To write this algebraically: 7(y^3)|y = 2
Substitute 2 for y: 7(2^3)
2^3 = 8, so substitute 8 for (2^3): 7*8 = 56
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β 12y agoThere needs to be some operator (+,*,/ etc) between "third power" and "7".
It's not 7 to the power of 2
8^3 (8x8x8) = 512 7^3 (7x7x7) = 343 9^3 (9x9x9) = 729
7
7x7x7x7=2401the product is 2401
There needs to be some operator (+,*,/ etc) between "third power" and "7".
56
7 times z reduced by a third of the product
If you mean: 2^3 times 2^7 times 2^3 then it equals 8192
2 to the power of 3 equals 8.
56 ^ 2 28 ^ 2 14 ^ 2 7 pf:2x2x2x7 2 to the third power 7
It's not 7 to the power of 2
Let's see in a prime factorization tree. 2, 98 2, 2, 49 2, 2, 7, 7. The answer is:2 to the power of 2 and 7 to the power of 2 ;/or 2 x 2 x 7 x 7
(5.9x^3 + 3.4x^2 - 7) - (2.9x^3 - 9.6x^2 + 3) = 3x^3 +13x^2 -10
7 to the third power or 343
8^3 (8x8x8) = 512 7^3 (7x7x7) = 343 9^3 (9x9x9) = 729
7