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maksim [4K]
3 years ago
14

Find the value of y, given that m_ KLM = 137º.

Mathematics
2 answers:
mr Goodwill [35]3 years ago
7 0

Answer:

y≈ 5.63

Step-by-step explanation:

m<KLM=137

137= 47+16y

137= 47 +16y    Subtract 47 on both sides.

90=16y   Divide 16 on both sides.

5.625=y   Round

5.63≈ y

Musya8 [376]3 years ago
6 0

Answer:

y ≈ 5.63

Step-by-step explanation:

m∠KLN and m∠NLM have to add up to m∠KLM

m∠KLN = 47°

m∠NLM = 16y°

m∠KLM = 137°

47 + 16y = 137

16y = 90

y = 45/8

y = 5.625

y ≈ 5.63

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What’s the perpendicular slope of 4/3
never [62]

Answer:

-3/4

Step-by-step explanation:

Perpendicular slopes are reciprocal opposites of the original slope. This means that the reciprocal of 4/3 is 3/4, and the opposite of 3/4 is -3/4.

3 0
3 years ago
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Find the 12th term of the geometric sequence 5, -25, 125, ...5,−25,125,...
katovenus [111]

Answer:

  • a_{12}=-244140625

Step-by-step explanation:

Considering the geometric sequence

5,-25,\:125,\:...

a_1=5

As the common ratio 'r' between consecutive terms is constant.

\mathrm{Compute\:the\:ratios\:of\:all\:the\:adjacent\:terms}:\quad \:r=\frac{a_{n+1}}{a_n}

r=\frac{-25}{5}=-5

r=\frac{125}{-25}=-5

The general term of a geometric sequence is given by the formula:  

a_n=a_1\cdot \:r^{n-1}

where a_1 is the initial term and r the common ratio.

Putting n = 12 , r = -5 and a_1=5 in the general term of a geometric sequence to determine the 12th term of the sequence.

a_n=a_1\cdot \:r^{n-1}

a_n=5\left(-5\right)^{n-1}

a_{12}=5\left(-5\right)^{12-1}

      =5\left(-5^{11}\right)

\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a

       =-5\cdot \:5^{11}

\mathrm{Apply\:exponent\:rule}:\quad \:a^b\cdot \:a^c=a^{b+c}

        =-5^{1+11}     ∵ 5\cdot \:5^{11}=\:5^{1+11}

        =-244140625

Therefore,

  • a_{12}=-244140625
6 0
3 years 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.

The Lagrangian is

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

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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}.
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3 years ago
The table represents some points on the graph of a linear function.
labwork [276]

Answer:

Step-by-step explanation:

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2 years ago
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Answer:

-11

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