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8090 [49]
4 years ago
5

Determine which justification best fits step 3 for the following proof

Mathematics
2 answers:
devlian [24]4 years ago
7 0

Answer: it's actually angle addition postulate

explanation: angle addition postulate is the one that allows you to say that the individual parts of a bigger angle can add up to equal that bigger angle

fgiga [73]4 years ago
6 0

Answer:

D

Step-by-step explanation:

Since ABC is collinear and angle 3 is a right angle, adding 1&2 together should equal 3, which comes out to being equal to line ABC

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Trava [24]

Answer:

X=18

Step-by-step explanation:

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3 years ago
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What is t²=81. thanks!!
seraphim [82]

Answer:

t= 9, t= -9

Step-by-step explanation:

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3 years ago
Determine the length of side BD in the triangle below.<br> 12<br> 15<br> 13<br> D<br> B<br> с
Romashka [77]

9514 1404 393

Answer:

  BD = 14

Step-by-step explanation:

The Pythagorean theorem can be used to find the lengths of CD and BC.

  CD² +12² = 13²

  CD = √(169 -144) = √25 = 5

__

  BC² +12² = 15²

  BC = √(225 -144) = √81 = 9

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The length BD is then ...

  BD = BC +CD

  BD = 9 + 5 = 14

The length of side BD is 14 units.

7 0
3 years ago
What is the answer please
ZanzabumX [31]

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centre

Step-by-step explanation:

3 0
3 years ago
Help me about this integral
NemiM [27]

The gradient theorem applies here, because we can find a scalar function <em>f</em> for which ∇ <em>f</em> (or the gradient of <em>f</em> ) is equal to the underlying vector field:

\nabla f(x,y,z)=\langle2xy,x^2-z^2,-2yz\rangle

We have

\dfrac{\partial f}{\partial x}=2xy\implies f(x,y,z)=x^2y+g(y,z)

\dfrac{\partial f}{\partial y}=x^2-z^2=x^2+\dfrac{\partial g}{\partial y}\implies\dfrac{\partial g}{\partial y}=-z^2\implies g(y,z)=-yz^2+h(z)

\dfrac{\partial f}{\partial z}=-2yz=-2yz+\dfrac{\mathrm dh}{\mathrm dz}\implies\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=C

where <em>C</em> is an arbitrary constant.

So we found

f(x,y,z)=x^2y-yz^2+C

and by the gradient theorem,

\displaystyle\int_{(0,0,0)}^{(1,2,3)}\nabla f\cdot\langle\mathrm dx,\mathrm dy,\mathrm dz\rangle=f(1,2,3)-f(0,0,0)=\boxed{-16}

5 0
3 years ago
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