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This is all explained on Solving Inequalities Instead, bring "2" to the left 3x−10 x−4 − 2 > 0 Then multiply 2 by (x−4)/ (x−4) 3x−10 x−4 − 2 x−4 x−4 > 0 Now we have a common denominator, let's bring it all together 3x−10 − 2 (x−4) x−4 > 0 Simplify x−2 x−4 > 0In mathematics, 1 − 2 3 − 4 ··is an infinite series whose terms are the successive positive integers, given alternating signs Using sigma summation notation the sum of the first m terms of the series can be expressed as ∑ n = 1 m n ( − 1 ) n − 1 {\displaystyle \sum _{n=1}^{m}n(1)^{n1}} The infinite series diverges, meaning that its sequence of partial sums, (1, −1, 2, −2,), does not So, if we currently know #5^x# and want to determine #5^(3x)#, we see their powers are off by ^b# this becomes #5^(3x)=1/3^3=1/27#

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3−5 x 3−4 is equal to

3−5 x 3−4 is equal to-Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with stepbystep explanations, just like a math tutor1) {4} 2) {−5} 3) {4,5} 4) {−5,4} 17 Which is a graph of y =cotx?

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 (3/4)5 ÷(5/3)5 is equal to (a) (3/4÷5/3)5 (b) (3/4 ÷ 5/3)1 (c) (3/4 ÷ 5/3)0 (d) (3/4 ÷ 5/3)10 Login Remember Register;(2 x − 5) 2 = 2 (2 x − 7 2) l e t 2 x = t ⇒ t 2 − 10 t 25 = 2 t − 7 t 2 − 12 t 32 = 0 (t − 8) (t − 4) = 0 ⇒ x = 2, 3The absolute minimum of f over the interval 1, 3 is −2, and it occurs at x = 3 as shown in the following graph Figure 419 This function has both an absolute maximum and an absolute minimum Step 1 Evaluate f at the endpoints x = 0 and x = 2 f(0) = 0andf(2) = 4 − 33√4 ≈ − 0762 Step 2

 Step 3 Write the solution in interval notation − 3, 2) All the numbers that make both inequalities true are the solution to the compound inequality Example 272 Solve the compound inequality Graph the solution and write the solution in interval notation 4x − 7 < 9 and 5x 8 ≥ 3 53/4 is equal to what2 Chapter 6 Applications of the Integral (b) Compute the area of the region below the graph of f and above the graph of g (a) f (x) = g(x) gives 2 − x2 = x which simplifies to 0 = x2 x −2 = (x 2)(x −1) and thus f (x) and g(x) intersect at x =−2andx = 1 (b) We have 1 − 2 2 − x2 −(x) dx = 2x − 1 3 x3 − 1 2 x2 − = 452 2 2 12 (A) (B) y = x y = 2 x22 Figure 2 4

Example x−32 < −5 First, let us clear out the "/2" by multiplying both sides by 2 Because we are multiplying by a positive number, the inequalities will not change x−32 ×2 < −5 ×2 x−3 < −10 Now add 3 to both sides x−3 3 < −10 3 x < −7 And that is our solution x < −7The minus sign, −, has three main uses in mathematics The subtraction operator a binary operator to indicate the operation of subtraction, as in 5 − 3 = 2 Subtraction is the inverse of addition The function whose value for any real or complex argument is the additive inverse of that argument For example, if x = 3, then −x = −3, but if x = −3, then −x = 3The modulus sign means that the absolute value of x−4 is less than 3 This means that −3 < x− 4 < 3 This is what is called a double inequality We must treat it as two separate inequalities From the left we get −3 < x− 4 and by adding 4 to both sides we obtain 1 < x On the right we have x− 4 < 3, and by adding 4 to both sides we

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Algebra II Formula Sheet Geometric Formulas Pi 22 7 V = lwh SA= 2(lw lh wh) l h w 1 3 h l r p = 4s A = s2 s s A = h(b1 b2) 1 2 b 1 b 2 h r h p = 2(l w) A = lw l w r c2 = a2 b2 a c b Abbreviations milligram mg gram gMATH 142 Integration by Partial Fractions Joe Foster Example 3 Compute ˆ −2x 4 (x2 1)(x −1)dx The process follows as before The most common mistake here is to not choose the right numerator for the term with theHere x = 3 is not a solution because it solves only one of the inequalities Answer Interval notation − 1, 3) Alternatively, we may interpret − 1 ≤ x < 3 as all possible values for x between or bounded by −1 and 3 on a number line For example, one such solution is x = 1 Notice that 1 is between −1 and 3 on a number line, or that

