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3 years ago

What is the equation of the line that is parallel to the line x = -2 and passes through the point (-5, 4)?

Mathematics

1 answer:

Nastasia [14]3 years ago

8 0

Answer:

The equation of the line that is parallel to the line x = -2 and passes through the point (-5, 4) is x=-5

Option A is correct.

Step-by-step explanation:

We need to find equation of the line that is parallel to the line x = -2 and passes through the point (-5, 4)

We need a line parallel to x=-2 or x+2=0 it should be of form x+k=0

We need to find k, by putting value of x=-5 as given in question the point(-5,4)

-5+k=0

k=5

So, the equation of line will be found by putting k=5:

x+k=0

x+5=0

x=-5

So, the equation of the line that is parallel to the line x = -2 and passes through the point (-5, 4) is x=-5

Option A is correct.

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3 years ago

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Given:

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To find:

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Solution:

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$\Delta MNO\cong \Delta PQR$

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3 years ago

What is -5/8 times 2/3 fraction

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3 years ago

Read 2 more answers

A student S is suspected of cheating on exam, due to evidence E of cheating being present. Suppose that in the case of a cheatin

Kay [80]

Answer:

Step-by-step explanation:

Suppose we think of an alphabet X to be the Event of the evidence.

Also, if Y be the Event of cheating; &

Y' be the Event of not involved in cheating

From the given information:

$P(\dfrac{X}{Y}) = 60\% = 0.6$

$P(\dfrac{X}{Y'}) = 0.01\% = 0.0001$

$P(Y) = 0.01$

Thus, $P(Y') \ will\ be = 1 - P(Y)$

P(Y') = 1 - 0.01

P(Y') = 0.99

The probability of cheating & the evidence is present is = P(YX)

$P(YX) = P(\dfrac{X}{Y}) \ P(Y)$

$P(YX) =0.6 \times 0.01$

$P(YX) =0.006$

The probabilities of not involved in cheating & the evidence are present is:

$P(Y'X) = P(Y') \times P(\dfrac{X}{Y'})$

$P(Y'X) = 0.99 \times 0.0001 \\ \\ P(Y'X) = 0.000099$

(b)

The required probability that the evidence is present is:

P(YX or Y'X) = 0.006 + 0.000099

P(YX or Y'X) = 0.006099

(c)

The required probability that (S) cheat provided the evidence being present is:

Using Bayes Theorem

$P(\dfrac{Y}{X}) = \dfrac{P(YX)}{P(Y)}$

$P(\dfrac{Y}{X}) = \dfrac{P(0.006)}{P(0.006099)}$

$P(\dfrac{Y}{X}) = 0.9838$

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3 years ago

How do you graph a possible situation?

Svet_ta [14]

To graph a situation that would involve a linear graph, first determine your x and y axes.
The x-axis will be the independent variable, one that does not change based on other variables. An example is time.

The y-axis, the dependent variable, depends on the independent variable.

The model equation for a linear line is y = mx + b.
"m" is the slope, and the "b" is the y-intercept (where the graph crosses the x-axis at x=0).

For example, a situtation could be that Joe starts with $10 in his account and adds $5 every day to his account.
The x-axis is time in days.
The y-axis is amount of money in his account.
The slope, or rate of change is 5.
The y-intercept, the amount of money he has at x=0 (0 days) is $10.

The equation would be y = 5x + 10
To draw this, plot the y-intercept at (0, 10), and the next point would be 5 units up and one unit to the right because the slope is 5, or 5/1 (remember slope is rise over run: "rise" up 5 and "over" to the right 1).

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