answer:To prove the given identities involving hyperbolic functions, we will use the definitions of operatorname{sh}(x) and operatorname{ch}(x) as follows: [ operatorname{sh}(x) = frac{e^x - e^{-x}}{2} ] [ operatorname{ch}(x) = frac{e^x + e^{-x}}{2} ] For convenience, we will also use properties of exponential functions. Proving the first identity: (operatorname{sh}(x pm y)=operatorname{sh} x operatorname{ch} y pm operatorname{ch} x operatorname{sh} y) 1. **Expression for (operatorname{sh}(x pm y)):** [ operatorname{sh}(x pm y) = frac{e^{x pm y} - e^{-(x pm y)}}{2} ] 2. **Expand using exponent properties:** [ e^{x + y} = e^x e^y quad text{and} quad e^{x - y} = e^x e^{-y} ] [ e^{-(x + y)} = e^{-x} e^{-y} quad text{and} quad e^{-(x - y)} = e^{-x} e^{y} ] 3. Substitute these into the definition of (operatorname{sh}(x + y)): [ operatorname{sh}(x + y) = frac{e^x e^y - e^{-x} e^{-y}}{2} ] 4. Distribute the division by 2: [ operatorname{sh}(x + y) = frac{e^x e^y - e^{-x} e^{-y}}{2} ] 5. **Express in terms of (operatorname{sh}) and (operatorname{ch}):** [ operatorname{sh} x operatorname{ch} y = left( frac{e^x - e^{-x}}{2} right) left( frac{e^y + e^{-y}}{2} right) ] [ operatorname{ch} x operatorname{sh} y = left( frac{e^x + e^{-x}}{2} right) left( frac{e^y - e^{-y}}{2} right) ] 6. Multiply and then combine these: [ begin{aligned} operatorname{sh} x operatorname{ch} y & = frac{1}{4} left( e^x e^y + e^x e^{-y} - e^{-x} e^{y} - e^{-x} e^{-y} right) operatorname{ch} x operatorname{sh} y & = frac{1}{4} left( e^x e^y - e^x e^{-y} + e^{-x} e^{-x} - e^{-x} e^{-y} right) end{aligned} ] 7. Combine these results: [ operatorname{sh} x operatorname{ch} y + operatorname{ch} x operatorname{sh} y = frac{1}{4} left( 2e^x e^y - 2e^{-x} -y right) = frac{e^x e^y - e^{-x} e^{-y}}{2} ] Thus, we get: [ operatorname{sh}(x + y) = operatorname{sh} x operatorname{ch} y + operatorname{ch} x operatorname{sh} y ] For the negative: [ operatorname{sh}(x - y) = operatorname{sh} x operatorname{ch} y - operatorname{ch} x operatorname{sh} y ] Hence, [ boxed{operatorname{sh}(x pm y)=operatorname{sh} x operatorname{ch} y pm operatorname{ch} x operatorname{sh} y} ] # Proving the second identity: (operatorname{ch}(x pm y)=operatorname{ch} x operatorname{ch} y pm operatorname{sh} x operatorname{sh} y) 1. **Expression for (operatorname{ch}(x pm y)):** [ operatorname{ch}(x pm y) = frac{e^{x pm y} + e^{-(x pm y)}}{2} ] 2. **Expand using exponent properties:** [ e^{x + y} = e^x e^y quad text{and} quad e^{x - y} = e^x e^{-y} ] [ e^{-(x + y)} = e^{-x} e^{-y} quad text{and} quad e^{-(x - y)} = e^{-x} e^{y} ] 3. Substitute these into the definition of (operatorname{ch}(x + y)): [ operatorname{ch}(x + y) = frac{e^x e^y + e^{-x} e^{-y}}{2} ] 4. Distribute the division by 2: [ operatorname{ch}(x + y) = frac{e^x e^y + e^{-x} e^{-y}}{2} ] 5. **Express in terms of (operatorname{sh}) and (operatorname{ch}):** [ operatorname{sh} x operatorname{sh} y = left( frac{e^x - e^{-x}}{2} right) left( frac{e^y - e^{-y}}{2} right) ] [ operatorname{ch} x operatorname{ch} y = left( frac{e^x + e^{-x}}{2} right) left( frac{e^y + e^{-y}}{2} right) ] 6. Multiply and then combine these: [ begin{aligned} operatorname{sh} x operatorname{sh} y & = frac{1}{4} left( e^x e^y - e^x e^{-y} - e^{-x} e^{y} + e^{-x} e^{-y} right) operatorname{ch} x operatorname{ch} y & = frac{1}{4} left( e^x e^y + e^x e^{-y} + e^{-x} e^{y} + e^{-x} e^{-y} right) end{aligned} ] 7. Combine these results: [ operatorname{sh} x operatorname{sh} y + operatorname{ch}x operatorname{ch} y = frac{1}{4} left( 2e^x e^y + 2e^{-x} -y right) = frac{e^x e^y + e^{-x} e^{-y}}{2} ] Thus, we get: [ operatorname{ch}(x + y) = operatorname{ch} x operatorname{ch} y + operatorname{sh} x operatorname{sh} y ] For the negative: [ operatorname{ch}(x - y) = operatorname{ch} x operatorname{ch} y - operatorname{sh} x operatorname{sh} y ] Hence, [ boxed{operatorname{ch}(x pm y)=operatorname{ch} x operatorname{ch} y pm operatorname{sh} x operatorname{sh} y} ]

