• Suppose f(x)=2x−8. Determine, in a simplified form, each of the following: f(a), f(a+h), [f(a+h)−f(a)]/h?

    f(x) = 2x - 8 Anything in the parenthesis gets substituted for x, and then simplify. So for f(a), you just substitute a for x and you're done: f(a) = 2a - 8 For f(a + h), substitute the sum of the variables and simplify if you can: f(a + h) = 2(a + h) - 8 f(a + h) = 2a + 2h - 8 Now that we have expressions for both f(a) and f(a + h), we... show more
    f(x) = 2x - 8 Anything in the parenthesis gets substituted for x, and then simplify. So for f(a), you just substitute a for x and you're done: f(a) = 2a - 8 For f(a + h), substitute the sum of the variables and simplify if you can: f(a + h) = 2(a + h) - 8 f(a + h) = 2a + 2h - 8 Now that we have expressions for both f(a) and f(a + h), we can substitute those expressions in this and simplify: [f(a + h) - f(a)] / h We get: [(2a + 2h - 8) - (2a - 8)] / h Now simplify: (2a + 2h - 8 - 2a + 8) / h The 2a and 8 terms cancel out, leaving: 2h / h And then the h's cancel out leaving: 2
    3 answers · Mathematics · 16 hours ago
  • How can adding doubles help you multiply by 2?

    x + x = 2x adding a number to itself is the same as multiplying that number by 2.
    x + x = 2x adding a number to itself is the same as multiplying that number by 2.
    4 answers · Mathematics · 17 hours ago
  • How many months have gone by?

    How many months are in 2 years? That will answer your question.
    How many months are in 2 years? That will answer your question.
    3 answers · Mathematics · 23 hours ago
  • When fully simplified, The rational expression (15v/v-4) - (60/v-4) becomes?

    I think you have: 15v / (v - 4) - 60 / (v - 4) So as the answers suggest, v cannot be 4 otherwise you are dividing by zero. You are subtracting fractions with the same denominator, so you can do so by subtracting the numerators: (15v - 60) / (v - 4) Now we can factor out the 15 in the numerator: 15(v - 4) / (v - 4) The (v - 4)'s cancel... show more
    I think you have: 15v / (v - 4) - 60 / (v - 4) So as the answers suggest, v cannot be 4 otherwise you are dividing by zero. You are subtracting fractions with the same denominator, so you can do so by subtracting the numerators: (15v - 60) / (v - 4) Now we can factor out the 15 in the numerator: 15(v - 4) / (v - 4) The (v - 4)'s cancel out, leaving: 15 So while v still cannot be 4, if v is any other value, this simplifies to 15 (answer B)
    2 answers · Mathematics · 1 day ago
  • Answer the following function?

    You could have: g(x) = x - 5 h(x) = ∛(-x² - 4)
    You could have: g(x) = x - 5 h(x) = ∛(-x² - 4)
    2 answers · Mathematics · 1 day ago
  • How is "x^8" read in English?

    Anything that has "exponent" or "power" or even "x to the eighth" would be understood enough. "x eight" from one of the other answers I would disagree with
    Anything that has "exponent" or "power" or even "x to the eighth" would be understood enough. "x eight" from one of the other answers I would disagree with
    4 answers · Mathematics · 2 days ago
  • How do you distribute an exponent to a fraction such that it is (a^i/b^j)^x/y?

    If you have two factors (or a quotient) that is taken to a power, then every factor (or part of the quotient) gets that exponent. A few examples: (xy)^a and (x/y)^a becomes: x^a y^a and x^a / y^a If you have an exponent of an exponent, you get the product of the exponents, example: (a^x)^y = a^(xy) And in the case that the exponent is a... show more
    If you have two factors (or a quotient) that is taken to a power, then every factor (or part of the quotient) gets that exponent. A few examples: (xy)^a and (x/y)^a becomes: x^a y^a and x^a / y^a If you have an exponent of an exponent, you get the product of the exponents, example: (a^x)^y = a^(xy) And in the case that the exponent is a fraction, doesn't change anything. You still apply it to all factors and multiply by any pre-existing exponents. so: (5² / 3²)^(3/2) We apply the 3/2 power to both the 5² and the 3². Both cases we multiply the 3/2 by the pre-existing 2 to get a new exponent of 3: 5³ / 3³
    2 answers · Mathematics · 3 days ago
  • Calc limit question?

