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3 years ago

Can u please help me with this question?

Mathematics

1 answer:

Masteriza [31]3 years ago

3 0

Answer:

B) $y=\frac{-2}{3}x-\frac{-13}{3}$

Step-by-step explanation:

2x +3y = 5

3y = -2x + 5

Divide both sides by 3

$\frac{3y}{3}=\frac{-2x}{3}+\frac{5}{3}\\\\y =\frac{-2}{3}x+\frac{5}{3}$

Parallel lines have same slope.

m = (-2/3) ; (-2 , -3)

y - y1 = m(x -x1)

$y-[-3]=\frac{-2}{3}(x-[-2])\\\\y + 3 = \frac{-2}{3}(x +2)\\\\y + 3 =\frac{-2}{3}x + 2*\frac{-2}{3}\\\\y+3=\frac{-2}{3}x-\frac{4}{3}\\\\y=\frac{-2}{3}x-\frac{4}{3}-3\\\\y=\frac{-2}{3}x-\frac{4}{3}-\frac{3*3}{1*3}\\\\y=\frac{-2}{3}x-\frac{4}{3}-\frac{9}{3}\\\\y=\frac{-2}{3}x-\frac{-13}{3}$

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It takes 2/3 of an hour to paint 2/5 of a room. How long does it take to paint 2/5 of a room?

Lubov Fominskaja [6]

It will take 40 minutes.

To solve, divide 60 by 3. 60 is how many minutes there are in a minute. 3 is how long it takes to paint 2/5 of the room. We do not need to use the number 2/5 in this situation. Now back to 60 divided by 3. The answer is 20 minutes. Now multiply 20 by 2 to get 2/3 of an hour. Therefore, it takes 40 minutes to paint 2/5 of a room.

8 0

3 years ago

10. Two lighthouses are located on a north-south line.

Kamila [148]

The question is an illustration of bearing (i.e. angles) and distance (i.e. lengths)

The distance between both lighthouses is 5783.96 m

I've added an attachment that represents the scenario.

From the attachment, we have:

$\mathbf{\angle A = 180^o - 120^o\ 43'}$

Convert to degrees

$\mathbf{\angle A = 180^o - (120^o +\frac{43}{60}^o)}$

$\mathbf{\angle A = 180^o - (120^o +0.717^o)}$

$\mathbf{\angle A = 180^o - (120.717^o)}$

$\mathbf{\angle A = 59.283^o}$

$\mathbf{\angle B = 39^o43'}$

Convert to degrees

$\mathbf{\angle B = 39^o + \frac{43}{60}^o}$

$\mathbf{\angle B = 39^o + 0.717^o}$

$\mathbf{\angle B = 39.717^o}$

So, the measure of angle S is:

$\mathbf{\angle S = 180 - \angle A - \angle B}$ ---- Sum of angles in a triangle

$\mathbf{\angle S = 180 - 59.283 - 39.717}$

$\mathbf{\angle S = 81}$

The required distance is distance AB

This is calculated using the following sine formula:

$\mathbf{\frac{AB}{\sin(S)} = \frac{AS}{\sin(B)} }$

Where:

$\mathbf{AS = 3742}$

So, we have:

$\mathbf{\frac{AB}{\sin(81)} = \frac{3742}{\sin(39.717)}}$

Make AB the subject

$\mathbf{AB= \frac{3742}{\sin(39.717)} \times \sin(81)}$

$\mathbf{AB= 5783.96}$

Hence, the distance between both lighthouses is 5783.96 m

Read more about bearing and distance at:

brainly.com/question/19017345

5 0

2 years ago

Help me please !!!

Nat2105 [25]

Answer:

For 3x^2+4x+4=0

Discriminant= = -32

The solutions are

(-b+√x)/2a= (-2+2√-2)/3

(-b-√x)/2a= (-2-2√-2)/3

For 3x^2+2x+4=0

Discriminant= -44

The solutions

(-b+√x)/2a= (-1+√-11)/3

(-b-√x)/2a= (-1-√-11)/3

For 9x^2-6x+2=0

Discriminant= -36

The solutions

(-b+√x)/2a= (1+√-1)/3

(-b-√x)/2a= (1-√-1)/3

Step-by-step explanation:

Formula for the discriminant = b²-4ac

let the discriminant be = x for the equations

The solution of the equations

= (-b+√x)/2a and = (-b-√x)/2a

For 3x^2+4x+4=0

Discriminant= 4²-4(3)(4)

Discriminant= 16-48

Discriminant= = -32

The solutions

(-b+√x)/2a =( -4+√-32)/6

(-b+√x)/2a= (-4 +4√-2)/6

(-b+√x)/2a= (-2+2√-2)/3

(-b-√x)/2a =( -4-√-32)/6

(-b-√x)/2a= (-4 -4√-2)/6

(-b-√x)/2a= (-2-2√-2)/3

For 3x^2+2x+4=0

Discriminant= 2²-4(3)(4)

Discriminant= 4-48

Discriminant= -44

The solutions

(-b+√x)/2a =( -2+√-44)/6

(-b+√x)/2a= (-2 +2√-11)/6

(-b+√x)/2a= (-1+√-11)/3

(-b-√x)/2a =( -2-√-44)/6

(-b-√x)/2a= (-2 -2√-11)/6

(-b-√x)/2a= (-1-√-11)/3

For 9x^2-6x+2=0

Discriminant= (-6)²-4(9)(2)

Discriminant= 36 -72

Discriminant= -36

The solutions

(-b+√x)/2a =( 6+√-36)/18

(-b+√x)/2a= (6 +6√-1)/18

(-b+√x)/2a= (1+√-1)/3

(-b-√x)/2a =( 6-√-36)/18

(-b-√x)/2a= (6 -6√-1)/18

(-b-√x)/2a= (1-√-1)/3

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3 years ago

If I have a Pentagon shape and it's 33 centimeters long, what would be my Perimeter?

Oksana_A [137]

The answer is 165. If you add 33 five times you get 165. Or you could do what i did and do 33 times 5.

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