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# I don't understand line: 12*e^ln(x)=1.5*e^x the result for this is approx. x=0.1444213531375 ?

I have attempted to understand the line in your solution above but I need more information. Thanks, Peter

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## Suggested Questions And Answer :

### I don't understand line: 12*e^ln(x)=1.5*e^x the result for this is approx. x=0.1444213531375 ?

e^ln=1 so e^ln(x)=x 12*e^ln(x)...12*x

### area between curves Xto the fourth+ysquared=12 and x=y

The line x=y bisects the enclosed area of the other curve x^4+y^2=12. The x coords of the points of intersection are given by x^4+x^2-12=0=(x^2+4)(x^2-3), so x=±sqrt(3). The y coords are given by y=x, so the points are (sqrt(3),sqrt(3)) and (-sqrt(3),-sqrt(3)).  The picture shows how the area is bisected, so all we need to do is to find the area of the whole figure and halve it. But x^4+y^2=12 is symmetrical so to find the area of the whole figure, we just look at a quadrant and find its area, then we double the result to find the area enclosed by the line and the curve. When y=0 we have x intercept: x^4=12, so the area of a quadrant is integral of ydx between x=0 and x=12^(1/4), where y=sqrt(12-x^2) and the integrand is sqrt(12-x^2)dx. Let x=sqrt(12)sin(z); dx=sqrt(12)cos(z)dz, so the integrand becomes: sqrt(12-12sin^2(z))sqrt(12)cos(z)dz=12cos^2(z)dz. cos(2z)=2cos^2(z)-1, so 2cos^2(z)=1+cos(2z) and 12cos^2(z)=6+6cos(2z). But x=sqrt(12)sin(z), so if x=0, z=0 and if x=12^(1/4), sin(z)=12^(1/4)/12^(1/2)=12^(-1/4), z=sin^-1(12^(-1/4)). sin(z)=12^(-1/4); cos(z)=sqrt(1-12^(-1/2)). The result of integration is: 6z+3sin(2z) between the limits 0 and sin^-1(12^(-1/4)). Plugging in values we get the area of a quadrant is 3.4033 + 2.7189 = 6.1222 approx., making half the area of the figure12.2443 approx. It doesn't matter that the line x=y splits the area obliquely, it is still half the area.

### volume of a solid generated by revolving area bounded by the curves about the indicated axis?

(a) The picture shows the given curve and line. The line PQ shows the height of a cylinder where P is a point on the curve. The cylinder is the result of rotating the line PQ about the y-axis as the axis of rotation (x=0). The line and curve intersect when x=6x-x^2, that is, x(5-x)=0 at x=0 and 5. The cylinder is hollow with an infinitesimally thin wall, thickness dx. The radius of the cylinder is x, the height of the cylinder is 5x-x^2 as can be seen by the geometry, and the area of this cylindrical shell is found by "rolling out" the cylinder into a rectangular lamina. So the length of the rectangle is 2πx, the circumference of the cylinder, and the height 5x-x^2, making the area 2πx^2(5-x). The volume of the cylinder is the area multiplied by the thickness dx: 2πx^2(5-x)dx. The sum of the volumes of the cylindrical shells gives us the volume of rotation of the shape bounded by the line and curve. This sum is ∫(2πx^2(5-x)dx) between the limits x=0 to 5. The integral evaluates to 2π(5x^3/3-x^4/4) and, applying the limits: 2π(625/3-625/4)=625π/6=327.25 cu units approx. (b) This time we have a cylinder radius 4-x and height PR=2√(4-x)-4+x. The volume of the cylindrical shell is 2π(4-x)(2√(4-x)-4+x)dx and the integral ∫(2π(4-x)(2√(4-x)-4+x)dx). Since y=4-x we can replace 4-x with y and dx by -dy. The integral becomes -2π∫(y(2√y-y)dy). Since x=4-y and x=(16-y^2)/4 we can write 4-y=(16-y^2)/4 and solve for y. 16-4y=16-y^2 so y(4-y)=0 and y=0 and 4 as evidenced in the picture. The limits are 4≥y≥0. The minus on the integral is changed to plus if we reverse the limits for y. So we need to evaluate the definite integral: 2π∫((2y^(3/2)-y^2)dy)=2π[(4/5)z^(5/2)-z^3/3] for 0≤y≤4. This evaluates to 2π(128/5-64/3)=128π/15=26.81 cu units approx. (c) The intersection point between the parabola and line y=4 is given by 16=8x, so x=2 and y=4: (2,4). The parabola meets the y-axis (x=0) at the origin (0,0). For a point P(x,y) on the parabola 4-y is the radius of a disc of thickness dx and therefore the volume of the disc is π(4-y)^2dx=π(16-8y+y^2)dx. Since y^2=8x we can evaluate the integral π∫((16-16√(2x)+8x)dx) for 0≤x≤2 to find the volume of revolution around the line y=4. This evaluates to 8π[2x-4√2x^(3/2)/3+4x^2] for 0≤x≤2=8π(4-16/3+16)=352π/3=368.61 cu units.

