If the radius of the circle is 3 in, what is the area of the circle?
1 answer:
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Answer:
Number of restaurant-purchased meals eaten in a restaurant = 70
Number of restaurant-purchased meals eaten in a car = 22
Number of restaurant-purchased meals eaten in a home = 60
Step-by-step explanation:
Let the number of people that will eat in a restaurant be R.
Let the number of people that will eat in a car be C.
Let the number of people that will eat in a home be H.
From the information given in the problem we have,
R+C+H = 152 ____equation (1)
C+H-R = 12 ____equation (2)
R = H+10 ____equation (3)
1) Plugging in R=H+10 from equation 3 into the equation 2, we get
C+H-R=12
=> C+H-(H+10)=12
=> C+H-H-10=12
2) Cancelling out +H and -H, we get
C-10=12
3) Add 10 to both sides
C-10+10=12+10
4) Cancelling out -10 and +10, we get
C=22
5) Plugin C=22 in equation 1, we have
R+C+H = 152
=> R+22+H=152
Subtract 22 from both sides,
R+22+H-22=152-22
Cancelling out +22 and -22 from the left side, we get
R+H=130 ____let it be equation (4)
6) Plugin C=22 in equation 2, we have
C+H-R = 12
22+H-R = 12
Subtracting 22 from both the sides, we get
22+H-R-22 = 12-22
Cancelling out +22 and -22 from the left side,
H-R = -10 ____let it be equation (5)
7) Adding equation 4 and equation 5, we get
(R+H)+(H-R) = 130 + (-10)
=> R+H+H-R = 130-10
8) Cancelling out R and -R from the left side, we get
2H = 120
9) Dividing both sides by 2, we get
10) Cancelling out the 2's from the left side, we have
H=60
11) Plugging in C=22 and H=60 in the equation 1, we have
R+C+H = 152
=> R+22+60=152
=> R + 82 = 152
12) Subtracting 82 from both the sides, we get
R + 82 -82 = 152 -82
13) Cancelling out +82 and -82 from the left side, we get
R = 70
<em>So, C=22, H=60, R=70</em>
Answer:
1) Real and same.
2) Real and distinct
3) Real and distinct
4) Real and distinct
5) Real and distinct
6) Real and distinct
7) Real and distinct
8) Real and distinct
Step-by-step explanation:
If a quadratic equation is in the form of ax² + bx + c = 0, then Discriminant of the equation D = b² - 4ac, which governs the nature of roots the equation has.
If D > 0, then there will be two different real roots.
If D = 0, then two same and real roots.
If D < 0, then two distinct but imaginary roots.
Now, 1) x² + 6x + 9 = 0 has D = 6² - 4 × 9 × 1 = 0. So, there will be two same and real roots.
2) 5x² - x = 4x² + 2
⇒ x² - x - 2 = 0
It has D = (-1)² - 4 × 1 × (-2) = 9. Therefore, the roots will be two distinct and real.
3)
⇒ 5 = x² - 4x
⇒ x² - 4x - 5 = 0.
So, D = (-4)² - 4 × (1) × (- 5) = 36. So, the equation has two real and distinct rools.
4) x(x - 5) = 4(5 - x)
⇒ x² - x - 20 = 0
Hence, D = (-1)² - 4 × 1 × (-20) = 81
So, the roots will be real and distinct.
5) x² + 7x + 1 = 0 has D = 7² - 4 × 1 × 1 = 45
So, the roots will be real and distinct.
6) 2x² + 9x + 3 = 0, has D = 9² - 4 × 2 × 3 = 57.
Hence, the roots will be real and distinct.
7) 5x² - 6 = 13x
⇒ 5x² - 13x - 6 = 0
So, D = (-13)² - 4 × 5 × (-6) = 289
So, the roots will be real and distinct.
8) x² - x = 3(x + 7)
⇒ x² - 4x - 21 = 0
It has D = (-4)² - 4 × 1 × (-21) = 100
So, the roots will be real and distinct. (Answer)