What is the interquartile range of the pumpkin weights?
2 answers:
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The complete question in the attached figure
we know that
sin40° = opposite side / hypotenuse
opposite side =AC
hypotenuse = AB-------> 10 in
so
sin 40=AC/AB---------> AC=AB*sin 40-------> AC=10*sin 40
the answer is
AC=10*sin 40
we know that
sin40° = opposite side / hypotenuse
opposite side =AC
hypotenuse = AB-------> 10 in
so
sin 40=AC/AB---------> AC=AB*sin 40-------> AC=10*sin 40
the answer is
AC=10*sin 40
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.
Answer: The answer is (C) 324 cubic cm.
Step-by-step explanation: As given in the question, given a solid oblique pyramid with a regular hexagonal base and area 54√3 cm². Also, the edge length of the base is 6cm and ∠BAC = 60°.
We are to find the volume of the pyramid.
The formula for finding the volume of a pyramid is given by
where, 'b' is the base area and 'h' is the perpendicular height of the pyramid.
Here, b = 54√3 cm², h = ?
Now, from the right-angled triangle ABC, we have
Therefore, the volume of the pyramid is
Thus, the required volume is This makes (C) as the correct option.