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kiruha [24]
3 years ago
6

Points J and K lie in plane H. Plane H contains two points. Point J is to the lower left and point K is to the uppercase right.

How many lines can be drawn through points J and K? 0 1 2 3
Mathematics
2 answers:
FrozenT [24]3 years ago
9 0

Answer:

The correct option is B) 1

Step-by-step explanation:

Consider the provided information.

It is given that there is a plane H on which two pints J and K are located.

We need to find how many lines can be drawn through the points J and K.

The Axiom of Euclid's geometry says that "Through any given two points X and Y, only one line can be drawn "

Therefore by Axiom in Euclid's geometry we can concluded that for the given points J and K in plane H , only one line can be drawn through points J and K.

Hence, the correct option is B) 1

xxMikexx [17]3 years ago
4 0

Answer:

1

Step-by-step explanation:

There's a theorem that says "through 2 points it only cross one and only one line". Having this as our start point, we can guarantee that there's only one line that can be drawn through those points

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If B is Jonas and his sister is 4 years older, that would be 4 plus B
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2 years ago
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Find cot theta. (-6,-3sqrt(5)
cupoosta [38]

0  Pretend that is theta XD


cot 0 = \frac{1}{tan0}


Find tan 0.

tan 0 = \frac{y}{x}

tan 0 = \frac{-3\sqrt{5}}{-6}

tan 0 = \frac{\sqrt{5}}{2}


cot 0 = \frac{1}{tan0}

cot 0 = \frac{1}{\frac{\sqrt{5} }{2} }    

Multiply 2/√5 to the numerator and the denominator (or multiply the reciprocal of the denominator to the top and bottom of the fraction)

cot 0 = \frac{1}{\frac{\sqrt{5}}{2} } (\frac{\frac{2}{\sqrt{5}} }{\frac{2}{\sqrt{5}}} )

cot 0 = \frac{2}{\sqrt{5}}  

Multiply √5 to the numerator and the denominator

cot 0 = \frac{2}{\sqrt{5}} (\frac{\sqrt{5} }{\sqrt{5} })

cot 0 = \frac{2\sqrt{5}}{5}


5 0
3 years ago
PLEASE HELP! (04.02 MC) Write an equation of a line parallel to line EF in slope-intercept form that passes through the point (2
Tanya [424]

Answer:

  C.  y = -2/3x +22/3

Step-by-step explanation:

You can choose the correct answer by realizing that the line must have a y-intercept greater than 6. The given point is (2, 6) and the line goes up and to the left from there. The y-intercept is obviously more than 6.

The only reasonable answer choice is ...

  y = -2/3x +22/3 . . . . . a y-intercept of 7 1/3

_____

There are a lot of ways to work problems like these. One is to write the equation of the line, then match that to the answer choices.

Another is to eliminate all the "impossible" answer choices, then choose the appropriate one from what's left. (The given line has a negative slope, eliminating choices A and B.)

I like to use the simplest method that will determine the correct answer. Here, that involves reading and understanding the question to obtain some idea of where the desired line must appear on the graph.

Of course, it helps to know that "slope-intercept form" is ...

  y = mx +b . . . . . . m represents the slope; b is the y-intercept

5 0
3 years ago
Write the equation of a line that is parallel to y =1/2x + 9 and goes through the point (-2,2)
dsp73

Answer:

y = 1/2x + 3

Step-by-step explanation:

The slope of a parallel line is the same as the original.

Find the new y-intercept.

y = 1/2x + b

2 = 1/2(-2) + b

2 = -1 + b

3 = b

y = 1/2x + 3

Best of Luck!

3 0
3 years ago
Help pleaseee , don’t really know how to do this :)
yawa3891 [41]

Answer:

x= -p+q+y

Step-by-step explanation:

-x on each side

y-x=p-q

-y on each side

-x=p-q-y

÷everything on both sides by ‐1

We have to divide by -1 because x can't be negative

x = -p+q+y

I really hope this is right because I don't want to let you down.

-LampteyJ

8 0
3 years ago
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