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3 years ago

Which of these equations is for a line perpendicular to y-3=5(x-2)?

Mathematics

1 answer:

olasank [31]3 years ago

6 0

Answer:

Step-by-step explanation:

y-3=5(x-2) (rearrange this to be in slope- intercept from) (add 3 to both sides)

y = 5(x-2) + 3 (distribute parentheses)

y = x(5) - 2(5) + 3

y = 5x -10 + 3

y = 5x - 7

recall that for a line with gradient m, the gradient of the perpendicular line will be - (1/m)

hence in our case, our gradient of the original line is 5, hence the gradient of the perpendicular line is -1/5

From the choices, the only one that is consistent with this is C

i.e choice C:

5y + x = 25

5y = -x + 25

y = -(1/5) x + 5 ===> gradient of -1/5

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What is the slope of the line through the points (-3, -5) and (-4, -2)

Ede4ka [16]

Answer:

slope= -3

Step-by-step explanation:

the slope formula is $\frac{y_{2} -y_{1} }{x_{2} -x_{1} }$

plug in your numbers: $\frac{(-2)-(-5)}{(-4)-(-3)}$

solve: $\frac{3}{-1}$

simplify: -3

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3 years ago

A box contains 5 red and 5 blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win $1.10; if

topjm [15]

Answer:

a) The expected value is $\frac{-1}{15}$

b) The variance is $\frac{49}{45}$

Step-by-step explanation:

We can assume that both marbles are withdrawn at the same time. We will define the probability as follows

#events of interest/total number of events.

We have 10 marbles in total. The number of different ways in which we can withdrawn 2 marbles out of 10 is $\binom{10}{2}$ .

Consider the case in which we choose two of the same color. That is, out of 5, we pick 2. The different ways of choosing 2 out of 5 is $\binom{5}{2}$ . Since we have 2 colors, we can either choose 2 of them blue or 2 of the red, so the total number of ways of choosing is just the double.

Consider the case in which we choose one of each color. Then, out of 5 we pick 1. So, the total number of ways in which we pick 1 of each color is $\binom{5}{1}\cdot \binom{5}{1}$ . So, we define the following probabilities.

Probability of winning: $\frac{2\binom{5}{2}}{\binom{10}{2}}= \frac{4}{9}$

Probability of losing $\frac{(\binom{5}{1})^2}{\binom{10}{2}}\frac{5}{9}$

Let X be the expected value of the amount you can win. Then,

E(X) = 1.10*probability of winning - 1 probability of losing = $1.10\cdot \frac{4}{9}-\frac{5}{9}=\frac{-1}{15}$

Consider the expected value of the square of the amount you can win, Then

E(X^2) = (1.10^2)*probability of winning + probability of losing = $1.10^2\cdot \frac{4}{9}+\frac{5}{9}=\frac{82}{75}$

We will use the following formula

$Var(X) = E(X^2)-E(X)^2$

Thus

Var(X) = $\frac{82}{75}-(\frac{-1}{15})^2 = \frac{49}{45}$

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3 years ago

7-3(10 divided by 2) divided by 1

Misha Larkins [42]

Answer:

-8

Step-by-step explanation:

Remember pemdas

7 - 3*(10:2):1 =

7-3*5:1=

7-15:1=

7- 15=

-8

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1 year ago

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salantis [7]

Answer:

Answer:

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Answer:

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