Exponents and division?
1. Factor a 3 out of the constants and cancel out x^5 answer 3x^3/4 2. To make an exponent positive put a move it to the denominator of the fraction. answer 1/n^5 3. Factor a 3 out of the constants, cancel x and y answer y^3/3 4. Factor a 5 out of the constants cancel an h and a k^2 answer 3h^5k 5. cancel f^5 answer 2f^5 1. cancel the 12, m, n 1/m^2n^4 to get rid of the fraction make the exponents negative answer m^-2 n^-4 2. cancel x^3 and Y^2 y^2/x^6 to get rid of the fraction make x's exponent negative y^2 x^-6 3. cancel 2, x^2 2y/x to get rid of the fraction make x's exponent negative 2y x^-1
Simlify. when appropriate, write in standard form. All exponents should be positive?
1) (-3x^2 + 4x - 7) + (2x^2 - 7x + 8) Remove the parentheses and then combine the "similar" terms: = -3x^2 + 2x^2 + 4x - 7x - 7 + 8 Simplify by adding and subtracting: = -x^2 - 3x + 1 2) (39a^4 - 4a^3 + 2a^2 - a - 7) - (10a^4 + 3a^3 - 2a^2 - a + 8) Multiply - 1 by each term inside each parenthesis and this becomes: = 39a^4 - 4a^3 + 2a^2 - a - 7 + ( - 10a^4 - 3a^3 + 2a^2 + a - 8) Now combine all similar terms: = 29a^4 - 7a^3 + 4a^2 - 15 3) - 3xy^3(x - 2y) Multiply - 3xy^3 by each term: = - 3x^2y^3 + 6xy^4 4) (8a^3b^2)(2a^ - 4b^ - 5) Remove all negative exponents in the term 2a^( - 4)b^( - 5): = (8a^3b^2)((2)/(a^4b^5)) Multiply 8a^3b^2: = 16/(ab^3) 4) (3x^3y^2)/(6x^ - 2y^5) Reduce the expression by removing a factor of 3x^( - 2)y^2 from both the numerator and denominator: = (x^5)/(2y^3) 5) (3x^2 + x - 1)(2x - 3) Multiply each term in the first polynomial by each term in the second polynomial: = 6x^3 - 7x^2 - 5x + 3 6) ( - 3x^2y^3z)^3 Multiply through the terms and this becomes: = - 27x^6y^(9)z^3 6) (3x + 7)(2x - 5) Multiply through the terms and this becomes: = 6x^2 - x - 35 7) 64x^3y^2 - 16x^2y^3 + (32x^5y^5)/ (8x^2y^2) Reduce this expression by removing a factor of 8x^2y^2 from the numerator and denominator: = 64x^3y^2 - 16x^2y^3 + 4x^3y^3 Next, reorder this alphabetically from left to right, starting with the highest order term: = 4x^3y^3 + 64x^3y^2 - 16x^2y^3 8) 2x^2z(3x - 2z) Multiply through the terms and this becomes: = 6x^3z - 4x^2z^2 9) (10a^3b^2c^7)/(5a^5bc^7) Reduce this by removing a factor of 5a^3bc^(7) from the numerator and denominator: = (2b)/(a^2) 10) (15a^4b^2c)^0 When you see anything raised to the 0th power, in other words ^0, it is simply 1. In this problem, notice that the entire thing is to the 0th power, so the answer is: = 1 11) (x + 6)^2 = (x + 6) (x + 6) Multiply each term in the first group by each term in the second group using the FOIL (First Outer Inner Last): = x^2 + 12x + 36
Simplify each in the space provided, showing all steps. Answers should have positive exponents.?
1. (2x^2y)^0 (3xy) 2. a^-2 b^3 a3 3. 4^-5 4^6 / 4^2 4. (2x)^-2 (2y)^3 (4x) 5. (a^-3 b^2 c)^-2 / (ab^-c3)^-1 6. 2^4 8^3 16^-2 / 32^-1 7. (5u^2v / 2uv^2) x (-3uv / 2u^2v0) ^3 8. (3^-1 + 2^-1)^2 9. a^-1 - 3a^-2 / 2^-2
Write the following with positive exponents, and then simplify, when possible?
1) (2x)^-2 2) (x^6/x^-8)^3 3) The number 0.23 104 is not in scientific notation because 0.23 is less than 1. Write 0.23 104 in scientific notation. Helppppppppppppppppppppppppppppppppppp...
How do you write expressions without a fraction bar?
