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3 years ago

Write an equation of the line passing through the point A(2, 0) that is parallel to the line y = 3x−5 .

1 answer:

4 0

Hello,

In order for lines to be parallel to each other, they need to have the same slope. So, let's plug in the points and slope for point-slope form.

The equation for point-slope form is: y-y1 = m (x-x1)

Let's plug in the points into the equation.
y-5 = m (x-0), and then let's plug in the slope for "m" : y-5 = 3 (x-0).

After plugging the numbers in, the equation is simplified to y-5 = 3 (x-0). Then distribute the 3 to the parenthesis. y-5 = 3x. To get y by itself, you need to add 5 to both sides. After that, the equation parallel to that equation is: y = 3x + 5.

Hope this helps! =)

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Answer:

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Answer:

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The probability that a randomly selected student from the class has brown eyes , given they are male

$P(\frac{B}{M} ) = 0.3030$

Step-by-step explanation:

Explanation:-

Given data

Brown Blue Hazel Green

Females 13 4 6 9

Males 10 2 9 12

Let 'B' be the event of brown eyes

Total number of males n(M) = 33

Let B/M be the event of randomly selected student from the class has brown eyes given they are male

The probability that a randomly selected student from the class has brown eyes , given they are male

$P(\frac{B}{M} ) = \frac{n(B)}{n(M)}$

From table the brown eyes from males = 10

$P(\frac{B}{M} ) = \frac{10}{33}$

$P(\frac{B}{M} ) = 0.3030$

Final answer:-

The probability that a randomly selected student from the class has brown eyes , given they are male

$P(\frac{B}{M} ) = 0.3030$

8 0

3 years ago

Convert 1 cal/(m^2 * sec * °C) into BTU/(ft^2 * hr * °F)

Crazy boy [7]

Answer:

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=0.03926\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

Step-by-step explanation:

To find : Convert $1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}$ into $\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

Solution :

We convert units one by one,

$1\text{ m}^2=10.7639\text{ ft}^2$

$1\text{ sec}=\frac{1}{3600}\text{ hour}$

$1\text{ cal}=0.003968\text{ BTU}$

Converting temperature unit,

$^\circ C\times \frac{9}{5}+32=^\circ F$

$1^\circ C\times \frac{9}{5}+32=33.8^\circ F$

So, $1^\circ C=33.8^\circ F$

Substitute all the values in the unit conversion,

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=\frac{0.003968}{10.7639\times \frac{1}{3600}\times 33.8}\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=\frac{0.003968}{0.101061}\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=0.03926\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

Therefore, The conversion of unit is $1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=0.03926\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

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3 years ago

Find the mean of the integers -16,-27,21,-19,14,-3

babunello [35]

To find the mean all you have to do is add all the numbers...
you should get -30
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3 years ago

In triangle EFG, m of E = 30 degrees, m of F = 60 degrees, and m of G = 90 degrees. Which of the following statements about tria

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Answer:

A. EG = √3 × FG

D. EG = √3/2 × EF

E. EF = 2 × FG

Step-by-step explanation:

∵ tan 60 = √3

∵ tan60 = EG/GF

∴ EG/GF = √3

∴ EG = √3 × GF ⇒ A

∵ m∠F = 60°

∵ sin60 = √3/2

∵ sin 60 = EG/EF

∴ √3/2 = EG/EF

∴ EG = √3/2 × EF ⇒ D

∵ cos60 = 1/2

∵ cos60 = GF/EF

∴ GF/EF = 1/2

∴ EF = 2 × GF ⇒ E

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3 years ago

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