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CaHeK987 [17]
3 years ago
8

Solve this equation for P n=p-k/j

Mathematics
1 answer:
ololo11 [35]3 years ago
5 0
Just subtract add the K/J to both sides. This isolates P by cancelling out the -K/J because -K/J+K/J=0. Your final equation should read: P=N+K/J
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3 years ago
Combine like terms.
Alex Ar [27]
7 + 4x - 10x + 15 = 22 - 6x

You just need to sum terms of polynomial with the same  stage of x and not to mix them out. Note that terms without x are actually terms with x^0 because x^0 = 1

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4 years ago
Find the equation of the tangent of x^2- xy + y^2 =7 at (-1, 2)
Citrus2011 [14]

Answer:

<u>It</u><u> </u><u>is</u><u> </u><u>3</u><u>y</u><u> </u><u>=</u><u> </u><u>4</u><u>x</u><u> </u><u>+</u><u> </u><u>1</u><u>0</u>

Step-by-step explanation:

Let's first get the slope of the curve.

[ slope is the derivative of the equation ]

{x}^{2}  - xy +  {y}^{2}  = 7

introduce dy/dx :

\frac{d}{dx} ( {x}^{2}  - xy +  {y}^{2} ) =  \frac{d}{dx} (7) \\  \\ 2x - (y +  \frac{dy}{dx} ) + 2y \frac{dy}{dx}  = 0

make dy/dx the subject:

2x - y -  \frac{dy}{dx}  + 2y \frac{dy}{dx}  = 0 \\  \\ 2y \frac{dy}{dx}  -  \frac{dy}{dx}  = y - 2x \\  \\  \frac{dy}{dx} (2y - 1) = y - 2x \\  \\  \frac{dy}{dx}  =  \frac{y - 2x}{2y - 1}

At point (-1, 2):

\frac{dy}{dx}  =  \frac{2 - 2( - 1)}{2(2) - 1}  \\  \\ slope =  \frac{4}{3}

but a tangent has the same slope as the curve:

y = mx + c

m is the slope

c is the y-intercept

At (-1, 2):

2 = ( - 1 \times  \frac{4}{3} ) + c \\  \\ c = 2 +  \frac{4}{3}  \\  \\ c =  \frac{10}{3}

equation:

y =  \frac{4}{3} x +  \frac{10}{3}  \\  \\ { \boxed{3y = 4x + 10}}

8 0
3 years ago
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erma4kov [3.2K]

Answer:

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4 years ago
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Partial product method of 475x4
vazorg [7]

Answer:

(400x4)+(70x4)+(5x4)=?

Step-by-step explanation:

Divide the greater number into hundreds tens and ones then add the products to get your answer

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2 years ago
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