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 Alternatively, we may interpret \(−1≤xSOLUTION Given 2 sinx2 sin2x−1 =1 Or, sin2xsinx−1=0 ∴ sinx = −1±√14 2 = −1√5 2 ⇒ sin2x= 3−√5 2 ⇒cos2x = √5−1 2 ∴ cos2x(1cos2x)= √5−1 2 × √51 2 =1January 7 Shift 1 Maths Question 1 The area of the region, enclosed by the circle x 2 y 2 = 2 which is not common to the region bounded by the parabola y 2 = 𝑥 and the straight line 𝑦 = 𝑥, is Question 2 Total number of sixdigit numbers in which only and all the five digits 1,3,5

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 (x 3)(y − 4) = 0 −12 A)The quantity in Column A is greater B)The quantity in Column B is greater C)The two quantities are equal D)The relationship cannot be determined from the information given Practice Questions Question 6Combine 3x and 4x to get 7x Combine 3 x and 4 x to get 7 x 7x98=2\left (x3\right)\left (x2\right) 7 x − 9 8 = 2 ( x − 3) ( x 2) Add 9 and 8 to get 1 Add − 9 and 8 to get − 1 7x1=2\left (x3\right)\left (x2\right) 7 x − 1 = 2 ( x − 3) ( x 2) Use the distributive property to multiply 2 by x3∫ x 3 2 x 4 − 2 x 2 1 x 2 − 1 d x ⇒ ∫ x 5 x 3 2 x 4 − 2 x 2 1 x 5 x 2 − 1 d x ⇒ ∫ x 2 2 x 4 − 2 x 2 1 x 3 1 − x 5 1 d x

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Five more than x x is equal to −3 −3 Translate x 5 = −3 x 5 = −3 Subtract 5 5 from both sides x 5 − 5 = −3 − 5 x 5 − 5 = −3 − 5 Simplify x = −8 x = −8Steps for Solving Linear Equation 3 (2x3)4 (x5)=x5 3 ( 2 x 3) − 4 ( x − 5) = x − 5 Use the distributive property to multiply 3 by 2x3 Use the distributive property to multiply 3 by 2 x 3 6x94\left (x5\right)=x5 6 x 9 − 4 ( x − 5) = x − 5 Use the distributive property to multiply 4 by x51) 2) 3) 4)

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Problem 324 Solution For x Misc 33Evaluate the definite integral ∫_1^4 𝑥−1𝑥−2𝑥−3 𝑑𝑥 I=∫_1^4 𝑥−1𝑥−2𝑥−3 𝑑𝑥 I=∫_1^4 𝑥−1 𝑑𝑥∫_1^4 𝑥−2 𝑑𝑥∫_1^4 𝑥−3 𝑑𝑥Solving 𝐈𝟏 I1=∫_1^4 𝑥−1 𝑑𝑥 We kow that 𝑥−1= { ( (𝑥−1) 𝑓𝑜𝑟 𝑥≥1@−(𝑥−1) 𝑓𝑜𝑟X 3 −1/4 −1/4 1 for some real number x If all entries of the matrix are 1 or −1, then each L i is equal to p (±1)2 (±1)2 (±1)2 (±1)2 = √ 4 = 2, and so detA ≤ 24 = 16 In fact, the following is such a matrix with determinant exactly equal to 16 1 1 1 −1 1 1 −1 1 −1 1 1 1 1 −1 1 1 6 Problem 518 Show that the determinant equals the product of the eigenvalues

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3(−1)4n = X∞ n=0 3, which diverges Therefore, the interval of convergence is (−1,1) 10 Find a power series representation for the function f(x) = x2 a3 −x3 and determine the interval of convergence Answer Rewriting f as f(x) = x2 1 a3 −x3 = x2 a3 1 1− x3 a3!, we can use the geometric series to see that f(x) = x2 a3 X∞ n=0 x3Less than or equal to, ≤ x ≤ 7 The value of x is less than or equal to 7 − 5 ≤ x ≤ 3 The value of x should lie between − 5 and 3, inclusive of both values Greater than or equal to, ≥ x ≥ 2 The value of x is greater than or equal to 2 2 ≥ x ≥ − 1 The value of x should lie between −🔴 Answers 1 🔴🔴 question Is 3 and 4/5 of 3/4 equal to 2 and 17/ ?

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If 5 x 3/4 is equal to 2 X then x is equal to2x5=3 , It is a Modulus function From the modulus function graph it is clear that it exhibits 2 values as its graph is straight line in both quadrants So it has 2 values one is ve value of given function and other is its additive inverse ieAs with floor functions, the best strategy with integrals or sums involving the ceiling function is to break up the interval of integration (or summation) into pieces on which the ceiling function is constant Find ∫ − 2 2 ⌈ 4 − x 2 ⌉ d x \displaystyle \int_ {2}^2 \big\lceil 4x^2 \big\rceil \, dx ∫ −22

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 Step 1 The given polynomial is P ( x) = 3 x 5 − 5 x 4 37 x 3 − x 2 − 176 x − 48 Step 2 Now x = 4 i is a zero of the given polynomial, therefore x = − 4 i is also zero of the given polynomial Therefore ( x − 4 i) ( x 4 i) divides the given polynomialIf 2f (X) − 3 F ( 1 X ) = X 2 (X ≠ 0), Then F(2) is Equal to (A) − 7 4 (B) 5 2 −1 (D) None of These CBSE CBSE (Commerce) Class 11 Textbook Solutions 7904 Important Solutions 14 Question Bank Solutions 6907 Concept Notes & Videos 365 Syllabus Advertisement Correct answer Which of the following tables represents a function?

