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yanalaym [24]
3 years ago
11

Pleaseee Help. What is the value of x in this simplified expression?

Mathematics
1 answer:
Oduvanchick [21]3 years ago
7 0

9514 1404 393

Answer:

  x = 7

  y = 5

Step-by-step explanation:

The applicable rule of exponents is ...

  a^-b = 1/a^b

__

For a=-j and b=7,

  (-j)^-7 = 1/(-j)^7   ⇒   x = 7

For a=k and b=-5,

  k^-5 = 1/k^5   ⇒   y = 5

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Solve for X (Also plz tell me what Y is)
erastova [34]
Since there is no value for either X or Y given, you can only change the formula around to equal X.

y=45x+38

First move the 38 to the left side by canceling it out (subtract 38 from both sides).

y-38=45x+38-38

So now we have:
y-38=45x

Now isolate the x on the right side by dividing both sides by 45.

(y-38) / 45 = x

So,

x = (y-38) / 45

(If the answer you’re looking for is a specific number, then the value for X or Y needs to be given!)
3 0
3 years ago
Find a vector equation and parametric equations for the line through the point (7,4, 5) and parallel to the vector 3i 2j-k .
aleksley [76]

Answer: vector equation r = (7+3t)i + (4+2t)j + (5 - 5t)k

parametric equations: x = 7 + 3t; y = 4 + 2t; z = 5 - 5t

Step-by-step explanation: The vector equation is a line of the form:

r = r_{0} + t.v

where

r_{0} is the position vector;

v is the vector;

For point (7,4,5):

r_{0} = 7i + 4j + 5k

Then, the equation is:

r = 7i + 4j + 5k + t(3i + 2j - k)

<u><em>r = (7 + 3t)</em></u><u><em>i</em></u><u><em> + (4 + 2t)</em></u><u><em>j </em></u><u><em>+ (5 - 5t)</em></u><u><em>k</em></u>

The parametric equations of the line are of the form:

x = x_{0} + at

y = y_{0} + bt

z = z_{0} + ct

So, the parametric equations are:

<em><u>x = 7 + 3t</u></em>

<em><u>y = 4 + 2t</u></em>

<em><u>z = 5 - 5t</u></em>

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3 years ago
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Andrew [12]
It can be a yes because the line cross and make a perpendicular
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Find the area of the composite area
QveST [7]
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Total area= 91+24.5= 115.5 in^2
8 0
3 years ago
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Hoochie [10]

Answer:

i belive it is y=-4x+5

Step-by-step explanation:

i just did the test

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3 years ago
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