This means that you have to move the decimal point 4 decimal places to the left, as denoted by the negative sign. Thus, this would add 3 zeros before 5. The answer would be 0.00059.
I hope the explanation was clear to you. Have a good day.
The correct answer to where the point(1, StartRoot 7 EndRoot) lies on the circle is A. Yes, the distance from (–2, 0) to (1, StartRoot 7 EndRoot) is 4 units
<h3>What is a Coordinate Plane?</h3>
This refers to the tool that is used to graph points on a two-dimensional plane that intersects two number lines.
Hence, given the information in the given diagram, there is a coordinate plane that shows a circle at the center that has certain values on the y and x-axis.
We can state that the point (1, StartRoot 7 EndRoot) lies on the circle shown because the distance from (–2, 0) to (1, StartRoot 7 EndRoot) is 4 units.
Read more about coordinate planes here:
brainly.com/question/7038052
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Answer:
a) F
b) B, E, D
Step-by-step explanation:
a) The segment with the greatest gradient has the largest change in y-values per unit change in x-values
From the given option, the rate of change of the <em>y </em>to the<em> </em>x-values of B = the gradient = (4 units)/(2 units) = 2
The gradient of F = (-3units)/(1 unit) = -3
The gradient of A = 4/4 = 1
The gradient of C = -2/5
The gradient of D = 2/6 = 1/3
The gradient of E = 3/4
The segment with the greatest gradient is F
b) The steepest segment has the higher gradient
From their calculated we have;
The gradient of segment B = 2 therefore, B is steeper than E that has a gradient of 3/4, and E is steeper than D, as the gradient of D = 1/3
Therefore, we have;
B, E, D.
L ≥ 5w + 4
2(L+w) ≥ 32
L+w ≥ 16
6w + 4 ≥ 16
6w ≥ 12
w ≥ 2
L ≥ 5w + 4
L ≥ 14
If L = 14 and w = 2 then the perimeter is 32.
These are the smallest dimensions that fit the criteria.
Note: If we raise the width by 1 to w = 3
then the length is at least 19. At 19 by 3 we have P = 44
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Divide both sides by -3, and replace
with
. Then

Factorize the quadratic in
to get

which in turn means

But
for all real
, so we can ignore the first solution. This leaves us with

If we allow for any complex solution, then we can continue with the solution we ignored:
