A rectangular garden 50 feet long and 10 feet wide is enclosed by a fence. To make the garden larger, while using the same fence
2 answers:
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Given:
Radius of the cone = 5 ft
Slant height of the cone = 21 ft
To find:
The surface area of the cone
Solution:
Surface area of cone formula:
where r is the radius and l is the slant height.
The surface area of the cone is 130π square feet.
Answer:
Before we graph we know that the slope, mx, could be read as
. To graph the the equation of the line, we begin at the point (0,0). From that point, because our rise is negative (-1), instead of moving upwards or vertically, we will move downards. Therefore, from point 0, we will vertically move downwards one time. Now, our point is on point -1 on the y-axis. Now, we have 2 as our run. From point -1, we move to the right two times. We land on point (2,-1). Because we need various points to graph this equation, we must continue on. In the end, the graph will look like the first graph given.
For the equation y = 2, the line will be plainly horizontal. Why? Because x has no value in the equation. The variable does not exist in this linear equation. Therefore, it will look like the second graph below. We graph this by plotting the point, (0,2), on the y-axis.
Answer:
D: 0.75
Step-by-step explanation:
<u>Explanation</u>:-
<u><em>Independent events:</em></u>
If the occurrence of the event 'B' is not effected by the occurrence or non- occurrence of the event 'A' then the event 'B' is said to be independent of 'A'
or
<em>The two events are independent if the incidence of one event is not affect the probability of other event.</em>
<em>P(A∩B) = P(A) P(B)</em>
Given data A and B are independent events
Given data
we know that A and B are independent events are
P(A∩B) = P(A) P(B)
Now calculation we get
<u><em>Final answer:</em></u>-
Answer:
Solutions are (3,1) and (4,2)
Step-by-step explanation:
Graph is shown in the attached sheet
Given are two systems of equations and we have to solve them using graph
For graphing let us first prepare table for x and y.
1)
I line II line
x 0 4.5 3 x 0 2.5 3
y 3 0 1 y -5 0 1
The two lines intersect at (3,1)
Hence solution is (3,1)
--------------------------------------------
2)
I line II line
x 0 2 4 x 0 6 4
y 0 1 2 y 3 0 2
The two lines intersect at (4,2)
Hence solution is (4,2)