    Let's start with this function: f(x) = x³ The first derivative can be defined as the limit has 0 approaches zero of: [f(x - h) - f(x)] / h When h = 0, you end up with 0 / 0 which isn't defined. If x = 3, we get the expression you have above in your question. Let's keep this in terms of x and h, but keep this in mind so we can return... show more
    Let's start with this function: f(x) = x³ The first derivative can be defined as the limit has 0 approaches zero of: [f(x - h) - f(x)] / h When h = 0, you end up with 0 / 0 which isn't defined. If x = 3, we get the expression you have above in your question. Let's keep this in terms of x and h, but keep this in mind so we can return to it later. We have f(x) defined, so let's get an expression for f(x - h): f(x) = x³ f(x - h) = (x - h)³ Let's expand that out: f(x - h) = (x - h)(x - h)(x - h) f(x - h) = (x² - 2hx + h²)(x - h) f(x - h) = x³ - hx² - 2hx² + 2h²x + h²x - h³ f(x - h) = x³ - 3hx² + 3h²x - h³ Let's substitute this expression back into the expression above for f(x - h) and x³ for f(x): [f(x - h) - f(x)] / h (x³ - 3hx² + 3h²x - h³ - x³) / h x³ terms cancel out leaving: (-3hx² + 3h²x - h³) / h We can factor out an h from the numerator and cancel it out with the denominator to get: h(-3x² + 3hx - h²) / h -3x² + 3hx - h² Now that we no longer have a denominator, we can set h = 0 and simplify: -3x² + 3(0)x - 0² -3x² + 0 - 0 -3x² And finally, we said your expression works when my expression is x = 3, so substitute that to get your answer: -3(3)² -3(9) -27 And to use Wolfram to check my answer, it gives me: -27
    3 answers · Mathematics · 3 days ago
  • Find the value of 19-x whenx=9 guys help me with this dumb math .?

    just substitute the value of 9 in for x and simplify: 19 - x 19 - 9 10
    just substitute the value of 9 in for x and simplify: 19 - x 19 - 9 10
    1 answer · Mathematics · 4 days ago
  • What are the restricted values for 1/(ax-b)?

    The restrictions come in when the denominator is zero. So set that to zero and solve for x to get: ax - b = 0 ax = b x = b / a That's your restriction. x can be any value except for (b / a) to have a real value.
    The restrictions come in when the denominator is zero. So set that to zero and solve for x to get: ax - b = 0 ax = b x = b / a That's your restriction. x can be any value except for (b / a) to have a real value.
    1 answer · Mathematics · 4 days ago
  • Arithmetic sequence Question?

    Best answer: The general for for the n'th term of an arithmetic sequence is: a(n) = a + b(n - 1) Where a is the first term and b is the common difference. Then we have an equation that can give us the sum of the first "n" terms of a sequence: S(n) = [ a + a(n)] n / 2 We are told that the sum of the first 5 terms is 85 and the sum... show more
    Best answer: The general for for the n'th term of an arithmetic sequence is: a(n) = a + b(n - 1) Where a is the first term and b is the common difference. Then we have an equation that can give us the sum of the first "n" terms of a sequence: S(n) = [ a + a(n)] n / 2 We are told that the sum of the first 5 terms is 85 and the sum of the first 6 terms is 123, so let's find expressions for a(5) and a(6), first: a(n) = a + b(n - 1) a(5) = a + b(5 - 1) and a(6) = a + b(6 - 1) a(5) = a + 4b and a(6) = a + 5b We don't yet know what a(5) and a(6) are, so leave them at that. Now we can find expressions for S(5) and S(6): S(n) = [ a + a(n)] n / 2 S(5) = [ a + a(5)] * 5 / 2 and S(6) = [ a + a(6)] * 6 / 2 S(5) = 5[ a + a(5)] / 2 and S(6) = 3[ a + a(6)] We do have expressions for a(5) and a(6), so: S(5) = 5(a + a + 4b) / 2 and S(6) = 3(a + a + 5b) S(5) = 5(2a + 4b) / 2 and S(6) = 3(2a + 5b) S(5) = (10a + 20b) / 2 and S(6) = 6a + 15b S(5) = 5a + 10b and S(6) = 6a + 15b And we do know the values for S(5) and S(6): 85 = 5a + 10b and 123 = 6a + 15b Now we have a system of two equations and two unknowns so we can solve for "a" and "b" (the first term and the common difference). Once we know those, we can find the values of the first 4 terms. Solving the first equation for a in terms of b: 85 = 5a + 10b 85 - 10b = 5a 17 - 2b = a Substitute into the other equation to solve for b: 123 = 6a + 15b 123 = 6(17 - 2b) + 15b 123 = 102 - 12b + 15b 123 = 102 + 3b 21 = 3b b = 7 Now we can solve for a: a = 17 - 2b a = 17 - 2(7) a = 17 - 14 a = 3 So the first four terms of this sequence is: 3, 10, 17, 24
    2 answers · Mathematics · 4 days ago
  • Nintendo Switch won't turn on?

