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aleksley [76]
3 years ago
8

Graph the line with slope -1 passing through the point (-1,-4)

Mathematics
1 answer:
Lyrx [107]3 years ago
4 0

Answer: look at pic

Step-by-step explanation:

y=mx+b

m:slope

b:y-intercept

m=-1 (slope) given in the question

y=-1x+b

Plug in the given point (-1,-4) then solve for b

-4=-1(-1)+b

b=-5

y=-x-5

then using the x-intercept and y-intercept you could graph the line

y-intercept = b = -5 (0,-5)

x-intercept (plug in y=0)

0=-x-5

x=-5 (-5,0)

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Answer :

(1) \frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}

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Step-by-step explanation :

(1) The given expression is: \frac{1}{x}+\frac{1}{y}

\frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}

(2) The given expression is: \frac{1}{5x}+\frac{1}{3y}

\frac{1}{5x}+\frac{1}{3y}=\frac{3y+5x}{(5x)\times (3y)}=\frac{3y+5x}{15xy}

(3) The given expression is: \frac{1}{7x}-\frac{1}{2y}

\frac{1}{7x}-\frac{1}{2y}=\frac{2y-7x}{(7x)\times (2y)}=\frac{2y-7x}{14xy}

(4) The given expression is: \frac{2}{5x}+\frac{3}{7y}

\frac{2}{5x}+\frac{3}{7y}=\frac{(2\times 3y)+(3\times 5x)}{(5x)\times (7y)}=\frac{6y+15x}{21xy}

(5) The given expression is: \frac{7}{11x}-\frac{1}{33y}

\frac{7}{11x}-\frac{1}{33y}=\frac{(7\times 33y)-(1\times 11x)}{(11x)\times (33y)}=\frac{231y-11x}{363xy}

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\frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}=\frac{(x\times 7y)+(3\times 2x\times 7y)-(2x\times x)}{(2x)\times (x)\times (7y)}=\frac{7xy+42xy-2x^2}{14x^2y}=\frac{7y+42y-2x}{14xy}

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