The best buy is the in. I really hope this helped
Answer:
9,11,16. 4,6,7. 6,8,12
Right. Acute. obtuse
Step-by-step explanation:
First substitute y=5x+7 so that would turn into 5x+7=6x
Then, isolate/solve for x for 5x + 7 = 6x, in this problem x=7
Substitute x=7 into y=5x+7
y=42
x=7, y=42
(7,42)
Answer:
f(x) = 4x^2 + 2x - 4.
Step-by-step explanation:
Let the quadratic function be y = f(x) = ax^2 + bx + c.
For the point (-2, 8) ( x = -2 when y = 8) we have:
a(-2)^2 + (-2)b + c = 8
4a - 2b + c = 8 For (0, -4) we have:
0 + 0 + c = -4 so c = -4. For (4, 68) we have:
16a + 4b + c = 68
So we have 2 systems of equations in a and b ( plugging in c = -4):
4a - 2b - 4 = 8
16a + 4b - 4 = 68
4a - 2b = 12
16a + 4b = 72 Multiplying 4a - 2b = 12 by 2 we get:
8a - 4b = 24
Adding the last 2 equations:
24a = 96
a = 4
Now plugging a = 4 and c = -4 in the first equation:
4(4) - 2b - 4 = 8
-2b = 8 - 16 + 4 = -4
b = 2.
There are 10,000 total four-digit numbers (1000 through 9999).
Multiples of 2 end in 0, 2, 4, 6, and 8. There are 9*10*10*5 = 4500 four-digit multiples of 2.
Multiples of 5 end in 0 or 5. There are 9*10*10*2 = 1800 four-digit multiples of 5.
There is redundancy between the two sets of numbers, namely those that end in 0, which are both multiples of 2 and 5. There are 9*10*10*1 = 900 four-digit multiples of both 2 and 5.
Then there are 4500 + 1800 - 900 = 5400 total four-digit numbers that are either multiples of 2 or 5, which means the remaining 4600 numbers are neither multiples of 2 nor 5.
