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2 years ago

What is an equation of the line that passes through the point(-1,-3) and is perpendicularly to the line x-2y=14

Mathematics

2 answers:

Nitella [24]2 years ago

8 0

Answer:

the equation will be

Step-by-step explanation:

2x+y+5=0

IRINA_888 [86]2 years ago

6 0

Answer:

2x + y + 5 = 0

Step-by-step explanation:

The equation of line is x-2y = 14 and we need to find the equⁿ of line which passes through (-1,-3) and is perpendicular to given line.

Find the slope of given line by converting it into slope intercept form.

=> x - 2y = 14

=> 2y = x - 14

=> y = x-14/2

=> y = x/2 - 7

Comparing to y = mx + c ,

=> Slope (m) = 1/2 .

Now the slope of line perpendicular to it , will have a slope ,

=> $m_{perp}$ = -1/m

=> $m_{perp}$ = -1/½

=> $m_{perp}$ = -2

Using point slope form ,

=> m = (y - y₁) / ( x - x₁)

=> -2 = { y - (-3) } / { x - (-1) }

=> -2 = y + 3/x + 1

=> -2(x + 1) = y + 3

=> -2x -2 = y + 3

=> y + 2x +3+2 = 0

=> 2x + y + 5 = 0

<h3>Hence the equation of line perpendicular to it is 2x + y + 5 = 0 .</h3>

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The temperature, H, in °F, of a cup of coffee t hours after it is set out to cool is given by the following equation. H = 70 + 1

Alex73 [517]

Answer:

The temperature a t = 0 is 190 °F

The temperature a t = 1 is 100 °F

The temperature a t = 2 is 77.5 °F

It takes 1.5 hours to take the coffee to cool down to 85°F

It takes 2.293 hours to take the coffee to cool down to 75°F

Step-by-step explanation:

We know that the temperature in °F, of a cup of coffee t hours after it is set out to cool is given by the following equation:

$H(t)=70+120(\frac{1}{4})^t$

a) To find the temperature a t = 0 you need to replace the time in the equation:

$H(0)=70+120(\frac{1}{4})^0\\H(0)=70+120\cdot 1\\H(0) = 70+120\\H(0)=190 \:\°F$

b) To find the temperature after 1 hour you need to:

$H(1)=70+120(\frac{1}{4})^1\\H(1)=70+120(\frac{1}{4})\\H(1) = 70+30\\H(1)=100 \:\°F$

c) To find the temperature after 2 hours you need to:

$H(2)=70+120(\frac{1}{4})^2\\H(2)=70+120(\frac{1}{16})\\H(2) = 70+\frac{15}{2} \\H(2)=77.5 \:\°F$

d) To find the time to take the coffee to cool down $85 \:\°F$ , you need to:

$85 = 70+120(\frac{1}{4})^t\\70+120\left(\frac{1}{4}\right)^t=85\\70+120\left(\frac{1}{4}\right)^t-70=85-70\\120\left(\frac{1}{4}\right)^t=15\\\frac{120\left(\frac{1}{4}\right)^t}{120}=\frac{15}{120}\\\left(\frac{1}{4}\right)^t=\frac{1}{8}$

$\mathrm{If\:}f\left(x\right)=g\left(x\right)\mathrm{,\:then\:}\ln \left(f\left(x\right)\right)=\ln \left(g\left(x\right)\right)$

$\ln \left(\left(\frac{1}{4}\right)^t\right)=\ln \left(\frac{1}{8}\right)$

$\mathrm{Apply\:log\:rule}=\log _a\left(x^b\right)=b\cdot \log _a\left(x\right)$

$t\ln \left(\frac{1}{4}\right)=\ln \left(\frac{1}{8}\right)$

$t=\frac{\ln \left(\frac{1}{8}\right)}{\ln \left(\frac{1}{4}\right)}\\t=\frac{3}{2} = 1.5 \:hours$

e) To find the time to take the coffee to cool down $75 \:\°F$ , you need to:

$75=70+120\left(\frac{1}{4}\right)^t\\70+120\left(\frac{1}{4}\right)^t=75\\70+120\left(\frac{1}{4}\right)^t-70=75-70\\120\left(\frac{1}{4}\right)^t=5\\\left(\frac{1}{4}\right)^t=\frac{1}{24}$

$\ln \left(\left(\frac{1}{4}\right)^t\right)=\ln \left(\frac{1}{24}\right)\\t\ln \left(\frac{1}{4}\right)=\ln \left(\frac{1}{24}\right)\\t=\frac{\ln \left(24\right)}{2\ln \left(2\right)} \approx = 2.293 \:hours$

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3 years ago

What is an equation of the line that passes through the point￼(-1,-3) and is perpendicularly to the line x-2y=14

What is an equation of the line that passes through the point(-1,-3) and is perpendicularly to the line x-2y=14