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3 years ago

Suppose that the functions s and t are defined for all real numbers x as follows.

1 answer:

8 0

$s(x)=2x+4;\ D_s=\mathbb{R}\\\\t(x)=5x;\ D_t=\mathbb{R}\\\\D_s=D_t\\\\(s+t)(x)=s(x)+t(x)=2x+4+5x=7x+4\\\\(s-t)(x)=s(x)-t(x)=2x+4-5x=-3x+4\\\\(s\times t)(x)=(2x+4)\times5x=10x^2+20x$

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2 years ago

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3 years ago

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telo118 [61]

Answer:

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Step-by-step explanation:

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3 years ago

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tigry1 [53]

Answer:

y=17/5 ,x= 38

Step-by-step explanation:

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5 0

2 years ago

1 pt) If a parametric surface given by r1(u,v)=f(u,v)i+g(u,v)j+h(u,v)k and −4≤u≤4,−4≤v≤4, has surface area equal to 1, what is t

natta225 [31]

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$1=\displaystyle\int_{-4}^4\int_{-4}^4\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

By multiplying each component in $\vec r_1$ by 5, we have

$\dfrac{\partial\vec r_2(u,v)}{\partial u}=5\dfrac{\partial\vec r_1(u,v)}{\partial u}$

and the same goes for the derivative with respect to $v$ . Then the area of the surface given by $\vec r_2(u,v)$ is

$\displaystyle\int_{-4}^4\int_{-4}^425\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\boxed{25}$

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3 years ago