solutions
Answer:
A) 2 miles, 5 miles, 8 miles
Step-by-step explanation:
Let's get some details cleared.
We're interested in finding how many miles he walked at the end of each hour in a period of 3 hours.
We know what is, so we can use that as our starting point.
Your answer is A.