The value of x in the equation is 26.7
<h3>How to solve an equation?</h3>
0.75x + 1.5y = 40
−1.5x − 1.5y = −60
add both equation to eliminate y .
Therefore,
−1.5x + 0.75x = -0.75x
1.5y + (- 1.5y) = 0
40 + (-60) = -20
Hence,
-0.75x = - 20
divide both sides by -0.75
-0.75x / -0/75 = - 20 / -0.75
x = - 20 / -0.75
x = 26.6666666667
x = 26.7
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Answer:
2 seconds
Step-by-step explanation:
Given
h = - 32t² + 64t
When the frisbee lands the height h = 0, that is
- 32t² + 64t = 0 ( multiply through by - 1 )
32t² - 64t = 0 ← factor out 32t from each term
32t(t - 2) = 0
Equate each factor to zero and solve for t
32t = 0 ⇒ t = 0
t - 2 = 0 ⇒ t = 2
t = 0 represents when the frisbee was thrown
and t = 2 when it lands
Hence it takes 2 seconds for the frisbee to land.
Answer and Step-by-step explanation:
(a) Given that x and y is even, we want to prove that xy is also even.
For x and y to be even, x and y have to be a multiple of 2. Let x = 2k and y = 2p where k and p are real numbers. xy = 2k x 2p = 4kp = 2(2kp). The product is a multiple of 2, this means the number is also even. Hence xy is even when x and y are even.
(b) in reality, if an odd number multiplies and odd number, the result is also an odd number. Therefore, the question is wrong. I assume they wanted to ask for the proof that the product is also odd. If that's the case, then this is the proof:
Given that x and y are odd, we want to prove that xy is odd. For x and y to be odd, they have to be multiples of 2 with 1 added to it. Therefore, suppose x = 2k + 1 and y = 2p + 1 then xy = (2k + 1)(2p + 1) = 4kp + 2k + 2p + 1 = 2(kp + k + p) + 1. Let kp + k + p = q, then we have 2q + 1 which is also odd.
(c) Given that x is odd we want to prove that 3x is also odd. Firstly, we've proven above that xy is odd if x and y are odd. 3 is an odd number and we are told that x is odd. Therefore it follows from the second proof that 3x is also odd.
We use P = i•e^rt for exponential population growth, where P = end population, i = initial population, r = rate, and t = time
P = 2•i = 2•15 = 30, so 30 = 15 [e^(r•1)],
or 30/15 = 2 = e^(r)
ln 2 = ln (e^r)
.693 = r•(ln e), ln e = 1, so r = .693
Now that we have our doubling rate of .693, we can use that r and our t as the 12th hour is t=11, because there are 11 more hours at the end of that first hour
So our initial population is again 15, and P = i•e^rt
P = 15•e^(.693×11) = 15•e^(7.624)
P = 15•2046.94 = 30,704
