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The OutsidersBased on this quote, what do Dally and Two Bit want to learn in The Outsiders? Page 122: "Sixteen...Dally and Two Bit have grown up on the streets. They have learned how to physically fight, to use switchblades, to always watch for Socs, to hide their emotions (especially fear), and to trust no...

MathA function `f(x, y)` is called homogeneous if for some integer `n` `f(tx, ty) = t^n f(x, y).` The given function satisfies this condition with `n = 0:` `f(tx, ty) = tan((tx)/(ty)) = tan(x/y) = f(x,...

MathThe equation of the tangent line at the point `(t, f(t))` is `y  f(t) = (x  t)f'(t).` For the given point of intersection `(x + 2, 0)` this equation becomes `f(t) = 2f'(t),` or `f'(t) = 1/2...

MathTo solve this equation, multiply by `y` and integrate: `yy' = 9/16 x,` `int yy' dx = int (9/16 x) dx,` `y^2/2 = 9/32 x^2 + C,` or `y = +sqrt(C  9/16 x^2),` where `C` is an arbitrary...

MathWe need to find the equation of the graph that passes through the point (0,2). The slope of the tangent line to the graph of y(x) is the derivative y' and it is given by the equation `y' = x/(4y)`...

MathThis differential equation can be solved by separating the variables. `(dr)/(ds) = e^(r  2s)` Dividing by e^r and multiplying by ds results in the variables r and s on the different sides of the...

MathAn ordinary differential equation (ODE) has differential equation for a function with single variable. A first order ODE follows . In the given problem: `(du)/(dv)=uvsin(v^2)` , we may apply...

MathThe given problem: `ysqrt(1x^2)y' xsqrt(1y^2)=0` is written in a form of first order "ordinary differential equation" or first order ODE. To evaluate this, we can apply variable separable...

MathThe given problem: `ysqrt(1+x^2)y' xsqrt(1+y^2)=0` is written in a form of first order ODE (ordinary differential equation). To evaluate this, we can apply variable separable differential...

MathThe problem: `2xy'ln(x^2)=0 ` is as first order ordinary differential equation that we can evaluate by applying variable separable differential equation: `N(y)y'=M(x)` `N(y)(dy)/(dx)=M(x)` `N(y)...

MathI suppose `y(x+1)` means `y(x)*(x+1),` not `y` at the point `x+1.` This differential equation may be expressed as `(y')/y = x  1` and then integrated: `lny = x^2/2  x + C, or y = Ce^(x^2/2...

MathFor the given problem: `sqrt(x)+sqrt(y)y' =0,` we may rearrange this to `sqrt(y)y' = sqrt(x)` Recall that `y'` is denoted as `(dy)/(dx)` then it becomes: `sqrt(y)(dy)/(dx) = sqrt(x)` Apply the...

MathFor the given problem:` yy'2e^x=0` , we can evaluate this by applying variable separable differential equation in which we express it in a form of `f(y) dy = f(x)dx` . Then, `yy'2e^x=0` can be...

MathFor the given problem: `12yy'7e^x=0` , we can evaluate this by applying variable separable differential equation in which we express it in a form of `f(y) dy = f(x)dx` . Then, `12yy'7e^x=0` can...

MathFor the given problem: `yln(x)xy'=0` , we can evaluate this by applying variable separable differential equation in which we express it in a form of `f(y) dy = f(x)dx` . to able to apply direct...

MathTo be able to evaluate the problem: `sqrt(14x^2)y'=x` , we express in a form of `y'=f(x)` . To do this, we divide both sides by `sqrt(14x^2)` ....

MathThe general solution of a differential equation in a form of` f(y) y'=f(x)` can be evaluated using direct integration. We can denote `y'` as `(dy)/(dx) ` then, `f(y) y'=f(x)` `f(y)...

MathThe general solution of a differential equation in a form of `y' = f(x) ` can be evaluated using direct integration. The derivative of y denoted as` y'` can be written as `(dy)/(dx)` then `y'=...

MathAn ordinary differential equation (ODE) has differential equation for a function with single variable. A first order ODE follows y' =f(x,y). The `y'` can be denoted as `(dy)/(dx) ` to be able to...

MathRecall that `y'` is the same as `(dy)/(dx)` . Then in the given problem: `(2+x)y'=3y` , we may write it as: `(2+x) (dy)/(dx) = 3y.` This will help to follow the variable separable differential...

Math`(dr)/(ds)=0.75s` This differential equation is separable since it has a form `N(y) (dy)/dx=M(x)` And, it can be rewritten as `N(y) dy = M(x) dx` So separating the variables, the equation...

MathThe general solution of a differential equation in a form of can be evaluated using direct integration. The derivative of y denoted as `y'` can be written as `(dy)/(dx)` then `y'= f(x) ` can be...

Math`(dy)/dx = (6x^2)/(2y^3)` This differential equation is separable since it can be rewritten in the form `N(y) dy=M(x) dx` So separating the variables, the equation becomes `2y^3dy = (6x^2)dx`...

MathRecall that an ordinary differential equation (ODE) has differential equation for a function with single variable. A first order ODE follows `(dy)/(dx) = f(x,y)` . In the given problem:...

Math`(dy)/dx = 3x^2/y^2` This differential equation is separable since it can be rewritten in the form `N(y)dy = M(x) dx` So separating the variables, the equation becomes `y^2dy = 3x^2dx` Taking the...

