4y
To find the product of (4x^4y^2) and (5y^4), multiply the coefficients and combine the like variables. The coefficients (4) and (5) multiply to give (20). For the variables, (y^2) and (y^4) combine to (y^{2+4} = y^6). Therefore, the product is (20x^4y^6).
To find the product of (-5x^2y^4) and (-8x^6y^4), multiply the coefficients and add the exponents of the like bases. The coefficients give (-5 \times -8 = 40), and for (x), (x^2 \times x^6 = x^{2+6} = x^8). For (y), (y^4 \times y^4 = y^{4+4} = y^8). Therefore, the product is (40x^8y^8).
When multiplying powers, you add them! y4 times y6 = y10. Try it with y = 2: 2 to the fourth = 16, 2 to the sixth = 64 16 x 64 = 1024 = 2 to the tenth.
The two whole numbers that have a difference of 4 and a product of 77 are 11 and 7. You can find these numbers by setting up the equations ( x - y = 4 ) and ( x \cdot y = 77 ). Solving these equations, you can express ( x ) as ( y + 4 ) and substitute it into the product equation, leading to ( (y + 4)y = 77 ). This results in the quadratic equation ( y^2 + 4y - 77 = 0 ), which factors to yield the solutions ( y = 7 ) and ( x = 11 ).
The product is (xy - 3x + 4y -12).I got that by performing the multiplication of (x + 4) (y - 3) .
To find the product of (4x^4y^2) and (5y^4), multiply the coefficients and combine the like variables. The coefficients (4) and (5) multiply to give (20). For the variables, (y^2) and (y^4) combine to (y^{2+4} = y^6). Therefore, the product is (20x^4y^6).
abs(y^3 - 4) < 8*y
To find the product of (-5x^2y^4) and (-8x^6y^4), multiply the coefficients and add the exponents of the like bases. The coefficients give (-5 \times -8 = 40), and for (x), (x^2 \times x^6 = x^{2+6} = x^8). For (y), (y^4 \times y^4 = y^{4+4} = y^8). Therefore, the product is (40x^8y^8).
When multiplying powers, you add them! y4 times y6 = y10. Try it with y = 2: 2 to the fourth = 16, 2 to the sixth = 64 16 x 64 = 1024 = 2 to the tenth.
The two whole numbers that have a difference of 4 and a product of 77 are 11 and 7. You can find these numbers by setting up the equations ( x - y = 4 ) and ( x \cdot y = 77 ). Solving these equations, you can express ( x ) as ( y + 4 ) and substitute it into the product equation, leading to ( (y + 4)y = 77 ). This results in the quadratic equation ( y^2 + 4y - 77 = 0 ), which factors to yield the solutions ( y = 7 ) and ( x = 11 ).
The product is (xy - 3x + 4y -12).I got that by performing the multiplication of (x + 4) (y - 3) .
(4y)-2
You multiply the unit cost by the number of units. The product of y and 4 is 4y.
39 * 406 = 15834
xy = 20 x+y=9 x=9-y y(9-y) = 20 9y-y(squared) = 20 0= y(squared)-9y+20 0= (y-5)(y-4) y= 5 or 4, x= 5 or 4 The two numbers are 5 and 4.
-2
x + y = 5xy=41 and 4 works. ■