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4 years ago

The co-ordinates of the point where the line y=x+4 crosses the y-axis

Mathematics

1 answer:

OverLord2011 [107]4 years ago

4 0

Answer:

Therefore the required point where it crosses y-axis is (0 , 4)

Step-by-step explanation:

Given:

$y=x+4$ ...............Equation of line

Intercepts:

The line which intersect on x-axis and y-axis are called intercepts.

There are two intercepts:

x-intercept: The line which intersect at x-axis. So when the line intersect at x-axis Y coordinate is zero.

y-intercept: The line which intersect at y-axis. So when the line intersect at y-axis X coordinate is zero.

For a line crosses Y-axis we have,

$y=x+4$

Put x = 0 in above equation we get

$y=0+4=4\\y=4$

Therefore the required point where it crosses y-axis is (0 , 4)

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1. Using the algebraic expression 6p + 5 = c. Where p represents packs of cards and c represents cost. Solve for the number of p

Scilla [17]

1.) 8
2.) 7
3.) 16
4.) 4
5.) -6

7 0

3 years ago

I need help on answering this question

liubo4ka [24]

Answer:

<h2>The answer is 72°</h2>

Step-by-step explanation:

Since < RQS = < QLK and < RQS = x

< QLK is also x

< QLK and < KLM lie on a straight line

Angles on a straight line add up to 180°

To find x add < QLK and < KLM and equate them to 180°

That's

< QLK + < KLM = 180°

x + x - 36 = 180

2x = 180 - 36

2x = 144

Divide both sides by 2

We have the final answer as

Hope this helps you

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4 years ago

Suppose that receiving stations X, Y, and Z are located on a coordinate plane at the points (4,5), (-6,-6), and (-14,2

Lilit [14]

Answer:

(-2, -3)

Step-by-step explanation:

A careful graph shows the point (-2, -3) is at the intersection of the circles whose radii are the given distances from the receiving stations.

_____

The simultaneous equations for the circles can be solved algebraically.

The epicenter is 10 units from X, so lies on the circle ...

(x -4)^2 +(y -5)^2 = 10^2

x^2 -8x +16 +y^2 -10y +25 = 100

x^2 +y^2 -8x -10y = 59

The epicenter is 5 units from Y, so lies on the circle ...

(x +6)^2 +(y +6)^2 = 5^2

x^2 +12x +36 +y^2 +12y +36 = 25

x^2 +y^2 +12x +12y = -47

The epicenter is 13 units from Z, so lies on the circle ...

(x +14)^2 +(y -2)^2 = 13^2

x^2 +28x +196 +y^2 -4y +4 = 169

x^2 +y^2 +28x -4y = -31

Subtracting the second equation from each of the other two, we get ...

(x^2 +y^2 -8x -10y) -(x^2 +y^2 +12x +12y) = (59) -(-47)

-20x -22y = 106 . . . . eq1 -eq2

(x^2 +y^2 +28x -4y) -(x^2 +y^2 +12x +12y) = (-31) -(-47)

16x -16y = 16 . . . . . . . .eq3 -eq2

These simultaneous linear equations can be solved a variety of ways. We might use substitution:

x = y+1 . . . . . from eq3 -eq2 divided by 16

10(y +1) +11y = -53 . . . . . from eq1 -eq2 divided by -2

21y = -63 . . . . . . . . . . . . simplify, subtract 10

y = -3

x = y+1 = -2

The epicenter is located at (x, y) = (-2, -3).

8 0

3 years ago

What is f(–3) for the function f(a) = –2a2 – 5a + 4?

tino4ka555 [31]

Follow the order of operation (PEMDAS) giving you 31

4 1

3 years ago

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If events X and Y are independent, what must be true? Check all that apply. P(Y | X) = 0 P(X | Y) = 0 P(Y | X) = P(Y) P(Y | X) =

lana [24]

if events X and Y are independent, then for intersection we multiply the probability

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We know that

$P(Y|X) =\frac{P(YintersectionX)}{P(X)}$

Now we replace P(Y) * P(X) for P(Y∩X)

$P(Y|X) =\frac{P(Y)*P(X)}{P(X)}$

Cancel out P(X)

So $P(Y|X) = P(Y)$

Like that

$P(X|Y) =\frac{P(XintersectionY)}{P(Y)}$

Now we replace P(X) * P(Y) for P(X∩Y)

$P(X|Y) =\frac{P(X)*P(Y)}{P(Y)}$

Cancel out P(Y)

So $P(X|Y) = P(X)$

P(Y | X) = P(Y) and P(X | Y) = P(X) are true

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3 years ago

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The co-ordinates of the point where the line y=x+4 crosses the y-axis​

The co-ordinates of the point where the line y=x+4 crosses the y-axis