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3 years ago

Can you help me with this please

Mathematics

1 answer:

marshall27 [118]3 years ago

5 0

Answer:

a) y = 2 x - 1

b) y = - x + 7

c) y = 7 x + 2

Step-by-step explanation:

Use a pair of (x1, y1) and (x2, y2) points to find the equation of the line for each tables:

a) (5, 9) and (10, 19)

slope: (y2 - y1) / (x2 - x1) = (19 - 9) / (10 - 5) = 10 / 5 = 2

Then y = 2 x + b

solve for b in: 9 = 2 (5) + b

b = 9 - 10 = -1

Then y = 2 x - 1

b) (2, 5) and (5, 2)

slope: (y2 - y1) / (x2 - x1) = (2 - 5) / (5 - 2) = -3 / 3 = -1

Then y = - x + b

solve for b in: 5 = - (2) + b

b = 5 + 2 = 7

Then y = - x + 7

c) (3, 23) and (6, 44)

slope: (y2 - y1) / (x2 - x1) = (44 - 23) / (6 - 3) = 21 / 3 = 7

Then y = 7 x + b

solve for b in: 23 = 7 (3) + b

b = 23 - 21 = 2

Then y = 7 x + 2

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Step-by-step explanation:

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3 years ago

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s344n2d4d5 [400]

Answer:

$\left(\dfrac{2}{3},\dfrac{1}{5}\right)$

Step-by-step explanation:

Given the system of two equations:

$\left\{\begin{array}{l}\dfrac{1}{2}x-\dfrac{3}{4}y=\dfrac{11}{60}\\ \\\dfrac{2}{5}x+\dfrac{1}{6}y=\dfrac{3}{10}\end{array}\right.$

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$\left\{\begin{array}{l}30x-45y=11\\ \\24x+10y=18\end{array}\right.$

Now multiply the first equation by 4 and the second equation by 5:

$\left\{\begin{array}{l}120x-180y=44\\ \\120x+50y=90\end{array}\right.$

Subtract them:

$(120x-180y)-(120x+50y)=44-90\\ \\120x-180y-120x-50y=-46\\ \\-230y=-46\\ \\y=\dfrac{46}{230}=\dfrac{1}{5}$

Substitute it into the first equation:

$30x-45\cdot \dfrac{1}{5}=11\\ \\30x-9=11\\ \\30x=11+9\\ \\30x=20\\ \\x=\dfrac{2}{3}$

The solution is $\left(\dfrac{2}{3},\dfrac{1}{5}\right)$

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3 years ago

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Answer:

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Step-by-step explanation:

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What is the length of line PS

stiks02 [169]

$PQ^2 = PR \times PS$

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$(x - 8)(x + 1) = 0$

$x - 8 = 0~~~or~~~x + 1 = 0$

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x = -1 makes the lengths of segments PQ and PR equal zero.
That is not possible, so the solution x = -1 is discarded.

x = 8

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Can you help me with this please​

Can you help me with this please