    I would contact Nintendo Support and see if they have any suggestions that you can do, otherwise, you may have to arrange for a repair/replacement. If the system is under a year old, this should be a covered repair and won't cost you anything (except maybe shipping costs). If it's older than a year, expect the repair to be about $150 give... show more
    I would contact Nintendo Support and see if they have any suggestions that you can do, otherwise, you may have to arrange for a repair/replacement. If the system is under a year old, this should be a covered repair and won't cost you anything (except maybe shipping costs). If it's older than a year, expect the repair to be about $150 give or take (usually is around half the cost of a new system).
    2 answers · Video & Online Games · 4 days ago
  • Does Good gpu with i5 cpu or i7 lag on csgo ofline deathmatch with bots answer from real experience not personal theory?

    What generation of i5 or i7 are you referring to? I have a gen1 i7 PC that's garbage when you try to play anything modern on it, then I have a 4th gen i5 that I can play most games on just fine.
    What generation of i5 or i7 are you referring to? I have a gen1 i7 PC that's garbage when you try to play anything modern on it, then I have a 4th gen i5 that I can play most games on just fine.
    1 answer · Video & Online Games · 4 days ago
  • Determine the area of the region?

    First, let's find the x value that is greater than zero where these two lines meet so we know the values of x in our area: x² + x + 5 = 3x + 13 x² - 2x - 8 = 0 (x - 4)(x + 2) = 0 x = -2 and 4 So our region we need the area of is between x = 0 and x = 4. Next, we can find the area under the curves in these ranges by getting the first... show more
    First, let's find the x value that is greater than zero where these two lines meet so we know the values of x in our area: x² + x + 5 = 3x + 13 x² - 2x - 8 = 0 (x - 4)(x + 2) = 0 x = -2 and 4 So our region we need the area of is between x = 0 and x = 4. Next, we can find the area under the curves in these ranges by getting the first anti-derivative of each function, then subtract the smaller from the larger to get what's left, which is the area we are wanting: f'(x) = x² + x + 5 and g'(x) = 3x + 13 First integral: f(x) = (1/3)x³ + (1/2)x² + 5x + c and g(x) = (3/2)x² + 13x + k To get the area under the curve from x = 0 to x = 4, solve for: f(4) - f(0) and g(4) - g(0) [(1/3)(4)³ + (1/2)(4)² + 5(4) + c] - [(1/3)(0)³ + (1/2)(0)² + 5(0) + c] and [(3/2)(4)² + 13(4) + k] - [(3/2)(0)² + 13(0) + k] [(1/3)(64) + (1/2)(16) + 20 + c] - [(1/3)(0) + (1/2)(0) + 0 + c] and [(3/2)(16) + 52 + k] - [(3/2)(0) + 0 + k] (64/3 + 8 + 20 + c) - (0 + 0 + 0 + c) and (24 + 52 + k) - (0 + 0 + k) (64/3 + 24/3 + 60/3 + c) - c and (76 + k) - k 148/3 + c - c and 76 + k - k 148/3 and 76 The 76 is the larger of the two, so if we subtract out the other, we have the area between the curves left: 76 - 148/3 228/3 - 148/3 80/3 unit²
    2 answers · Mathematics · 5 days ago
  • What is negative 12 over 5 plus 13 over 6 plus negative 3 and 2 over 3−12/5 + 13/6 +(−3 2/3)?

    Presuming you mean: -12/5 + 13/6 + (-3 2/3) The first thing I'll do is turn the mixed number into an improper fraction, then turn the addition of a negative into a subtraction: -12/5 + 13/6 + (-11/3) -12/5 + 13/6 - 11/3 Now we need a common denominator. 30 is the LCD for this: -72/30 + 65/30 - 110/30 Now we can add/subtract the... show more
    Presuming you mean: -12/5 + 13/6 + (-3 2/3) The first thing I'll do is turn the mixed number into an improper fraction, then turn the addition of a negative into a subtraction: -12/5 + 13/6 + (-11/3) -12/5 + 13/6 - 11/3 Now we need a common denominator. 30 is the LCD for this: -72/30 + 65/30 - 110/30 Now we can add/subtract the numerators: (-72 + 65 - 110) / 30 -117 / 30 This reduces: -39 / 10 And if we want to put that back into a mixed number: -3 9/10 which then can reduce to
    2 answers · Mathematics · 5 days ago
  • Prove that the product of any two odd numbers is always odd.?

