標題:
數條中一英文數學題(未必答曬)
發問:
1. 62∨(10)=2. Represent each of the following by a binary number. 1×2^4+0×2^3+1×2^2+0×2^1+13. Write down the following numbers in expanded form. 11110∨(2)4. Simplify the following expressions. (a) (x^3/y^2)^4.(x^2y)^3 (b)... 顯示更多 1. 62∨(10)= 2. Represent each of the following by a binary number. 1×2^4+0×2^3+1×2^2+0×2^1+1 3. Write down the following numbers in expanded form. 11110∨(2) 4. Simplify the following expressions. (a) (x^3/y^2)^4.(x^2y)^3 (b) a^2×a^3×b^4/(a^2b^3) (c) (-4a)^3/(-2ab)^2 5. Convert 110110∨(2) into a denary number. 6. Convert 37∨ (10) into a binary number. 7. Convert 444∨(10) into a hexadecimal number. 8. Convert the following binary numbers into hexadecimal numbers. 111101110101∨2 9. Convert the following hexadecimal numbers into binary numbers. 9C∨16 10. There are mistakes in the shaded parts. Make necessary corrections. (a) 2^3×3^2=(2×3)^3+2=6^5(the shaded part is =(2×3)^3+2=6^5) (b) (-a^2)^3=a^2×3=a^6(the shaded part is =a^2×3=a^6) (c) (a^2)^3=a^5(the shaded part is =a^5) (d) a^10.a^2/a^5.a^6=a^10+2-5+6=a^13(the shaded part is=a^10+2-5+6=a^13 )
最佳解答:
1. 62(10) = 6x10 + 2 ...... (ans) 2. The binary number = 10101 ...... (ans) 3. Expanded form = 1x2^4 + 1x2^3 + 1x2^2 + 1x2^1 + 0 ...... (ans) 4. (a) (x^3/y^2)^4.(x^2y)^3 = (x^12/y^8).(x^6y^3) = x^(12+6).y^(-8+3) = x^18.y^(-5) = x^18/y^5 ...... (ans) (b) a^2×a^3×b^4/(a^2b^3) = a^(2+3+2).b^(4-3) = a^7b ...... (ans) (c) (-4a)^3/(-2ab)^2 = (-4)^3a^3/(-2)^2a^2b^2 = -64a^3/4a^2b^2 = (-64/4)a^(3-2)/b^2 = -16a/b^2 ...... (ans) 5. 110110(2) = 1x2^5 + 1x2^4 + 0x2^3 + 1x2^2 + 1x2^1 + 0 = 32 + 16 + 0 + 4 + 2 + 0 = 54(10) ...... (ans) 6. 37/2 = 18 ... 1 18/2 = 9 ... 0 9/2 = 4 ... 1 4/2 = 2 ... 0 2/2 = 1 ... 0 27(10) = 100101(2) ...... (ans) 7. 444/16 = 27 ... 12=C 27/16 = 1 ... 11=B 444(10) = 1BC(16) ...... (ans) 8. 111101110101(2) = 1111(2)x2^8 + 111(2)x2^4 + 101(2) = (2^3 + 2^2 + 2^1 + 1)x16^2 + (2^2 + 2^1 + 1)x16 + (2^2 + 1) = (8 + 4 + 2 + 1)x16^2 + (4 + 2 + 1)x16 + (4 + 1) = 15x16^2 + 7x16 + 5 = F75(16) ...... (ans) 9. 9/2 = 4 ... 1 4/2 = 2 ... 0 2/1 = 1 ... 0 12/2 = 6 ... 0 6/2 = 3 ... 0 3/2 = 1 ... 1 9C(16) = 9x16 + 12 = 1001(2)x2^4 + 1100(2) = 10011100(2) ...... (ans) 10. (a) Correction: 2^3 x 3^2 cannot be simplified by using index laws. ...... (ans) (b) Correct: (-a^2)^3 = [(-1).a^2]^3 = (-1)^3.a^(2x3) = -a^6 ...... (ans) (c) Correction: (a^2)^3 = a^(2x3) = a^6 ...... (ans) (d) Correction: a^10.a^2/a^5.a^6 = a^(10+2-5-6) = a ...... (ans)
其他解答:
數條中一英文數學題(未必答曬)
發問:
