answer:【Solution】Solution: 160 times 3-90,begin{array}{l}=480-90, =390 text { (meters), }end{array}Answer: The cave is 390 meters long.Therefore, the answer is: 390.

answer:1. Given the function ( f ) defined for all ( x in mathbb{R} ) with ( f(0) = 0 ), and the properties: [ f(9 + x) = f(9 - x) quad text{and} quad f(x - 10) = f(-x - 10) quad forall x in mathbb{R} ] we need to determine the least number of zeros ( f ) can have in the interval ([0, 2014]).2. From the first property ( f(9 + x) = f(9 - x) ), we can rewrite it as: [ f(9 + (x - 9)) = f(9 - (x - 9)) implies f(x) = f(18 - x) ] This shows that ( f ) is symmetric about ( x = 9 ).3. From the second property ( f(x - 10) = f(-x - 10) ), we can rewrite it as: [ f((x + 10) - 10) = f(-(x + 10) - 10) implies f(x) = f(-x - 20) ] This shows that ( f ) is symmetric about ( x = -10 ).4. Combining these symmetries, we get: [ f(x) = f(18 - x) = f(x - 38) ] This implies that ( f ) has a period of 38.5. Since ( f ) is periodic with period 38, we can find the zeros of ( f ) within the interval ([0, 2014]). Given ( f(0) = 0 ), we have: [ f(38n) = 0 quad text{for all integers } n ] and since ( f(x) = f(18 - x) ), we also have: [ f(38n + 18) = 0 quad text{for all integers } n ]6. To find the number of zeros in ([0, 2014]), we calculate: [ text{For } f(38n) = 0: quad 38n leq 2014 implies n leq leftlfloor frac{2014}{38} rightrfloor = 53 ] This gives us 54 zeros (including ( n = 0 )).7. Similarly, for ( f(38n + 18) = 0 ): [ 38n + 18 leq 2014 implies 38n leq 1996 implies n leq leftlfloor frac{1996}{38} rightrfloor = 52 ] This gives us 53 zeros.8. Therefore, the total number of zeros in ([0, 2014]) is: [ 54 + 53 = 107 ]9. To show that this minimum is achievable, consider the function: [ f(x) = cosleft(frac{pi(x - 9)}{19}right) - cosleft(frac{9pi}{19}right) ] This function has zeros precisely at ( x equiv 0 pmod{38} ) and ( x equiv 18 pmod{38} ).The final answer is (boxed{107}).

answer:## Solutiony^{prime}=left(operatorname{arctg}left(e^{x}-e^{-x}right)right)^{prime}=frac{1}{1+left(e^{x}-e^{-x}right)^{2}} cdotleft(e^{x}-e^{-x}right)^{prime}=begin{aligned}& =frac{1}{1+left(e^{x}-e^{-x}right)^{2}} cdotleft(e^{x}+e^{-x}right)=frac{e^{x}+e^{-x}}{1+e^{2 x}-2+e^{-2 x}}= & =frac{e^{x}+e^{-x}}{1+e^{2 x}-2+e^{-2 x}}=frac{e^{3 x}+e^{x}}{e^{4 x}-e^{2 x}+1}end{aligned}Kuznetsov Differentiation 7-26

answer:Given that the terminal side of angle α passes through the point (1, -1), we have x = 1 and y = -1. The distance r from the origin to the point (x, y) is given by:r = sqrt {x^{2}+y^{2}}= sqrt {2}Using the definition of the cosine function for any angle, we have:cosα = frac {x}{r}= frac { sqrt {2}}{2}Therefore, the correct answer is:boxed{text{C}}This problem requires the application of the definition of trigonometric functions for any angle. It is a basic problem that primarily tests the understanding of this concept.

answer:Solution. Let's find the density function:f(x)=F^{prime}(x)=left{begin{array}{ccc}0 & text { for } & x2end{array}right.Let's find the expected valueM(X)=int_{-2}^{2} x f(x) mathrm{d} x=int_{-2}^{2} x cdot frac{1}{4} mathrm{~d} x=0(the integrand function is odd, and the limits of integration are symmetric about the origin).Let's find the required variance, considering that M(X)=0:D(X)=int_{-2}^{2}[x-M(X)]^{2} f(x) mathrm{d} x=int_{-2}^{2} x^{2} cdot frac{1}{4} mathrm{~d} x=frac{2}{4} int_{0}^{2} x^{2} mathrm{~d} x=frac{4}{3}

answer:## Solution.Let's write the equation asbegin{aligned}& (sin 2 t-3) cdot 2 sin ^{2} tleft(sin ^{2} t-1right)-1=0 & -(sin 2 t-3) cdot 2 sin ^{2} t cos ^{2} t-1=0, (sin 2 t-3) cdot 4 sin ^{2} t cos ^{2} t+2=0 & (sin 2 t-3) sin ^{2} 2 t+2=0, sin ^{3} 2 t-3 sin ^{2} 2 t+2=0 & sin ^{3} 2 t-sin ^{2} 2 t-2 sin 2 t+2=0, sin ^{2} 2 t(sin 2 t-1)-2left(sin ^{2} 2 t-1right)=0 & sin ^{2} 2 t(sin 2 t-1)-2(sin 2 t-1)(sin 2 t+1)=0 & (sin 2 t-1)left(sin ^{2} 2 t-2 sin 2 t-2right)=0end{aligned}From this,1) sin 2 t-1=0, sin 2 t=1, 2 t=frac{pi}{2}+2 pi k, t_{1}=frac{pi}{4}+pi k=frac{pi}{4}(4 k+1) k in mathbb{Z};2) sin ^{2} 2 t-2 sin 2 t-2=0. Solving this equation as a quadratic in sin 2 t, we get sin 2 t=1+sqrt{3}>1, varnothing ; sin 2 t=1-sqrt{3},2 t=(-1)^{n} arcsin (1-sqrt{3})+pi n, t_{2}=(-1)^{n} frac{1}{2} arcsin (1-sqrt{3})+frac{pi n}{2}, n in mathbb{Z}Answer: t_{1}=frac{pi}{4}(4 k+1) ; t_{2}=(-1)^{n} frac{1}{2} arcsin (1-sqrt{3})+frac{pi n}{2}, where k and n in mathbb{Z}.