    You can prove it. Let's take any positive integer: n If we double it, we know it is even: 2n Then if we add 1 to it, we know it's odd: 2n + 1 For the second odd number, let's do the same with an "m" instead of an "n", so this odd number is: 2m + 1 So no matter what positive value you have for n or m, (2n + 1) and... show more
    You can prove it. Let's take any positive integer: n If we double it, we know it is even: 2n Then if we add 1 to it, we know it's odd: 2n + 1 For the second odd number, let's do the same with an "m" instead of an "n", so this odd number is: 2m + 1 So no matter what positive value you have for n or m, (2n + 1) and (2m + 1) will be odd. Now let's see what happens when we multiply them together: (2n + 1)(2m + 1) 4nm + 2n + 2m + 1 We can factor out a 2 from the first three terms: 2(2nm + n + m) + 1 And recall our general form. 2n + 1 .. n is any positive integer to make this odd. We have "2nm + n + m" which will also be a positive integer presuming n and m were, so if we make this substitution: z = 2nm + n + m We end up with: 2z + 1 So the result is an odd number.
    9 answers · Mathematics · 6 days ago
  • Solve by factorisation 2x^2+3x-20=0?

    2x² + 3x - 20 = 0 Since your high-degree coefficient is 2, we know that the factor has to contain 2x and x, so: (2x + ?)(x + ?) = 0 Then the constant term is -20, so the product of the two "question marks" has to be -20. With it being negative, one is negative the other is positive. Let's start with -4 and 5 and see where we go... show more
    2x² + 3x - 20 = 0 Since your high-degree coefficient is 2, we know that the factor has to contain 2x and x, so: (2x + ?)(x + ?) = 0 Then the constant term is -20, so the product of the two "question marks" has to be -20. With it being negative, one is negative the other is positive. Let's start with -4 and 5 and see where we go from there. So just a guess: (2x - 4)(x + 5) = 0 Now we need the middle term to be 3x, so if we multiply the 2x and 5 and add it to the -4 and x, it should be +3, if it's not, then we try other numbers: 2x * 5 + (-4)x 10x - 4x 6x Not what we want, so let's see what happens with -5 and 4: (2x - 5)(x + 4) = 0 checking: 2x * 4 - 5x 8x - 5x 3x This works out, so this is the factored form. Two numbers multiplied together to be zero can only work if one of the numbers is zero. So let's solve both factors for zero to get our two answers: 2x - 5 = 0 and x + 4 = 0 2x = 5 and x = -4 x = 5/2 and x = -4
    3 answers · Mathematics · 6 days ago
  • Standard form y = (-1/2)x² + 5x - (17/2) to a quadratic equation(Quadratic Formla). Please right it out for me!?

    It already is a quadratic equation in standard form. are you asking to have it solved for the zeroes using the quadratic formula? If so, using y = 0. we have: (-1/2)x² + 5x - (17/2) = 0 Let's get rid of the fractions and the negative leading coefficient, so let's multiply both sides by -2: x² - 10x + 17 = 0 Now we can use the... show more
    It already is a quadratic equation in standard form. are you asking to have it solved for the zeroes using the quadratic formula? If so, using y = 0. we have: (-1/2)x² + 5x - (17/2) = 0 Let's get rid of the fractions and the negative leading coefficient, so let's multiply both sides by -2: x² - 10x + 17 = 0 Now we can use the quadratic formula using: a = 1 b = -10 c = 17 x = [ -b ± √(b² - 4ac)] / (2a) x = [ -(-10) ± √((-10)² - 4(1)(17))] / (2 * 1) x = [ 10 ± √(100 - 68)] / 2 x = [ 10 ± √(32)] / 2 x = [ 10 ± √(16 * 2)] / 2 x = [ 10 ± 4√(2)] / 2 x = 5 ± 2√2
    1 answer · Mathematics · 6 days ago
  • How do you play pokemon go on pc?

    You'd need an android emulator. But without the ability to walk around, if you want to "simulate" that, you'd have to spoof your GPS, which is against EULA of the game and can get you banned. If you want to play the game, I'd suggest you get a compatible device with a compatible data plan and play it as intended. show more
    You'd need an android emulator. But without the ability to walk around, if you want to "simulate" that, you'd have to spoof your GPS, which is against EULA of the game and can get you banned. If you want to play the game, I'd suggest you get a compatible device with a compatible data plan and play it as intended.
    1 answer · PC · 6 days ago