-24=u/6 just times 6 to both sides -24*6=-144 rewrite
-144=u
Hope this helps have a nice day
By critically observing the cross-sections of the three-dimensional object I used, a cone is the cross-sectional shape I find most surprising.
<h3>The cross-section of a three-dimensional object?</h3>
In this exercise, you're required to use an online tool to investigate and determine the cross-sections of three-dimensional objects such as pyramids, cylinders, cones, etc., by passing different planes through them.
By critically observing the cross-sections of the three-dimensional object I used, a cone is the cross-sectional shape I find most surprising because rotating the slice around Y produced a circular curve that transitioned into a parabolic curve.
Read more on cross-sections here: brainly.com/question/1924342
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The area of a circle is A=πr².
Since π is about 3.14 we can substitute that into the equation.
So A=3.14×r²
We can substitute the given radius for r.
A= 3.14 × 5²
When a number is squared, or has the two over it, we multiply it by itself.
5 × 5= 25
A= 3.14 × 25
Simplify this and the area of the circle is about 78.5
A≈78.5
Answer:
x = 6
y = 13
Step-by-step explanation:
since DE and JK are congruent, they are both 18
to find 'y':
3y - 21 = 18
3y = 39
y = 13
DF and JL are diagonals that are congruent so:
9x - 23 = 7x - 11
2x = 12
x = 6
Analysis:
1) The graph of function f(x) = √x is on the first quadrant, because the domain is x ≥ 0 and the range is y ≥ 0
2) The first transformation, i.e. the reflection of f(x) over the x axis, leaves the function on the fourth quadrant, because the new image is y = - √x.
3) The second transformation, i.e. the reflection of y = - √x over the y-axis, leaves the function on the third quadrant, because the final image is - √(-x). This is, g(x) = - √(-x).
From that you have, for g(x):
* Domain: negative x-axis ( -x ≥ 0 => x ≤ 0)
* Range: negative y-axis ( - √(-x) ≤ 0 or y ≤ 0).
Answers:
Now let's examine the statements:
<span>A)The functions have the same range:FALSE the range changed from y ≥ 0 to y ≤ 0
B)The functions have the same domains. FALSE the doman changed from x ≥ 0 to x ≤ 0
C)The only value that is in the domains of both functions is 0. TRUE: the intersection of x ≥ 0 with x ≤ 0 is 0.
D)There are no values that are in the ranges of both functions. FALSE: 0 is in the ranges of both functions.
E)The domain of g(x) is all values greater than or equal to 0. FALSE: it was proved that the domain of g(x) is all values less than or equal to 0.
F)The range of g(x) is all values less than or equal to 0.
TRUE: it was proved above.</span>
