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Sergeeva-Olga [200]

3 years ago

10

What is the equation of the line that is perpendicular to f(x) and passes through the point (4, –6)? a.4x + 3y = –2 b. 3x + 4y =

–12 c.4x – 3y = 34 d.3x – 4y = 36

1 answer:

user100 [1]3 years ago

4 0

Answer:

8

Step-by-step explanation: because its even

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MAVERICK [17]

Answer:

y=-x-2

Step-by-step explanation:

To find the slope of the line, you can use the slope formula. I did this with (-3,1) and (2,-4) to get: (1+4)/(-3-2). This ends up as -5/5, which simplifies to -1.

next I plugged in the slope and one point into the line equation: y=mx+b. I plugged in (-4)=(-1)(2)+b. This equated to b=-2.

Check Work:

The equation -4=(-1)(2)-2 is true

The equation 1=(-1)(-3)-2 is true

5 0

3 years ago

Monique needs to buy a baseball for $20, a soccer ball for $15, and a set of tennis balls for $12.

MAXImum [283]

(1 - y)12 represents the price of the set of tennis balls after a coupon was applied.

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3 years ago

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Una tienda virtual ofreció el 15% de descuento el dia de su aniversario.ese dia Jorge compro una lavadora que regularmente cuest

makvit [3.9K]

Answer:

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Step-by-step explanation:

8 0

2 years ago

Approximately how much water does the average american use every day?

Sphinxa [80]

Answer: Estimates vary, but each person uses about 80-100 gallons of water per day

4 0

3 years ago

Find the laplace transform of f(t) = cosh kt = (e kt + e −kt)/2

iren2701 [21]

Hello there, hope I can help!

I assume you mean $L\left\{\frac{ekt+e-kt}{2}\right\}$
With that, let's begin

$\frac{ekt+e-kt}{2}=\frac{ekt}{2}+\frac{e}{2}-\frac{kt}{2} \ \textgreater \ L\left\{\frac{ekt}{2}-\frac{kt}{2}+\frac{e}{2}\right\}$

$\mathrm{Use\:the\:linearity\:property\:of\:Laplace\:Transform}$
$\mathrm{For\:functions\:}f\left(t\right),\:g\left(t\right)\mathrm{\:and\:constants\:}a,\:b$
$L\left\{a\cdot f\left(t\right)+b\cdot g\left(t\right)\right\}=a\cdot L\left\{f\left(t\right)\right\}+b\cdot L\left\{g\left(t\right)\right\}$
$\frac{ek}{2}L\left\{t\right\}+L\left\{\frac{e}{2}\right\}-\frac{k}{2}L\left\{t\right\}$

$L\left\{t\right\} \ \textgreater \ \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{t\right\}=\frac{1}{s^2} \ \textgreater \ L\left\{t\right\}=\frac{1}{s^2}$

$L\left\{\frac{e}{2}\right\} \ \textgreater \ \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{a\right\}=\frac{a}{s} \ \textgreater \ L\left\{\frac{e}{2}\right\}=\frac{\frac{e}{2}}{s} \ \textgreater \ \frac{e}{2s}$

$\frac{ek}{2}\cdot \frac{1}{s^2}+\frac{e}{2s}-\frac{k}{2}\cdot \frac{1}{s^2}$

$\frac{ek}{2}\cdot \frac{1}{s^2} \ \textgreater \ \mathrm{Multiply\:fractions}: \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d} \ \textgreater \ \frac{ek\cdot \:1}{2s^2} \ \textgreater \ \mathrm{Apply\:rule}\:1\cdot \:a=a$
$\frac{ek}{2s^2}$

$\frac{k}{2}\cdot \frac{1}{s^2} \ \textgreater \ \mathrm{Multiply\:fractions}: \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d} \ \textgreater \ \frac{k\cdot \:1}{2s^2} \ \textgreater \ \mathrm{Apply\:rule}\:1\cdot \:a=a$
$\frac{k}{2s^2}$

$\frac{ek}{2s^2}+\frac{e}{2s}-\frac{k}{2s^2}$

Hope this helps!

3 0

4 years ago