Since ( 1/2) is constant with respect to ( t), move ( 1/2) out of the integral.( 1/2∫ 1+(cos)(2t)dt)Split the single integral into multipleintegrals.( 1/2(∫ 1dt+∫ (cos)(2t)dt))Since ( 1) is constant with respect to ( t), move ( 1) out of the integral.( 1/...
求定义域和值域 y = square root of x-2+1y=√x−2+1y=x-2+1 将√x−2x-2 的被开方数设为大于或等于 00,以求使表达式有意义的区间。 x−2≥0x-2≥0在不等式两边同时加上 22。 x≥2x≥2定义域为使表达式有定义的所有值 xx。 区间计数法: [2,∞)[2,∞) 集合符号: {x|x≥2}{x|...
(x-2)1=y2 x-2=y2 x-2=y2 x-2=y2 x−2=y2x-2=y2 在等式两边都加上22。 x=y2+2x=y2+2 xx yy f (y) = y + 2 y=2√x−2y=x-22 ( ) | [ ] √ ≥ 7 8 9
【解析】 Rewrite $$ 1 4 y ^ { 1 4 } $$_zas $$ ( y ^ { 7 } ) ^ { 2 } $$·(14z). $$ \sqrt { ( y ^ { 7 } ) ^ { 2 } \cdot ( 1 4 z ) } $$ Pull terms out from under the radical. $$ y ^ { 7 } \sqrt { 1 4 z } $$ 反馈...
Answer to: Find the area of the region below the curve y = square root of x and y = 1/ 2 x . By signing up, you'll get thousands of step-by-step...
( x^(1/2)-y^(1/2)=1) Differentiate both sides of the equation. ( d/(dx)(x^(1/2)-y^(1/2))=d/(dx)(1)) Differentiate the left side of the equation. ( -d/(dx)[y]1/(2y^(1/2))+1/(2x^(1/2))) Since ( 1) is constant with respect to ( x), the derivative...
Answer to: Find the inverse function of y = square root x - 2 square root x - 1, x greater than or equal to 2. By signing up, you'll get thousands...
a) f' (x) = 1 + 3 square root x, where f (4) = 25. b) f'' (x) = 2 cos x where f (0) = -1 and f'' (pi / 2) = 0. Solve the given initial value problems a) (e^x+y) \,dx + (2+x+ye^y)\,dy = 0; \quad y(0)...
解析 Subtract ( √x) from both sides of the equation. (√y=√a-√x) To remove the radical on the left side of the equation, square both sides of the equation. ( (√y)^2=((√a-√x))^2) Simplify each side of the equation. ( y=((√a-√x))^2) ...
代数输入 三角输入 微积分输入 矩阵输入 y=arcsin(x) 求解x 的值 x=sin(y) ∣y∣≤2π 求解y 的值 y=arcsin(x) ∣x∣≤1 图表 共享 复制 已复制到剪贴板