Simplify each element of the matrix( [(array)(cc)-3⋅ x+1& -3⋅ 2 -3⋅ 0& -3⋅ 1(array)]).( [(array)(cc)6& y 18& z(array)]+[(array)(cc)-3x+1& -6 0& -3(array)]=[(array)(cc)18& 17 9w& 4z(array)])Add the corresponding elements of ( [(ar...
(1)2× 103 (2)72× 22 (3)23× 5 (4)3× 44 (5)0× 102 (6)53× 33 (7)24× 32 (8)32× 104 相关知识点: 试题来源: 解析 (1)Solution:2× 103 = 2 × 10 × 10 × 10 = 2000 (2)72× 22 = 7 × 7 × 2 × 2 = 196 (3)23× 5 = 2 × 2 × 2 ×...
Simplify: a.(4 iota^(3)-iota)/(2 iota+1) b.(sqrt(3)+iota)^(6) 05:56 Find the square root of : a.-24+10 iota b.5+12 iota 03:06 If (2+iota)(2+2 iota)(2+3 iota)...(2+9 iota)=x+iota y, then 5.8.13... 02:08 The modulus value of (3-2t)(3+2 iota)(1...
View Solution What is the simplified value of (x+1x)(x2+1x2)(x4+1x4)(x8+1x6)(x16+1x16) View Solution Free Ncert Solutions English Medium NCERT Solutions NCERT Solutions for Class 12 English Medium NCERT Solutions for Class 11 English Medium ...
若要简化模型,我们替换 regressors 变量 x 1 和 x 2 张力力 x,在每个测量时刻被计算作为平均 x 1 和 x 2。考虑到图 6 的理论 翻译结果4复制译文编辑译文朗读译文返回顶部 要简化模型,我们替代regressors可变物x1和x2用在每测量片刻被计算作为平均x1和x2的紧张力量x。考虑到图6,理论 翻译结果5复制译文编辑...
百度试题 结果1 题目Simplify:[7^(-1)+(3/2)]^(-1)÷[6^(-1)+(3/2)^(17-1) (34)/(35) z(35)/(34) (21)/(105) 4/(121)E)None of these 相关知识点: 试题来源: 解析 B 反馈 收藏
结果1 题目9. Simplify:(a)|-7|+|2| (b) |-9|+|5|-|-7| | (c)||-2||3|-|-3||-2|(d) (|5|-|-3|)*1-6| |(e) (|-3|+|2|)(|-4|-|-2|) (a) 9 (b) 7 (c)0 (d)12 (e)10 相关知识点: 试题来源: 解析 (a) 9 (b) 7 (c) 0 (d) ...
12⋅(6−10y+4z)12⋅(6-10y+4z) Apply thedistributive property. 12⋅6+12(−10y)+12(4z) Simplify. Tap for more steps... Cancel theof2. 2out of6. 12⋅(2(3))+12(-10y)+12(4z) 12⋅(2⋅3)+12(-10y)+12(4z) ...
2. 4^{\log_4 6} - \log_4{4^{-2}} {/eq} Logarithms: The logarithm of a number, {eq}y=\log_b x {/eq}, is the number {eq}y {/eq} such that, {eq}x=b^y {/eq}. The number {eq}b>0 {/eq} is the base of the logarithm. For {eq}x<1 {/eq}...
6by1. 12⋅(6+7)+11 Add6and7. 12⋅13+11 12and13. 132+11 132+11132+11 To write1111as afractionwith acommon denominator,multiplyby2222. 132+11⋅22132+11⋅22 Combine1111and2222. 132+11⋅22132+11⋅22 Combinethenumeratorsover thecommon denominator. ...