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3 years ago

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What is the slope of a line perpendicular to the line whose equation is 6x+y=66x+y=6. fully reduce your answer?

1 answer:

shepuryov [24]3 years ago

8 0

It looks like your equation repeated twice. I'm assuming it's 6x + y = 6.

I'm going to put your equation into slope-intercept form: y = - 6x + 6.

For the slope to be perpendicular, we do the opposite reciprocal of the slope given. So your slope of a perpendicular line is 1/6.

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F(x) = -9x + 2 and g(x) = -9x + 6, find (f - g)(7)

Lemur [1.5K]

Answer:

I think there is an error in the question because

(f-g) = -4

(f-g) (7) = NO SOLUTION

Step-by-step explanation:

$f(x) = -9x + 2 \\g(x) = -9x + 6\\(f - g)(7)\\(f - g) = -9x + 2 - (-9x+6)\\(f - g) = -9x +2 +9x-6\\(f - g) = -9x +9x+2-6\\(f - g) = -4$

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3 years ago

Find the value of x. <br><br> A. 60<br><br> B. 50 <br><br> C. 40 <br><br> D. 30

garri49 [273]

We can see that two sides of that triangle area equal

so, their corresponding angle will also be equal

now, we know that

sum of all angles in any triangle is always 180

so, we can get equation as

$x+100+x=180$

now, we can solve for x

$2x+100=180$

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$x=40$

so, option-C..............Answer

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3 years ago

Read 2 more answers

ben has 6 pounds of bolts in his auto shop he has to seperate them into boxeseach with 1/3 pound of bolts how many boxes can he

IgorLugansk [536]

Answer:

2 pounds in each

Step-by-step explanation:

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3 years ago

What is the midpoint of (-3,-7) and (9,8)

Sliva [168]

Hope this will help you. It is a little hard to explain.

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3 years ago

The voltage V in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance R is slowly increasin

hoa [83]

Answer:

$\frac{dI}{dt}=-0.00016 A/s$

Step-by-step explanation:

We are given that

By ohm's law

$V=IR$

R=389 ohm

I=0.03 A

$\frac{dV}{dt}=-0.06V/s$

$\frac{dR}{dt}=0.07ohm/s$

We have to find rate of change of current I means $\frac{dI}{dt}$

Differentiate the equation w.r.t t

$\frac{dV}{dt}=\frac{dI}{dt}R+I\frac{dR}{dt}$

Substitute the values then we get

$-0.06=\frac{dI}{dt}\times 389+0.03\times 0.07$

$-0.06=389\frac{dI}{dt}+0.0021$

$-0.06-0.0021=389\frac{dI}{dt}$

$-0.0621=389\frac{dI}{dt}$

$\frac{dI}{dt}=\frac{-0.0621}{389}=-0.000160 A/s$

Hence, the current I is changing at the rate=-0.00016A/s

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3 years ago