Answer:
a.) Between 0.5 and 3 seconds.
Step-by-step explanation:
So I just went ahead and graphed this quadratic on Desmos so you could have an idea of what this looks like. A negative quadratic, and we're trying to find when the graph's y-values are greater than 26.
If you look at the graph, you can easily see that the quadratic crosses y = 26 at x-values 0.5 and 3. And, you can see that the quadratic's graph is actually above y = 26 between these two values, 0.5 and 3.
Because we know that the quadratic's graph models the projectile's motion, we can conclude that the projectile will also be above 26 feet between 0.5 and 3 seconds.
So, the answer is a.) between 0.5 and 3 seconds.
I hope someone solves your problem :)
Answer:
Step-by-step explanation:
<em>Note:</em><em> You missed to add some of the details of the question.
</em>
<em>Hence, I am solving your concept based on an assumed graph which I have attached. It would anyways clear your concept.</em>
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Answer:
Please check the explanation.
Step-by-step explanation:
Given the right angled-triangle ABC as shown in the attached diagram
From the triangle:
Ф= ∠C = 30°
AB = 6 units
BC = y
tan Ф = opp ÷ adjacent
The opposite of ∠C = 30° is the length '6'.
The adjacent of ∠C = 30° is the length 'y'.
As Ф= ∠C = 30°
so
tan Ф = opp ÷ adjacent
tan 30 = 5 ÷ y
1 ÷ √3 = 5 ÷ y
y = 8.7 units
Therefore, the length of the unknown side length 'y' is 8.7 units.
10 degrees Celsius is 50 degrees Fahrenheit.
The equation for C to F is below.
C(9/5)+32= F
Multiply C by 1.8 (or 9/5) then add 32.
10(9/5)+32= 50 F