1. 62∨(10)=2. Represent each of the following by a binary number. 1×2^4+0×2^3+1×2^2+0×2^1+13. Write down the following numbers in expanded form. 11110∨(2)4. Simplify the following expressions. (a) (x^3/y^2)^4.(x^2y)^3 (b)... 顯示更多 1. 62∨(10)= 2. Represent each of the following by a binary number. 1×2^4+0×2^3+1×2^2+0×2^1+1 3. Write down the following numbers in expanded form. 11110∨(2) 4. Simplify the following expressions. (a) (x^3/y^2)^4.(x^2y)^3 (b) a^2×a^3×b^4/(a^2b^3) (c) (-4a)^3/(-2ab)^2 5. Convert 110110∨(2) into a denary number. 6. Convert 37∨ (10) into a binary number. 7. Convert 444∨(10) into a hexadecimal number. 8. Convert the following binary numbers into hexadecimal numbers. 111101110101∨2 9. Convert the following hexadecimal numbers into binary numbers. 9C∨16 10. There are mistakes in the shaded parts. Make necessary corrections. (a) 2^3×3^2=(2×3)^3+2=6^5(the shaded part is =(2×3)^3+2=6^5) (b) (-a^2)^3=a^2×3=a^6(the shaded part is =a^2×3=a^6) (c) (a^2)^3=a^5(the shaded part is =a^5) (d) a^10.a^2/a^5.a^6=a^10+2-5+6=a^13(the shaded part is=a^10+2-5+6=a^13 )
最佳解答:
1. 62(10) = 6x10 + 2 ...... (ans) 2. The binary number = 10101 ...... (ans) 3. Expanded form = 1x2^4 + 1x2^3 + 1x2^2 + 1x2^1 + 0 ...... (ans) 4. (a) (x^3/y^2)^4.(x^2y)^3 = (x^12/y^8).(x^6y^3) = x^(12+6).y^(-8+3) = x^18.y^(-5) = x^18/y^5 ...... (ans) (b) a^2×a^3×b^4/(a^2b^3) = a^(2+3+2).b^(4-3) = a^7b ...... (ans) (c) (-4a)^3/(-2ab)^2 = (-4)^3a^3/(-2)^2a^2b^2 = -64a^3/4a^2b^2 = (-64/4)a^(3-2)/b^2 = -16a/b^2 ...... (ans) 5. 110110(2) = 1x2^5 + 1x2^4 + 0x2^3 + 1x2^2 + 1x2^1 + 0 = 32 + 16 + 0 + 4 + 2 + 0 = 54(10) ...... (ans) 6. 37/2 = 18 ... 1 18/2 = 9 ... 0 9/2 = 4 ... 1 4/2 = 2 ... 0 2/2 = 1 ... 0 27(10) = 100101(2) ...... (ans) 7. 444/16 = 27 ... 12=C 27/16 = 1 ... 11=B 444(10) = 1BC(16) ...... (ans) 8. 111101110101(2) = 1111(2)x2^8 + 111(2)x2^4 + 101(2) = (2^3 + 2^2 + 2^1 + 1)x16^2 + (2^2 + 2^1 + 1)x16 + (2^2 + 1) = (8 + 4 + 2 + 1)x16^2 + (4 + 2 + 1)x16 + (4 + 1) = 15x16^2 + 7x16 + 5 = F75(16) ...... (ans) 9. 9/2 = 4 ... 1 4/2 = 2 ... 0 2/1 = 1 ... 0 12/2 = 6 ... 0 6/2 = 3 ... 0 3/2 = 1 ... 1 9C(16) = 9x16 + 12 = 1001(2)x2^4 + 1100(2) = 10011100(2) ...... (ans) 10. (a) Correction: 2^3 x 3^2 cannot be simplified by using index laws. ...... (ans) (b) Correct: (-a^2)^3 = [(-1).a^2]^3 = (-1)^3.a^(2x3) = -a^6 ...... (ans) (c) Correction: (a^2)^3 = a^(2x3) = a^6 ...... (ans) (d) Correction: a^10.a^2/a^5.a^6 = a^(10+2-5-6) = a ...... (ans)
其他解答:
此文章來自奇摩知識+如有不便請留言告知
1. 62(10) is a denary number , so the answer is 62 naturally . 2.1×2^4+0×2^3+1×2^2+0×2^1+1 = 10101(2) . 3.1×2^4+1×2^3+1×2^2+1×2^1+0 ( expanded form ) 5.110110(2) is a binary number , so the answer is 1*2^5+1*2^4+0*2^3+1*2^2+1*2^1+0 = 54(10) 6. 37(10) is a denary number , so the answer is 100101(2) . 7. 444(10) is a denary number , so the answer is 1BC(16) . 8.111101110101(2) is a binary number , so the answer is 1*2^11+1*2^10+1*2^9+1*2^8+0*2^7+1*2^6+1*2^5+1*2^4+0*2^3+1*2^2+0*2^1+1 = 3957(10) . 3957(10) = F75(16) 9.9C(16) is a hexadecimal numbers , so the answer is 10011100(2) 第4and10條未能幫你解答 . 好對不起 .
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