The answer is 11 and 14/33.
y = ln x , 1 <= x <= 3, about x axis and n = 10, dy/dx = 1/ x
S = (b a) ∫ 2π y √( 1 + (dy/dx) ^2) dx
so our f(x) is 2π y √( 1 + (dy/dx) ^2)
(b - a) / n = / 3 = (3-1) / 30 = 1/15
x0 = 1 , x1 = 1.2, x2 = 1.4, x3 = 1.6 ....... x(10) = 3
So we have , using Simpsons rule:-
S10 = (1/15) ( f(x0) + 4 f(x1) + 2 f)x2) +.... + f(x10) )
= (1/15) f(1) + f(3) + 4(f(1.2) + f(1.6) + f(2) + f(2.4) + f(2.8)) + 2(f(1.4) + f(1.8) + f(2.2) + f(2.6) )
( Note f(1) = 2 * π * ln 1 * √(1 + (1/1)^2) = 0 and f(3) = 2π ln3√(1+(1/3^2) = 7,276)
so we have S(10)
= 1/15 ( 0 + 7.2761738 + 4(1.4911851 +
There can't be a point-slope form until you have TWO points.
A single point doesn't have any other form.
There are an infinite number of lines that all go through a single point.
Every one of them has a different slope, and a different point-slope form.
Step-by-step explanation: Remember that the parent graph f(x) = x², is a parabola that opens up and passes through the origin.
Notice that g(x) = x² + 2 has 2 added to the parent term.
This means that g(x) = x² + 2 is the graph of f(x) = x² translated two units up.
<h3>Answer: angle T = 70</h3>
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Work Shown:
Quadrilateral RSTU is a kite. In geometry, any kite has two pairs of adjacent congruent sides. In this case, RU = RS is one pair of adjacent congruent sides (single tickmarks), while TU = TS is the other pair of adjacent congruent sides (double tickmarks).
Draw diagonal line segment TR. This forms triangles TUR and TSR.
Through the SSS (side side side) congruence theorem, we can prove that the two triangles TUR and TSR are congruent.
Then by CPCTC (corresponding parts of congruent triangles are congruent), we can say,
angle U = angle S = 90
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Re-focus back on quadrilateral RSTU (ignore or erase line segment TR). The four angles of any quadrilateral will always add to 360 degrees. Let x be the measure of angle T.
(angleU)+(angleR)+(angleS)+(angleT) = 360
90+110+90+x = 360
290+x = 360
290+x-290 = 360-290 ... subtract 290 from both sides
x = 70
<h3>angle T = 70</h3>
