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kiruha [24]
3 years ago
5

A line with a slope of -7 passes through the points (u,0) and (8,-7) what is the value of u?

Mathematics
1 answer:
laila [671]3 years ago
6 0

Answer:

u = 7

Step-by-step explanation:

Using the slope intercept formula:

\frac{y2 - y1}{x2 - x1}  = slope

Substitute our values from the points and the known slope:

\frac{-7 - 0}{8 - u} = -7

Now we multiply both sides by (8 - u) to isolate the x:

(8-u) x \frac{-7}{8 - u} = -7 (8 - u)

Next we add 56 to both sides:

-7 = -56 + 7u

Finish off by dividing both sides by 7:

7u = 49

u = 7

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10 months ago
Use lagrange multipliers to find the shortest distance, d, from the point (4, 0, −5 to the plane x y z = 1
Varvara68 [4.7K]
I assume there are some plus signs that aren't rendering for some reason, so that the plane should be x+y+z=1.

You're minimizing d(x,y,z)=\sqrt{(x-4)^2+y^2+(z+5)^2} subject to the constraint f(x,y,z)=x+y+z=1. Note that d(x,y,z) and d(x,y,z)^2 attain their extrema at the same values of x,y,z, so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.

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L(x,y,z,\lambda)=(x-4)^2+y^2+(z+5)^2+\lambda(x+y+z-1)

Take your partial derivatives and set them equal to 0:

\begin{cases}\dfrac{\partial L}{\partial x}=2(x-4)+\lambda=0\\\\\dfrac{\partial L}{\partial y}=2y+\lambda=0\\\\\dfrac{\partial L}{\partial z}=2(z+5)+\lambda=0\\\\\dfrac{\partial L}{\partial\lambda}=x+y+z-1=0\end{cases}\implies\begin{cases}2x+\lambda=8\\2y+\lambda=0\\2z+\lambda=-10\\x+y+z=1\end{cases}

Adding the first three equations together yields

2x+2y+2z+3\lambda=2(x+y+z)+3\lambda=2+3\lambda=-2\implies \lambda=-\dfrac43

and plugging this into the first three equations, you find a critical point at (x,y,z)=\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right).

The squared distance is then d\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)^2=\dfrac43, which means the shortest distance must be \sqrt{\dfrac43}=\dfrac2{\sqrt3}.
7 0
3 years ago
Which of the following shows the true solution to the logarithmic equation 3 log Subscript 2 Baseline (2 x) = 3 x = negative 1 x
sineoko [7]

The equation which shows the true solution to the given logarithmic equation is,

x=1

<h3>What is power rule of logarithmic function?</h3>

The power of the log rule says that the number written in the front of the logarithmic function can be raised to the power of the logarithmic function.

For example,

n\log_b(M)=\log_b(M)^n

Given information-

The logarithmic equation given in the problem is,

3\log_2(2x)=3

Using the power rule of the logarithmic function,

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Using the exponent rule of the logarithmic function the above equation can be written as,

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Solve it further,

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Hence, the equation which shows the true solution to the given logarithmic equation is,

x=1

Learn more about the rules of logarithmic function here;

brainly.com/question/13473114

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