Simplify (2x^3z-2)^-3 write answer without negative exponents SHOW WORK?
( 2 x³ z⁻² )⁻³ Negative exponents mean that the quantity is a fraction. So in this case, z⁻² means 1/z² So the amount in the parentheses is actually ( 2 x³ / z² )⁻³ Now the entire parentheses is raised to a negative power, so it becomes a fraction to a positive power, like this: ( z² / 2 x³ )³ Now you raise every exponent inside the parentheses to the power on the outside by multiplying each exponent by 3 z⁶ / 2³ x⁹ You can write 2³ as the number 8 z⁶ / 8 x⁹ <-- answer with no negative exponents Answer: z⁶ / 8 x⁹
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
Math Help! (5 Questions; 2 Answer Reviews Needed) Best Answer Will be Rewarded?
A. In your own words, explain how to find the degree of the following polynomial: 2x^3y^4 + 6xy^2z^5. B. Simplify: 32x^10y^3z^2/8x^6yz^5 C. Simplify: (-3x^3y^2z)^3 D. Simplify: (3x + 1)(2x^2 - 3x + 1) 3x • 2x^2 = 6x^3 3x • -3x = -9x^2 3x • 1 = 3x 1 • 2x^2 = 2x^2 1 • -3x = -3x 1 • 1 = 1 6x^3 - 9x^2 + 2x^2 + 3x - 3x + 1 6x^3 - 7x^2 + 1 (This is how far I got with part D. I am not sure if this is correct or complete.) E. Simplify: (2x + 3y)(5x - 3y) F. Simplify: 9x^5 + 3x^2 +6x / 3x G. Write in scientific notation: 2,130,000 ( I believe it is 2.13x10^6 , but I am not sure) I do not wish for any negative comments telling me to do my own homework; I clearly have been trying. I am just stuck on this portion of my work. Any help will be greatly appreciated. Thanks in advance!
Simplify 3x - 2y + 9 + 9y - x?
3x - 2y + 9 + 9y - x, it can be written as :- =(3x - x) + (9y - 2y) + 9 ( We take like terms together. eg. 3x,-x and 9y and -2y) = 2x + 7y + 9
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.
What is the binomial expansion of [math](1+x) ^{-2}[/math]?
Hope this will help to build fundamentals of binomial..
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!
Math home work help me plz quiz tomorow?
hello, this is actually very simple. when you have 2 bases that are the same ( the base is the number being raised to the exponent, so if its 4 to the 2nd, 4 is the base) you just subtract the bottom exponent from the top exponent, so you have: 4 to the second over 4 (any number without any exponent is always to the first power), you subtract the 2 from the 1 and you keep the base, so it becomes 4 to the first power or just 4 the second one would be 5-7= -2, and when you have any base raised to a negative number, it just means 1 over that number raised to that power but without the negative, so for example you have 6 to the -2= 1 over 6 to the second power i hope that helps, good luck on your test!