Math`(dy)/dx = x/y` This differential equation is separable since it can be rewritten in the form `N(y)dy = M(x)dx` So separating the variables, the equation becomes `ydy = xdx` Integrating both...

MathFormula for compounding n times per year: `A=P(1+r/n)^(nt)` Formula for compounding continuously: `A=Pe^(rt)` A=Final Amount P=Initial Amount r=rate of investment expressed as a percent...

MathFormula for compounding n times per year `A=P(1+r/n)^(nt)` Formula for compounding continuously `A=Pe^(rt)` A=Final Amount P=Initial Amount r=rate of investment expressed as a decimal n=number of...

MathThe formula in compounding interest is `A = P(1 + r/n)^(n*t)` where A is the accumulated amount P is the principal r is the annual rate n is the number of compounding periods in a year, and t is...

MathThe formula in compounding interest is `A = P(1 + r/n)^(n*t)` where A is the accumulated amount P is the principal r is the annual rate n is the number of compounding periods in a year, and t is...

MathThe formula in compounding interest is `A = P(1 + r/n)^(n*t)` where A is the accumulated amount P is the principal r is the annual rate n is the number of compounding periods in a year, and t is...

MathThe formula in compounding interest is `A = P(1 + r/n)^(n*t)` where A is the accumulated amount P is the principal r is the annual rate n is the number of compounding periods in a year, and t is...

MathFormula: `y=Ce^(kt)` `1/2C=Ce^(k*1599)` `1/2=e^(k*1599)` `ln(1/2)=1599klne` `ln(1/2)=1599k` `ln(1/2)/1599=k` `k=4.3349x10^4` `y=Ce^(kt)` `y=Ce^[(4.3349x10^4)(100)]` `y=.9576C` Final...

Math`(dy)/dx=58x` This differential equation is separable since it has a form `N(y) (dy)/dx=M(x)` And, it can be rewritten as `N(y) dy = M(x) dx` So separating the variables, the equation...

Math`(dy)/dx = x + 3` This differential equation is separable since it can be rewritten in the form `N(y)dy = M(x) dx` So separating the variables, the equation becomes `dy = (x+3)dx` Integrating both...

MathIn this differential equation, the independent variable `x` and the dependent variable `y` are at the different sides of the equation. We can integrate both parts with respect to `x` and find the...

MathIn order to use integration to solve this differential equation, multiply both sides of the equation by dx: `dy = (10x^4 2x^3)dx` . Now we can integrate both sides, using the formula for the...

MathFirst, determine the derivative of the given function: `y'(x) = cos(x).` Then substitute `y` and `y'` into the given equation: `x cos(x)  2sin(x) = x^3 e^x.` Is this a true equality for all x?...

MathIt is known that the derivative of `cosh^1(x)` is equal to `1/sqrt(x^2  1).` Our function is `cosh^1(3x),` and to find its derivative we need to use the chain property: `y'= (cosh^1(3x))' =...

MathRecall that first fundamental theorem of calculus indicates that `int_a^b f(x) dx = F(x)_a^b:` `f(x)` as the integrand function `F(x)` as the antiderivative of `f(x)` "`a` " as the lower boundary...

MathIt is known that `(cosh(x))' = sinh(x),` and this integral is simply speaking `int cosh^2(x1) d(cosh(x1)).` Now let's make a formal substitution: `u =cosh(x1),` then `du = sinh(x1) dx,` and the...

MathThe tangent line must go through the given point `(x_0, y_0)` and have the slope of `y'(x_0).` Thus the equation of the tangent line is `(y  y_0) = (x  x_0)*y'(x_0).` To find the derivative we...

MathFirst, we need the derivatives of `sinh(x)` and `cosh(x).` They are wellknown: `(sinh(x))' = cosh(x)` and `(cosh(x))' = sinh(x).` Linearity rule and the product rule are then sufficient to find...

Math`f(x)=cosh(8x+ 1)` Take note that the derivative formula of cosh is `d/dx[cosh(u)] = sinh(u) *(du)/dx` Applying this formula, the derivative of the function will be `f'(x) = d/dx [cosh(8x+1)]`...

MathA limit is the value that the function approach as x approaches "a". In the given problem, the `xgt` `oo` indicates that independent variable x approaches large negative number or for given...

MathBy the definition, the hyperbolic sine `sinh(x) = (e^x  e^(x))/2.` When `x > +oo,` we have `e^x > +oo` and `e^(x) = 1/e^x > 0.` Therefore `sinh(x) > +oo` when `x > +oo.`...

MathLet's recall the definitions: `tanh(x) = (sinh(x))/(cosh(x)),` `se ch(x) = 1/(cosh(x)).` Also, `cosh(x) = (e^x + e^(x))/2` and `sinh(x) = (e^x  e^(x))/2.` Now the left side of our identity...

MathBy the definition, hyperbolic cosine `cosh(x) = (e^x + e^(x))/2.` For `x=0` we have `e^x = e^(x) = 1` and therefore `cosh(0) = (1 + 1)/2 =` 1. This is the answer. By the way, this is the minimum...

Sinners in the Hands of an Angry GodIn the sermon, Jonathan Edwards asserts that damnation "does not slumber; it will come swiftly, and, in all probability, very suddenly" upon those who least expect it. So, the answer is that...