ASA, SAS, AAS,
if LA stands for leg-angle then you could do that too
Answer:
Step-by-step explanation:
Average Temperatures Suppose the temperature (degrees F) in a river at a point x meters downstream from a factory that is discharging hot water into the river is given by
T(x) = 160-0.05x^2
a. [0, 10]
For x = 0
T(0) = 160 - 0.05 × 0^2
T(0) = 160
For x = 10
T(10) = 160 - 0.05 × 10^2
T(10) = 160 - 5 = 155
The average temperature
= (160 + 155)/2 = 157.5
b. [10, 40]
For x = 10
T(10) = 160 - 0.05 × 10^2
T(10) = 160 - 5 = 155
For x = 40
T(10) = 160 - 0.05 × 40^2
T(10) = 160 - 80 = 80
The average temperature
= (80 + 155)/2 = 117.5
c. [0, 40]
For x = 0
T(0) = 160 - 0.05 × 0^2
T(0) = 160
For x = 40
T(10) = 160 - 0.05 × 40^2
T(10) = 160 - 80 = 80
The average temperature
= (160 + 80)/2 = 120
Answer:
-5x
Step-by-step explanation:
The quadratic equation for this would be f(x) = 5x^2 - 10x - 120.
In order to find that, we need to start by taking our x intercept values and setting them equal to zero.
x = 6 ----> subtract 6 from both sides
x - 6 = 0
x = -4 ----> add 4 to both sides
x + 4 = 0
Now that we have both of these zero terms, we can multiply them to get a standard form.
f(x) = (x - 6)(x + 4)
And while this will give us the zeros we need, it will no give us the lead coefficient. So we must multiply by the desired lead coefficient.
f(x) = 5(x - 6)(x + 4)
f(x) = 5(x^2 - 6x + 4x - 24)
f(x) = 5(x^2 - 2x - 24)
f(x) = 5x^2 - 10x - 120
Answer:
The answer is 4
Step-by-step explanation:
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