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Complete question :
The Venn diagram relating to the question can be found in the picture attached below :
Answer:
A.) 15 ; b.) 17 ; c.) 20 ; d.) 19 ; e.) 55 ; 67; 76 ;100 ; F.) 369
Step-by-step explanation:
Let :
Cars = C ; Motorcycle = M ; Tricycle = T ; Walking = W
a) How many students ride in a tricycle, motorcycle and car going to their school
Intersection of the 3 modes;
(C n M n T) = 15 ; it is the number which sits in between all the three circles.
B.) How many students ride in both a motorcycle and a tricycle?
(M n T) = 17 ; number in the middle of both circles representing motorcycle and tricycle
C.) How many students ride in both a motorcycle and a car?
(M n C) = 20 ; number in the middle of both circles representing motorcycle and Car
D) How many students ride in both a car and tricycle?
(C n T) = 19 ; number in the middle of both circles representing Car and tricycle
e.)How many students go to school
in a car only = 55
in a motorcycle only = 67
Tricycle only = 76
Walking = 100
F.) How many Grade Seven students of Koronadal National Comprehensive High School are there in all?
(100 + 67 + 76 + 55 + 19 + 20 + 17 + 15) = 369
Answer:
Step-by-step explanation:
The quadratic formula for a equation of form
ax²+bx + c = 0 is
For the first equation,
x²+3x-4=0,
we can match that up with the form
ax²+bx + c = 0
to get that
ax² = x²
divide both sides by x²
a=1
3x = bx
divide both sides by x
3 = b
-4 = c
. We can match this up because no constant multiplied by x could equal x² and no constant multiplied by another constant could equal x, so corresponding terms must match up.
Plugging our values into the equation, we get
as our possible solutions
Plugging our values back into the equation, x²+3x-4=0, we see that both f(-4) and f(1) are equal to 0. Therefore, this has 2 real solutions.
Next, we have
x²+3x+4=0
Matching coefficients up, we can see that a = 1, b=3, and c=4. The quadratic equation is thus
Because √-7 is not a real number, this has no real solutions. However,
(-3 + √-7)/2 and (-3 - √-7)/2 are both possible complex solutions, so this has two complex solutions
Finally, for
4x² + 1= 4x,
we can start by subtracting 4x from both sides to maintain the desired form, resulting in
4x²-4x+1=0
Then, a=4, b=-4, and c=1, making our equation
Plugging 1/2 into 4x²+1=4x, this works as the only solution. This equation has one real solution