I would start with the standard form equation of the parallel line.
.. 5x -2y = 5(-2) -2(3) = -16
Then solve for y.
.. 2y = 5x +16 . . . add 2y+16; then divide by 2 for the next equation
.. y = (5/2)x +8 . . . . . . . corresponds to selection (B)
Answer:
true
Step-by-step explanation:
no explanation but its true.
<h3>Answer is -9</h3>
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Work Shown:
(g°h)(x) is the same as g(h(x))
So, (g°h)(0) = g(h(0))
Effectively h(x) is the input to g(x). Let's first find h(0)
h(x) = x^2+3
h(0) = 0^2+3
h(0) = 3
So g(h(x)) becomes g(h(0)) after we replace x with 0, then it updates to g(3) when we replace h(0) with 3.
Now let's find g(3)
g(x) = -3x
g(3) = -3*3
g(3) = -9
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alternatively, you can plug h(x) algebraically into the g(x) function
g(x) = -3x
g( h(x) ) = -3*( h(x) ) ... replace all x terms with h(x)
g( h(x) ) = -3*(x^2 + 3) ... replace h(x) on right side with x^2+3
g( h(x) ) = -3x^2 - 9
Next we can plug in x = 0
g( h(0) ) = -3(0)^2 - 9
g( h(0) ) = -9
we get the same result.
Answer:
(a) It is an experiment.
(b) Group III has been tested to compare the effect of tea
Step-by-step explanation:
24 volunteers are randomly selected to one of the three groups.
Group I drinks two cups of hot black tea without milk
Group II drinks two cups of hot black tea with milk
Group III drinks two cups of hot water but no tea.
At the end of the month, the researchers measured the change in each of the participant's heart health.
The average change of health status of three Groups can be measured and let it be X1', X2' and X3'. We can set up the hypothesis of no difference of heart health i.e H0 : µ1 = µ2 = µ3 against the alternative hypothesis that they are not equal. As the sample size is less than 30, we can use t-statistic.
Under the above logic, (a) it is an experiment.
(b) Group III has been tested to compare the effect of tea or to have comparison between GroupI and Group III.