answer:If Jennifer will be 30 years old in ten years, that means she is currently 30 - 10 = 20 years old. In ten years, Jordana will be three times as old as Jennifer will be at that time. So, Jordana will be 3 * 30 = 90 years old in ten years. To find out how old Jordana is now, we subtract 10 years from her age in ten years: 90 - 10 = 80 years old. Therefore, Jennifer's sister Jordana is currently boxed{80} years old.

answer:To determine how much each person should pay for the rent of the pasture, we need to calculate the share of each person based on the number of oxen they put in and the duration for which they used the pasture. Let's calculate the share for each person: For a: Number of oxen = 10 Duration (months) = 7 Share of a = 10 oxen * 7 months = 70 ox-months For b: Number of oxen = 12 Duration (months) = 5 Share of b = 12 oxen * 5 months = 60 ox-months For c: Number of oxen = 15 Duration (months) = 3 Share of c = 15 oxen * 3 months = 45 ox-months Now, let's calculate the total ox-months used by all three: Total ox-months = Share of a + Share of b + Share of c Total ox-months = 70 + 60 + 45 Total ox-months = 175 ox-months The total rent for the pasture is Rs. 140. To find out how much c should pay, we need to calculate the proportion of the total ox-months that c used: Proportion of c's use = Share of c / Total ox-months Proportion of c's use = 45 / 175 Now, we can calculate the amount c should pay based on his proportion of use: Amount c should pay = Proportion of c's use * Total rent Amount c should pay = (45 / 175) * Rs. 140 Let's calculate the exact amount: Amount c should pay = (45 / 175) * 140 Amount c should pay = (45 * 140) / 175 Amount c should pay = 6300 / 175 Amount c should pay = Rs. 36 Therefore, c should pay Rs. boxed{36} as his share of the rent for the pasture.

answer:To solve the expression, combine like terms: - **Constant terms**: (3 + 9 + 15 = 27) - **Linear terms in (x)**: (-5x + 11x - 17x = -11x) - **Quadratic terms in (x^2)**: (-7x^2 - 13x^2 + 19x^2 = -1x^2) - **Cubic term in (x^3)**: (2x^3) Thus, the combined expression equates to: [ 27 - 11x - x^2 + 2x^3 ] Therefore, the simplified expression is: [ boxed{2x^3 - x^2 - 11x + 27} ]