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liq [111]
4 years ago
7

What is the length of line segment KJ?

Mathematics
1 answer:
Karolina [17]4 years ago
5 0

Please consider the attached file.

We can see that triangle JKM is a right triangle, with right angle at M. Segment KM is 6 units and segment MJ is 3 units. We can also see that KJ is hypotenuse of right triangle.

We will use Pythagoras theorem to solve for KJ as:

KJ^2=KM^2+MJ^2

KJ^2=6^2+3^2

KJ^2=36+9

KJ^2=45

Now we will take positive square root on both sides:

\sqrt{KJ^2}=\sqrt{45}

KJ=\sqrt{9\cdot 5}

KJ=3\sqrt{5}

Therefore, the length of line segment KJ is 3\sqrt{5} and option D is the correct choice.

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AlladinOne [14]

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To obtain an average (arithmetic mean) of exactly 2/3, what fraction must be added to 1/6, 1/2, and 1/3?
ollegr [7]

Answer:

D. 5/3

Step-by-step explanation:

\frac{1/6+1/2+1/3+x}{4} =2/3

1/6+1/2+1/3+x=8/3

x=8/3-1/6-1/2-1/3

x=8/3-1/6-3/6-2/6

x=8/3-6/6

x=16/6-6/6=10/6

simplified

x=5/3

Hope this helps

3 0
2 years ago
A car motercycle leaves at noon from the same location, heading in the same direction. The average speed of the car is 30mph slo
alina1380 [7]

Answer:

speed of motorcycle = 40 mph

speed of car = 50 mph

Step-by-step explanation:

Here is the complete question

A car and a motorcycle leave at noon from the same location, heading in the same direction. The average speed of the car is 30 mph slower than twice the speed of the motorcycle. In two hours, the car is 20 miles ahead of the motorcycle. Find the speed of both the car and the motorcycle, in miles per hour.

Speed = distance / time

This question would be solved using simultaneous equation

let m =  average speed of the motorcycle

c = average speed of the car

c = 2m - 30  equation 1

20 =(c - m) x 2 equation 2

insert equation 1 into equation 2 and divide through by 2

10 = (2m - 30) - m

solve for m

m = 40 mph

substitute for m in equation 1

2(40) - 20 = 50 mph

8 0
3 years ago
Check whether the function yequalsStartFraction cosine 2 x Over x EndFraction is a solution of x y prime plus yequalsnegative 2
Jobisdone [24]

The question is:

Check whether the function:

y = [cos(2x)]/x

is a solution of

xy' + y = -2sin(2x)

with the initial condition y(π/4) = 0

Answer:

To check if the function y = [cos(2x)]/x is a solution of the differential equation xy' + y = -2sin(2x), we need to substitute the value of y and the value of the derivative of y on the left hand side of the differential equation and see if we obtain the right hand side of the equation.

Let us do that.

y = [cos(2x)]/x

y' = (-1/x²) [cos(2x)] - (2/x) [sin(2x)]

Now,

xy' + y = x{(-1/x²) [cos(2x)] - (2/x) [sin(2x)]} + ([cos(2x)]/x

= (-1/x)cos(2x) - 2sin(2x) + (1/x)cos(2x)

= -2sin(2x)

Which is the right hand side of the differential equation.

Hence, y is a solution to the differential equation.

6 0
4 years ago
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