Guide :

Evaluate ;Find 3x2 - y3 - y3 - z if x = 3, y = -2, and z = -5.

please help with this Im really struggling with this

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Evaluate the following expression using the value... - OpenStudy


Evaluate the following expression using the values given: ... Find 3x2 - y3 - y3 - z if x = 3, y = -2, and z = -5.\[3x^2-y^3-y^3-z\] ... Find 3x2 - y3 - y3 - z if x ...
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Evaluate the following expression using the values given ...


Evaluate the following expression using the values given: Find 3x2 − y3 − y3 − z if x = 3, y = −2, and z = −5. Numerical Answers Expected!
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Evaluate the following expression using the values given ...


Evaluate the following expression using the values given: Find 3x2 - y3 - y3 - z if x = 3, y = -2, and z = -5. weegy; Answer; Search; More; ... Find 3x2 - y3 - y3 - z ...
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View question - Evaluate the following expression using the ...


Evaluate the following expression using the values given: Find 3x2 - y3 - y3 - z if x = 3, y = -2, and z = -5. 0 . ... Find 3x2 - y3 - y3 - z if x = 3, y = -2, ...
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Part 1 1. Evaluate the following expression using the values ...


Evaluate the following expression using the values ... Find 3x2 y3 y3 z if x = 3, y = 2, and z = 5. ... 2x3 + 3y2 − 17 when x = 3 and y = 4. 3. Evaluate the ...
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I just recieved this question on homework and hav... - OpenStudy


I just recieved this question on homework ... Evaluate the following expression using the values given: Find 3x2 - y3 - y3 - z if x = 3, y = -2, and z = -5 ...
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How do you find x - 3y if x = 3 and y = -2? | Socratic


9 substitute the values x = 3 and y = - 2 into the expression x - 3y rArr 3 -3 (- 2 ) = 3 + 6 = 9 -3( -2 ) ... How do you find x - 3y if x = 3 and y = -2? Algebra ...
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Suggested Questions And Answer :


Given the function defined by t(x) = x3 – 3x2 + 2x – 3, find t(10).

????????????????? yu wanna EVALUATE  a funkshun at x=10 ?????????????
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Evaluate ;Find 3x2 - y3 - y3 - z if x = 3, y = -2, and z = -5.

????????????? "3x2" ????? maebee that shood be 3*x ??? or maebee 3*x^2 ?????
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Find the values of A, B, C, D, and E by evaluating 3x2 + x + 1 at x = 2 based on the given algorithm.Working through each step of the algorithm is shown below:

http://web.cs.ucdavis.edu/~ma/ECS20/hw4-sol.pdf
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find the center of mass of the solid of uniform density biunded by the graphs of the equations

