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3 years ago

PLZ HELP...Enter in standard form the equation of the line passing through the given point and having the given slope.

Mathematics

1 answer:

Alenkinab [10]3 years ago

7 0

Standard form for the equation of the line is

$$y=-\frac{2}{3} x-\frac{25}{3}$

Solution:

Given point is (2, –5).

slope of the line m = $-\frac{2}{3}$

Here, $x_1=2, y_1=-5$

Equation of a line passing through the point:

$y-y_1=m(x-x_1)$

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

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

$$y+5=-\frac{2}{3} x-\frac{10}{3}$

Subtract 15 from both sides of the equation.

$$y+5-5=-\frac{2}{3} x-\frac{10}{3}-5$

$$y=-\frac{2}{3} x-\frac{25}{3}$

Standard form for the equation of the line is

$$y=-\frac{2}{3} x-\frac{25}{3}$

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a recipe calls for 2 2/3 cups of flour every 1/6 cup of cinnamon.If Richard has 1 cup of cinnamon already in the recipe, how man

Mashcka [7]

Step-by-step explanation:

2 2/3

=>If 8/3 = 1/6

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(By cross-multiplying the numerators and denominators)

=> 1×6 8×6

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Cancelling out , we have

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The final answer becomes 16

Hope it helps.

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3 years ago

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raketka [301]

Answer:

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Step-by-step explanation:

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3 years ago

Find out six rational numbers lying between -4/8 and -3/4

rjkz [21]

Answer:

-41/80, -45/80, -48/80, -51/80, -55/80, -59/80

Step-by-step explanation:

-4/8, -3/4 = -4/8, -6/8

= -40/80, -60/80

∴ Rational numbers between -40/80 and -60/80 = -41/80, -45/80, -48/80, -51/80, -55/80, -59/80.

4 0

3 years ago

An oil refinery is located on the north bank of a straight river that is 2 km wide. A pipeline is to be constructed from the ref

hichkok12 [17]

Answer:

P is exactly 3km east from the oil refinery.

Step-by-step explanation:

Let's d be the distance in km from the oil refinery to point P. So the horizontal distance from P to the storage is 3 - d and the vertical distance is 2. Hence the diagonal distance is:

$\sqrt{(3 - d)^2 + 2^2} = \sqrt{(3 - d)^2 + 4}$

So the cost of laying pipe under water with this distance is

$800000\sqrt{(3 - d)^2 + 4}$

And the cost of laying pipe over land from the refinery to point P is 400000d. Hence the total cost:

$800000\sqrt{(3 - d)^2 + 4} + 400000d$

We can find the minimum value of this by taking the 1st derivative and set it to 0

$800000\frac{2*0.5*(3-d)(-1)}{\sqrt{(3 - d)^2 + 4}} + 400000 = 0$

We can move the first term over to the right hand side and divide both sides by 400000

$1 = 2\frac{3 - d}{\sqrt{(3 - d)^2 + 4}}$

$\sqrt{(3 - d)^2 + 4} = 6 - 2d$

From here we can square up both sides

$(3 - d)^2 + 4 = (6 - 2d)^2$

$9 - 6d + d^2 + 4 = 36 - 24d + 4d^2$

$3d^2-18d+27 = 0$

$d^2 - 6d + 9 = 0$

$(d - 3)^2 = 0$

$d -3 = 0$

d = 3

So the cost of pipeline is minimum when P is exactly 3km east from the oil refinery.

3 0

3 years ago

mr. bird leaves for a walk at 7:50 a.m. she allows 25 minutes later what time does she arrive to work

Tatiana [17]

Answer: 8:15

Step-by-step explanation:

If she leaves at 7:50 and arrives to work 25 minutes later, then she will get there at 8:15.

Something easy you could use:

7:50 25

+ 10 -10

____ ___

8:00 15

+ 15 -15

____ ____

8:15 0

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3 years ago