Answer:
The solutions listed from the smallest to the greatest are:
x: 
y: -1 1 -1 1
Step-by-step explanation:
The slope of the tangent line at a point of the curve is:
The tangent line is horizontal when 
. Then:
, for all 
, for all
The first four solutions are:
x:
y: 1 -1 1 -1
The solutions listed from the smallest to the greatest are:
x:
y: -1 1 -1 1
<span>y = tan^−1(x2/4)</span>
tan(y) = x2/4
sec2(y) = x/2
y′ = xcos^2(y)/2
<span>cos^2(y) = <span>16x2+16</span></span>
<span>y′ = <span>8x/(<span>x2+16)
let u be x2+16
du is 2x dx
dy = 4 du / u
y = 4 ln (</span></span></span>x2 <span>+ 16)
y at x =0 = </span> 4 ln (<span>16) = 11.09</span>
Answer:
=38.192x
Step-by-step explanation:
I hope this helps you
Answer:
he made the error in step 2
Step-by-step explanation:
The correct answer is: [B]: "40 yd² " .
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First, find the area of the triangle:
The formula of the area of a triangle, "A":
A = (1/2) * b * h ;
in which: " A = area (in units 'squared') ; in our case, " yd² " ;
" b = base length" = 6 yd.
" h = perpendicular height" = "(4 yd + 4 yd)" = 8 yd.
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→ A = (1/2) * b * h = (1/2) * (6 yd) * (8 yd) = (1/2) * (6) * (8) * (yd²) ;
= " 24 yd² " .
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Now, find the area, "A", of the square:
The formula for the area, "A" of a square:
A = s² ;
in which: "A = area (in "units squared") ; in our case, " yd² " ;
"s = side length (since a 'square' has all FOUR (4) equal side lengths);
A = s² = (4 yd)² = 4² * yd² = "16 yd² "
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Now, we add the areas of BOTH the triangle AND the square:
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→ " 24 yd² + 16 yd² " ;
to get: " 40 yd² " ; which is: Answer choice: [B]: " 40 yd² " .
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