Exam 7: Arithmetic Sequence: Common Difference and First n Terms

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Graph the parabola. - x=3y22y+6x=3 y^{2}-2 y+6  Graph the parabola. - x=3 y^{2}-2 y+6

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Provide an appropriate response. -Consider the selections of seven apples from a barrel of 75 apples. Is this a combination, a permutation, or neither?

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Evaluate the sum. Round to two decimal places, if necessary. - k=14(1)k2k\sum _ { k = 1 } ^ { 4 } ( - 1 ) ^ { k } \cdot - 2 k

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Use mathematical induction to prove that the statement is true for every positive integer n. - 5+25+35++5n=5n(n+1)25 + 2 \cdot 5 + 3 \cdot 5 + \ldots + 5 n = \frac { 5 n ( n + 1 ) } { 2 }

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Find a formula for the nth term of the arithmetic sequence shown in the graph. -Find a formula for the nth term of the arithmetic sequence shown in the graph. -

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A student is told to work any 6 out of 10 questions on an exam. In how many different ways can he complete the exam? (The correctness of his answers has no bearing.)

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Find the common difference for the arithmetic sequence. -5, 6, 7, 8, . . .

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Suppose that a family has 5 children and that the probability of having a girl is 12\frac { 1 } { 2 } \text {. } . What is the probability of having at least three boys?

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Match the equation of the ellipse with the appropriate description. - x225=1y29\frac { x ^ { 2 } } { 25 } = 1 - \frac { y ^ { 2 } } { 9 }

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Use mathematical induction to prove that the statement is true for every positive integer n. - (43)n=43n\left( 4 ^ { 3 } \right) ^ { n } = 4 ^ { 3 n }

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The series of sketches below starts with an equilateral triangle having sides of length 1 (one). In the following steps, equilateral triangles are constructed by joining the midpoints of the sides of the preceding triangle. Develop a formula for the length of the sides of the new triangle. The series of sketches below starts with an equilateral triangle having sides of length 1 (one). In the following steps, equilateral triangles are constructed by joining the midpoints of the sides of the preceding triangle. Develop a formula for the length of the sides of the new triangle.

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In how many ways can the letters in the word PAYMENT be arranged if the letters are taken 6 at a time?

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Find the common difference for the arithmetic sequence. -8, 13, 18, 23, . . .

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Graph the ellipse. - 81x2+36y2=291681 x ^ { 2 } + 36 y ^ { 2 } = 2916  Graph the ellipse. - 81 x ^ { 2 } + 36 y ^ { 2 } = 2916

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Use mathematical induction to prove that the statement is true for every positive integer n. - 4n>14 ^ { n } > 1

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Evaluate the sum using the given information. - =1,=3,=-5,=4, and \Deltax=-0.3;f(x)= f \Deltax (Round to the nearest tenth, if necessary.)

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Write the series using summation notation. - 1+122+132+142+1521 + \frac { 1 } { 2 ^ { 2 } } + \frac { 1 } { 3 ^ { 2 } } + \frac { 1 } { 4 ^ { 2 } } + \frac { 1 } { 5 ^ { 2 } }

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Identify the type of conic section. -Identify the type of conic section consisting of the set of all points in the plane for which the distance from the point (0, 18) is twice the distance from the line y y=3y = 3

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When a satellite is near Earth, its orbital trajectory may trace out a hyperbola, a parabola, or an ellipse. The type of trajectory depends on the satellite's velocity VV in meters per second. It will be hyperbolic if V>kDV > \frac { k } { \sqrt { D } } , parabolic if V=kD\mathrm { V } = \frac { \mathrm { k } } { \sqrt { \mathrm { D } } } , and elliptical if V<kD\mathrm { V } < \frac { \mathrm { k } } { \sqrt { \mathrm { D } } } , where k=2.82×107\mathrm { k } = 2.82 \times 10 ^ { 7 } is a constant and D\mathrm { D } is the distance in meters from the satellite to the center of Earth. Solve the problem. -If a satellite is scheduled to leave Earth's gravitational influence, its velocity must be increased so that its trajectory changes from elliptical to hyperbolic. Determine the minimum increase in velocity necessary for a Satellite traveling at a velocity of 5696 meters per second to escape Earth's gravitational influence when D=167×106 m.\mathrm { D } = 167 \times 10 ^ { 6 } \mathrm {~m} .

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Evaluate the sum using the given information. - x1=0,x2=4, and x3=4x _ { 1 } = 0 , x _ { 2 } = - 4 \text {, and } x _ { 3 } = 4 i=13(xi32xi+1)\sum _ { i = 1 } ^ { 3 } \left( \frac { x _ { i } - 3 } { 2 x _ { i } + 1 } \right)

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