Answer:
Acute
Step-by-step explanation:
The answer is: 10 x¹³ y¹⁰ .
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1x^8 * 2y^(10) * 5x^5 =
1* 2* 5 * x^8 * x^5 * y^(10) =
10 * x^(8+5) * y^(10) =
10 * x^(13) * y^(10) = 10 x^(13) y^10 ; write as:
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10 x¹³ y¹<span>⁰ .
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To find the product of (4x-5y)^2,
we can rewrite the problem as:
(4x-5y)(4x-5y) (two times because it is squared)
Now, time to use that old method we learned in middle school:
FOIL. (Firsts, Outers, Inners, and Lasts)
FOIL can help us greatly in this scenario.
Let's start by multiplying the 'Firsts' together:
4x * 4x = <em>16x^2</em>
Now, lets to the 'Outers':
4x * -5y = <em>-20xy</em>
Next, we can multiply the 'Inners':
-5y * 4x = <em>-20xy</em>
Finally, let's do the 'Lasts':
-5y * -5y = <em>25y</em>^2
Now, we can take the products of these equations from FOIL and combine like terms. We have: 16x^2, -20xy, -20xy, and 25y^2.
-20xy and -20xy make -40xy.
The final equation (product of (4x-5y)^2) is:
16x^2 - 40xy + 25y^2
Hope I helped! If any of my math is wrong, please report and let me know!
Have a good one.
The slope of the given line is ∆y/∆x = (4-6)/(5-1) = -2/4 = -1/2. The slope of the desired (perpendicular) line is the negative reciprocal of this, -1/(-1/2) = 2.
The point-slope form of the equation for a line with slope m through point (h, k) is
... y - k = m(x - h)
For slope 2 and the given point, your equation can be
... y + 6 = 2(x + 2)
