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3 years ago

Rlly need help with Linear equations in any form ;-;

Mathematics

1 answer:

S_A_V [24]3 years ago

6 0

Answer:

y = -x - 2

Step-by-step explanation:

Find two points on the line

(0, -2)

(4, -6)

(Y2-Y1)/(X2-X1) = gradient of the line

Where (X1,Y1) = (0, -2) and (X2,Y2) = (4, -6)

(-6--2)/(4-0)

(-6+2)/4

(-4)/(4)

-1

Y = (-1)X + c

Substitute either set of coordinates in to find c.

(0, -2)

Y = (-1)X + c

-2 = (-1)0 + c

-2 = 0 + c

-2 = c

(4,-6)

Y = (-1)X + c

-6 = (-1)4 + c

-6 = -4 +c

-6 + 4 = c

-2 = c

Y = -X -2

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Select the proper order from least to greatest for 1/2 ,5/6 , 2/7 , 4/2

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3 years ago

3. Try It #3 Write the point-slope form of an equation of a line with a slope of -2 that passes through the point (-2,2). Then r

vova2212 [387]

Answer:

Point-slope form of equation given as $y-2=-2(x+2)$ .

Slope-intercept form of equation is given as $y=-2 x-2$ .

Step-by-step explanation:

In the question, it is given that the slope of a line is -2 and it passes from (-2,2).

It is asked to write the point-slope form of the equation and rewrite it as slope-intercept form.

To do so, first find the values which are given in the question and put it in the formula of point-slope form. Simplify the equation to rewrite as slope-intercept form.

Step 1 of 2

Passing point of the line is (-2,2).

Hence, $x_{1}=-2$ and

$y_{1}=2 \text {. }$

Also, the slope of the line is -2.

Hence, m=-2

Substitute the above values in point-slope form of equation given by $y-y_{1}=m\left(x-x_{1}\right)$

$\begin{aligned}&y-y_{1}=m\left(x-x_{1}\right) \\&y-2=-2(x-(-2) \\&y-2=-2(x+2)\end{aligned}$

Hence, point-slope form of equation given as y-2=-2(x+2).

Step 2 of 2

Solve y-2=-2(x+2) to write it as slope-intercept form given by y=mx+c.

$\begin{aligned}&y-2=-2(x+2) \\&y-2=-2 x-4 \\&y=-2 x-4+2 \\&y=-2 x-2\end{aligned}$

Hence, slope-intercept form of equation is given as y=-2x-2.

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2 years ago

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3 years ago

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let the number be x.

a number decreased means

and the sum of the number and 4 is

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to put the expression together, we get. put them in parentheses

x - ( x+4 )

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3 0

3 years ago

5. A yacht is cruising at 20 knots at bearing of 185° when a 15 knot wind starts to blow at a bearing of 290°. Find the directio

kicyunya [14]

Answer:

Direction of the boat is 43.05° and the speed is 21.66 knot

Step-by-step explanation:

yacht speed = 20 knots

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wind speed = 15 knots

wind bearing = 290° = 20° above the negative horizontal x-axis

we find the x and y components of the boat velocities

for yacht,

x component = -20 cos 85° = -1.74 knots

y component = -20 sin 85° = -19.92 knots

for the wind,

x component = -15 cos 20° = -14.09 knots

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