You need to know exponent laws. A POSITIVE exponent in the denominator is a NEGATIVE exponent in the numerator. Example of first one: a^7 ---- a^10 Note, the base is a in both numerator and denominator, so this allows us to raise the denominator into the numerator such that: a^(7-10) = a^-3, so now you have expressed it without a division or fraction bar as you put it. By exponent laws, a^-3 is the same thing as 1/a^3,...so a^-3 = 1/a^3 Number 2 Use cancelation of the terms 4x^2y/2x^3 = 4/2 x^2/x^3, and this gives us 2 and by exponent laws x^2/x^3 = x^(2 - 3) = x^-1 So our answer then is: 2x^-1 and as I said this is also equal to 2/x, note in the denominator, that the value of 1 becomes positive and that x is the same thing as x^1. Third one: x^3/x^9 3/9 y^4/y^2 = x^(3 - 9) 1/3 y^(4 - 2) = x^-6 1/3 y^2 = 1/3x^-6y^2 and 1/3 is still showing with a fraction so 1/3 = 0.333 approximately and so the final solution is: 0.333x^-6y^2
What is the number of terms in the expansion of [math](x - 3x² + 3x³)^{20}[/math]?
Taking ‘x’ common, we get (1 – 3x + 3x^2)^20.Now, in the expansion, highest and lowest powers of x will be 40 and 0 respectively.And since it has continuous powers of x (0,1 and 2), the expansion will have all the powers of x ranging from 0 to 40.Therefore, the no. of terms in the expansion will be 41.
Simplify the following expression and write your answer using positive exponents only.?
(a) [ (x^2 y^-4) / (x^3 y^2) ]^-3 (since you are dividing by a neg root use the reciprocal) (x^2 y^-4) (x^3 y^2)^3 (x^2 y^-4) (x^9 y^6) (x^11 y^2) (b) [ (-2x^6 y^-3) / 9z^7 ]^-2 (-2x^6 y^-3)( 81z^14) (81z^14) /4x^12 y^-6) (81z^14 y^6) / (4x^12)
How do I obtain constant terms in binomial expansion?
Let us try to view the general term of a binomial expansion in a slightly different way.Let us consider an example where we need to find the constant term in the expansion of [math](x - \frac{2}{x^2})^9[/math]General term for the above binomial is: [math]T_{r+1} = \ ^{9}C_{r}(x)^{9- r}(-\frac{2}{x^2})^{r}[/math][math]T_{r+1} = \ ^{9}C_{r}(x)^{9- r}(-2)^{r}(x)^{-2r}[/math][math]T_{r+1} = \ ^{9}C_{r}(-2)^{r}(x)^{9-3r}[/math]Now for a term in the expansion to be constant, the power of x should be 0.So, 9 - 3r = 0r =3Therefore, the 4th term in the expansion of [math](x - \frac{2}{x^2})^9[/math] is the constant term.[math]T_{3+1} = \ ^{9}C_{3}(-2)^{3}(x)^{9-3*3}[/math][math]T_{4} = \ 84 (-8) \ = \ - 672 [/math]Method: Step 1: Find the general term in the expansion of the binomial.Step 2: Collect all the powers of x terms and make it one entityStep 3: Set the power of x equal to 0 and find the value of rStep 4: Substitute back the value of r in the general term to get the constant term.Note: If r is fractional, then there is no constant term in the expansion.I hope it helps!
Put y= x2 + 4x - 6 in vertex form (y=a(x-h)2 + k) and state the vertex, axis of symmetry, y-intercept-PLZ HELP?
Hint: Use ^ to indicate exponents in the future if you do not know how to write the exponents. Example: y = x^2 + 4x - 6 You have to complete the square to convert it into vertex form. y = x² + 4x - 6 Group. y = (x² + 4x) - 6 Factor y = (x² + 4x) - 6 Add placeholders. y = (x² + 4x + ___) - 6 - 1(___) Notice that the second blank is multiplied by -1 to account for what you had to add to complete the square. Take the coefficient of the x term: 4 Divide it by 2: 4 / 2 = 2 Square it: (2)² = 4 Add 4 to both blanks. y = (x² + 4x + 4) - 6 - 1(4) x² + 4x + 4 is the expanded form of a perfect square binomial. Remember that (a + b)² = a² + 2ab + b². Apply this to what you have. y = (x² + 4x + 4) - 6 - 1(4) y = (x + 2)² - 6 - 1(4) Simplify the rest. y = (x + 2)² - 6 - 4 y = (x + 2)² - 10 Remember that the vertex form is: y = a(x - h)² + k CHECK: y = (x + 2)² - 10 y = [(x)² + 2(x)(2) + (2)²] - 10 y = (x² + 4x + 4) - 10 y = (x²) + 1(4x) + 1(4) - 10 y = x² + 4x + 4 - 10 y = x² + 4x - 6 TRUE ANSWER: y = (x + 2)² - 10 is the vertex form. Given: y = (x + 2)² - 10 Means: h = -2 Means: k = -10 Means: a = 1 ANSWER: The vertex is at (-2, -10). Since the equation is a function of x and a is positive, the parabola opens upwards. Parabolas that open upwards or downwards have an axis of symmetry that is the same as the h coordinate of the vertex. ANSWER: The axis of symmetry is at x = -2. Remember that y-intercepts have x = 0. Find the value of y when x = 0 by substitution x with 0 in the original equation. y = x² + 4x - 6 y = 0² + 4(0) - 6 y = 0 + 0 - 6 y = -6 ANSWER: The y-intercept is at (0, -6).