### Write an equation of the line passes through P and is parallel to the given equation. P(6,0) y=4

??????????????? "passesesesese thru" ??????????? a line GO THRU points. Me assume yu want a line tu go thru point=(6,0) Y=4 look line formula for a horizontal line. Horizontal line thru (6,0) be y=6

### graph -5x + 3y = -15

The first thing to do is to rearrange the equation into standard form: 3y=5x-15. Now divide through by 3: y=5x/3-5. We only need two points to draw a straight line graph. The normal process is to find the intercepts, that is, the points where the line cuts the axes. To find the y intercept, where the line cuts the y axis, we put x=0. That means we mark the point -5 on the y axis. If we put x=0 in the original equation we get: 3y=-15 and so y=-5, the same result. To find the x intercept we put y=0. If we do this in the original equation we get: -5x=-15 so x=3. That's probably easier than putting y=0 in the rearranged equation, but the result is the same: 5*3/3-5=0. So we mark the point 3 on the x axis.now we can draw a line through the two intercepts and that's the graph. The slope of the graph is 5/3, and the y intercept, as we've seen. The standardised equation exposes the slope and intercept whereas the original equation didn't. The slope is bigger than 1, so it's steeper than 45 degrees and it slopes to the right, a positive slope.

### Equations of the line.

I dont understand yer "quesshun" yu sae it form triangel with x & y axises That impli tu me a rite triangel with base along the x-axis & 1 side along y-axis. then yu want a point on the diagonal tu go thru (3,4) slope gotta be negativ, so line hit positiv side av both axisis Draw lines thru (3,4) giv y-intersept=4*3*tangent(theta) & x-intersept=3+4*sine(theta) Area=(1/2)(x-intersept)*(y-intersept) I'm tired now.   Dont wanna du no more werk on this

line 1: 3y+4x=-6...strate line: y=-(4/3)x -2 line 2: 4y-3x=12...strate line y=+(3/4)x+3 line 1 slope=-4/3  line 2 slope=+3/4...this =-1/slope av line1...that meen 2 lines 90 deg apart tu see if 2 lines hit, put y from line 1 intu line 2 4*[-(4/3)x -2] -3x=12 -(16/3)x -3x=12 +8 -(25/3)x=20 x=-(3/25)*20 x=-2.4 y=-(4/3)x -2=(4/3)*(2.4) -2  =3.2-2 y=1.2 2 lines hit at (-2.4,+1.2)

### what is the approximate slope of the tangent line to the curve x^(3) +y^(3) =xy at x=1 ? I'm a little stuck. I believe this involves implicit differentiation. Thanks

what is the approximate slope of the tangent line to the curve x^(3) +y^(3) =xy at x=1 ? I'm a little stuck. I believe this involves implicit differentiation. Thanks   Impicit differentiation it is! The rule is: when we differentiate f(y) = y^3, wrt x, then df/dx = (df/dy)*(dy/dx), i.e. df/dx = (d(y^3)/dy)*(dy/dx) df/dx = (3y*2)*(dy/dx) So, when differentiating f(y) wrt x, simply diffrentiate f(y) wrt y and then multiply that result by y' (=dy/dx) What we have now is: x^(3) +y^(3) =xy differentiating wrt x, 3x^2 + (3y^2).y' = x.y' + y (using the product rule on the lhs there) 3x^2 - y = (x - 3y^2).y' y' = (3x^2 - y) / (x - 3y^2) Using again x^(3) +y^(3) = xy. At x = 1, 1^3 + y^3 = 1.y y - y^3 = 1 Graphing this cubic and reading off the graph. At x = 1, y = -1.32 (approx) Using these values with y' = (3x^2 - y) / (x - 3y^2) y' = (3.(1^2) - (-1.32)) / (1 - 3.(-1.32)^2) y' = (3 + 1.32) /( 1 - 5.2272) y' = 4.32 / (-4.2272) y' = -1.0219 y = -1.022 (approx)

### (x-y)*10=200 and 5(x+y)=200 porportion.??

me dont no wot yu meen bi (15,5), but 15/5=3 line 1: x-y=20 line 2: x+y=40 add tugether...2x=60 x=30 x+y=40, so y=40-x=10 x/y=30/10=3

### how do i find the slope of the line parallel to the graph of 3x-4y=12?

how do i find the slope of the line parallel to the graph of 3x-4y=12? The slope of parallel lines is the same. So, manipulate the given equation into slope/intercept form and you will have the slope of any parallel line. 3x-4y=12 -4y = -3x + 12 y = (-3x + 12) / -4 y = 3/4 x - 3 The slope of this line is 3/4, and that is also the slope of any line parallel to it.