x^2+y^2=a^2 is a cylinder with its circular cross-section in the x-y plane; the cylinder is sliced at an angle by the plane z=cy, forming a wedge. y=0 is the x-z plane and z=0 is the x-y plane; these planes and the conditions that y and z are positive impose further constraints on the shape. The cylinder is sliced in half by the x-z plane so the cross-section will be a semicircle and only the upper half is required. The semicircular end rests on the z=0 plane, and doesn't extend beyond the vertex of the wedge. The length of the wedge is ac, because the slope of the angle of the wedge is tan^-1(c) to the y axis and the radius of the wedge is a. Viewed from the side, along the x axis, the wedge is a right-angled triangle with sides a (height), ac (length) and hypotenuse a√(1+c^2). In everyday terms, the wedge looks like the end of a lipstick that has been sliced across. Going back to the side view of the wedge, we can see that the flat end has a height a. As we move along the wedge towards the point the height changes and shrinks to zero when we reach the vertex. The length at a point P(y,z) is a-y. This height is the distance of a chord to the circumference, starting at height a. The first task is to work out the area of a segment. We do this by subtracting the area of a triangle from the area of a sector, given the distance of the chord from the circumference. This distance is a-y, so the distance from the centre of the circle is, conveniently, y. y/a=cosø where ø is half the angle subtended by the arc of the sector. So ø=cos^-1(y/a) and the area of the sector is given by 2ø/2π=ø/π=A/πa^2, or A=a^2ø (ø is measured in radians). The area of the isosceles triangle formed by joining the radial ends of the arc of the sector is B=aysinø=y√(a^2-y^2), and the area of the segment is A-B=a^2cos^-1(y/a)-y√(a^2-y^2). (Note that when y=0, this expression becomes πa^2/2, the area of the semicircle.) The volume of the segment is (a^2cos^-1(y/a)-y√(a^2-y^2))dz where infinitesimal dz is the thickness of the segment. The mass of the segment is directly proportional to its volume because the density is uniform, so the volume is sufficient to represent the mass. The centre of gravity (COG) of the wedge can be found by summing moments about a point G(x,y,z) which is the COG. This sum has to be zero. Since a semicircle is symmetrical in the x-y plane about the y axis, we know the x coord of G must be zero. We need G(0,g,h) where g is the height along the z axis of the COG and h the z position. We need to find the COG of a segment first. The length of the chord of a segment is 2√(a^2-y^2). Earlier we discovered this when we were finding out the areas of the sector and isosceles triangle. Consider just two dimensions. The length of the chord is one dimension and if we create a rectangle from this length and an infinitesimal width dy, the area will be 2√(a^2-y^2)dy. If we think of the rectangle as having mass, its mass is proportional to its area, so area is a sufficient measure for mass. Let's call the COG of the segment point C(x,y)=(0,q), then the moment of the rectangle about the COG=C(0,q) is mass times distance from C=2(q-y)√(a^2-y^2)dy. If we sum the rectangles over the interval y to a we have 2∫((q-y)√(a^2-y^2)dy)=0 for y≤y≤a. The interval looks strange, but it means that y is just a general value, which is determined when we return to the wedge. What we are trying to do is to find q relative to y. We need to evaluate the definite integral, so we split it: 2q∫(√(a^2-y^2)dy)-2∫(y√(a^2-y^2)dy. Call these (a) and (b). To evaluate (a), let y=asinu, then dy=acosudu; √(a^2-y^2)=acosu and the integral becomes 2a^2q∫(cos^2(u)du). Since cos(2u)=2cos^2(u)-1, cos^2(u)=½(cos(2u)+1), so we have a^2q∫((cos(2u)+1)du)=a^2q(sin(2u)/2+u). Since sin(2u)=2sinucosu, this becomes a^2q(sinucosu+u)=a^2q(y/a * √(1-(y/a)^2)+sin^-1(y/a))=qy√(a^2-y^2)+a^2qsin^-1(y/a). Now we apply the limits for y: πa^2q/2-qy√(a^2-y^2)+a^2qsin^-1(y/a). To evaluate (b), let u=a^2-y^2, then du=-2ydy and the integral is +∫√udu = (2/3)u^(3/2)=(2/3)(a^2-y^2)^(3/2). Applying the limits we have: -(2/3)(a^2-y^2)^(3/2). (a)+(b)=πa^2q/2-qy√(a^2-y^2)+a^2qsin^-1(y/a)-(2/3)(a^2-y^2)^(3/2)=0. Multiply through by 6: 3πa^2q-6qy√(a^2-y^2)+6a^2qsin^-1(y/a)-4(a^2-y^2)^(3/2)=0. From this q=4(a^2-y^2)^(3/2)/(3πa^2-6y√(a^2-y^2)+6a^2sin^-1(y/a)). Fearsome though this looks, when y=0 and z=0, q=4a^3/(3πa^2)=4a/(3π), which is in fact the COG of a semicircle. When y=a and z=ac, q=0. Strictly speaking, we should write q(y) instead of just q, showing q to be a dependent variable rather than a constant. To find the COG of the wedge, we need to find the moments of the COGs of the segments. For a segment distance y above and cy along the z axis, we have the above expression for q. So q(y) is the y coord of the COG of the segment and cy is the z coord. The x coord is 0. The mass of the segment is (a^2cos^-1(y/a)-y√(a^2-y^2))dz, as we saw earlier, and this mass can be considered to be concentrated or focussed at the COG of the segment (0,q(y),cy). The COG of the wedge is at G(0,g,h) so the distance of the mass of the segment from G is g-q(y) and h-cy in the y and z directions. The y-moment is (g-4(a^2-y^2)^(3/2)/(3πa^2-6y√(a^2-y^2)+6a^2sin^-1(y/a))(a^2cos^-1(y/a)-y√(a^2-y^2))dz and the z-moment is (h-cy)(a^2cos^-1(y/a)-y√(a^2-y^2))dz. The sums of these individual components are zero. These lead to rather complicated integrals, requiring considerably more space than is available to work out, even if I knew how to do so!
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Evaluate the derivative at the given value of x: If y = x^4 +4x^3 – 2x +2, find dy/dx |x = -1

Evaluate the derivative at the given value of x: If y = x^4 +4x^3 – 2x +2, find dy/dx |x = -1 dy/dx = 4x^3 + 12x^2 - 2 dy/dx(-1) = 4(-1)^3 + 12 (-1)^2 - 2                = -4 + 12 - 2 = 6
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evaluate the given funtion. f(x)=2x+4; find f(3x)-3f(x)

Question: evaluate the given funtion. f(x)=2x+4; find f(3x)-3f(x) . f(3x) = 2(3x) + 4 = 6x + 4 3f(x) = 3(2x + 4) = 6x + 12 f(3x) - 3f(x) = 6x + 4 - 6x - 12 = -8 Answer: f(3x) - 3f(x) = -8
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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. 
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can you please explain how to to find dy/dx for the function x^2 y+ Y^2 x = -2

I need to find dy/dx of this function and evaluate the derivative at the point (2,-1) x^2 y+y^2 x = -2 solve the equation to be  = 0 x^2y + y^2x +2 = 0 find partial derivatives for dy and dx, the first term has two parts x^2y has two partial derivatives 2xy dx + x^2 dy the second term has 2 parts y^2x has 2 partial derivatives y^2 dx + 2yx dy and 2 has no derivative 2xy dx + y^2 dx + x^2 dy + 2yx dy = 0 2xy + y^2 dx = (-x^2 - 2xy) dy dy/dx = (2xy + y^2)/(-x^2 - 2xy) use the point (2,-1) dy/dx = [2*2*(-1) + (-1)^2]/[-1(2)^2 -2*2*(-1)] dy/dx = (-4 + 1) / (-4 + 4)  there is a 0 in the de=nominator therefore it is undefined
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Evaluate the function for f(x) = 3x2 + 1 and g(x) = x − 3. (fg)(−4)   f(x) = 3x2 + 1 and g(x) = x − 3.   fg = 3(x – 3)2 + 1   (fg)(−4) fg(-4) = 3(-4 – 3)2 + 1 fg(-4) = 3(-7)2 + 1 fg(-4) = 3*49 + 1 fg(-4) = 147 + 1 fg(-4) = 148
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evaluate the logarithm log8 8

I think that what you mean is: evaluate the logarithm of 8 where the logarithm is taken to the base 8. Definition: The logarithm of a number, to a particular base, is the power to which the base must be taken in order to give the given number. i.e. Logb n = p  where   b^p = n Since we have to find the log of the number 8, where the log is taken to the base of 8, then we need to find a power, p, such that 8 (the base) taken to the power, p, will equal the given number, 8. i.e. we need to find p such that 8^p = 8 The answer is obviously p = 1  
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