We know that
the equation of the vertical parabola in the vertex form is
<span>y=a(x-h)²+k
</span>where
(h,k) is the vertex of the parabola
if a> 0 then
the parabola opens upwards
if a< 0
then the parabola open downwards
in this problem we have
f(x)=−5(x+7)²<span>+6
</span>a=-5
so
a< 0 -------> the parabola open downwards
the vertex is the point (-7,6) is a maximum
the answer is the option<span>
a = -5, opens down</span>
see the attached figure
Answer:
3x^2 + 3 --> 6, 15, 30, 51
2x^2 - 1 --> 1, 7, 17, 31
x^2 + 2 --> 3, 6, 11, 18
Step-by-step explanation:
Let's start with the first equation; 3x^2 + 3
Substitute 1 (the first digit in a sequence) for x.
3(1^2) + 3
3(1) + 3
3 + 3 = <u>6</u>
3(2^2) + 3 Then the second digit.
3(4) + 3
12 + 3 = <u>15</u>
Since the two numbers we have so far are 6 and 15, there is only one sequence this could match. 6, 15, 30, 51.
2(1^2) - 1
2(1) - 1
2 - 1 = <u>1</u>
This equation represents 1, 7, 17, 31.
These same steps apply to the other equation as well.
1^2 + 2, then 2^2 + 2, then 2^2 + 2, and so on. (But we don't need to do extra work to figure that out.)
<h3>
Answer: C) increase from 45 degrees to 50 degrees</h3>
===============================================
Explanation:
Let's calculate the angle B based off the arcs CDF and GHJ
B = (far arc - near arc)/2
B = (arc CDF - arc GHJ)/2
B = (130 - 40)/2
B = 90/2
B = 45
----------
Now let's change GHJ to 30 degrees, while keeping the other arc the same.
B = (far arc - near arc)/2
B = (arc CDF - arc GHJ)/2
B = (130 - 30)/2
B = 100/2
B = 50
Angle B has increased from 45 degrees to 50 degrees.
I divided the figure into two rectangles, found their areas separately and added them up to get the final area 168 cm squared.