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X 1 −1 4 −4 y 8 8 10 10 x 2 2 3 3 y 6 −6 7 7 x 5 −5 5 −5 y 10 11 12 13 x 2 2 7 7 ySolution Answer (c) 7 C 3 x 4 (x log 2 x) 3 = 4480 35x 4 (x log 2 x) 3 = 4480 x 4 (x log 2 x) 3 = 128 Take log wrt base 2 we get 4 log 2 x 3 log 2 (x log 2 x) = log 2 128 LetX 2 2 2 4 4 4 6 6 6 8 8 8 y –4 –4 –4 –2 –2 –2 2 2 2 0 0 0 6 A coordinate plane with a segment of a linear function with the slope of 3 over 2 between points (2,2) and (2,8)

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If 3^x = 4^x1 , then x is equal toA (6, 3), B (−3, 5), C (4, −2) and D (X, 3x) Are Four Points If ∆ Dbc ∆ Abc = 1 2, Then X is Equal to(x 4) 2 x (x − 3) ≤ 0 Step 4 Find the roots and asymptotes of the inequality by equating each factor to 0 x = 0 x − 3 = 0 x = 3 x 4 = 0 x = − 4 \begin{aligned} x & = 0\\ x 3 = 0 \implies x & = 3\\ x 4 = 0 \implies x & = 4 \end{aligned} x x − 3 = 0 x x 4 = 0 x = 0 = 3 = − 4 Step 5 Plot the points on the number line Now, start with the largest factor, ie 3 Initially, a curve from the positive region of the number line should intersect that point (here 3

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X therefore is equal to 2 or −2 b) x − 3 = 1 x is 1 unit away from 3 x therefore is equal to 2 or 4 c) x 3 = 1 x is 1 unit away from −3 x therefore is equal to −4 or −2 d) x − 5 ≤ 2 x is less than or equal to 2 units away from 5 x therefore may take any value in the closed interval between 3 and 7 A closed intervalExample 2 ()()− ×−775 3 is equal to (a) ()−7 8 (b) – ()7 8 (c) ()−7 15 (d) ()−7 2 Solution Correct answer is (a) Example 3 For any two nonzero integers x any y, x3 ÷ y3 is equal to (a) æö 0 ç÷èø x y (b) x 3 y (c) æö 6 ç÷èø x y (d) æö 9 ç÷èø x y Solution Correct answer is (b) In Examples 4 and 5, fill in the blanks to make the statements true Example 4Ask a Question (3/4)^5 ÷(5/3)^5 is equal to ← Prev Question Next Question → 0 votes 54 views asked in Exponents by Dev01 (517k points) closed by Dev01 (3/4) 5

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3x3 −5x2 −48x 80 is equivalent to 1) (x2 −16)(3x −5) 2) (x2 16)(3x −5)(3x 5) 3) (x 4)(x −4)(3x −5) 4) (x 4)(x −4)(3x −5)(3x −5) 16 What is the solution set for the equation 5x 29 =x 3?A ⃗ = (2, − 3, 4), b ⃗ = (7, 1, − 6) r ⃗ × Question 15 If the fourth term in the expansion of (x x log_{2} x) 7 is 4480, then the value of x where x ∈ N is equal to a 4;If any individual factor on the left side of the equation is equal to 0 0, the entire expression will be equal to 0 0 x−3 = 0 x 3 = 0 x4 = 0 x 4 = 0 Set the first factor equal to 0 0 and solve Tap for more steps Set the first factor equal to 0 0 x − 3 = 0 x 3 = 0 Add 3 3

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4 • (x7) Step5 Add up the four terms of step 4 (x4) • (x7) Which is the desired factorization Equation at the end of step 2 (x 4) • (x 7) = 0 Step 3 Theory Roots of a product 31 A product of several terms equals zero When a product of two or more terms equals zero, then at least one of the terms must be zero Transcript Ex 63, 18 For the curve 𝑦=4𝑥3 −2𝑥5, find all the points at which the tangent passes through the originLet (ℎ , 𝑘) be the Required Point on the Curve at which tangent is to be taken Given Curve is 𝑦=4𝑥^3−2𝑥^5 Since Point (ℎ , 𝑘) is on the Curve ⇒ (ℎ , 𝑘) will satisfy the Equation of Curve Putting 𝑥=ℎ , 𝑦=𝑘 𝑘=4ℎ^3−2ℎ^5Solve one equation for one variable, then plug the answer for that variable into the second equation to find the value of the second variable Now plug both variables into the last problem to get the answer Solve the first equation for the variab

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