3 years ago

(GIVING BRAINLIEST!!)

Mathematics

1 answer:

Ira Lisetskai [31]3 years ago

6 0

Answer:

IT IS THE ANSWER B) SNOW

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Sketch the curve with the given vector equation. Indicate with an arrow the direction in which t increases. r(t) = 4 cos(t)i + 4

slega [8]

Answer:

The equation to this question is " $x^2+y^2=16$ "

Step-by-step explanation:

Given value:

$\to r(t) = 4 \cos (t)i + 4 sin(t)j + k\\where \\\\\to x(t)= 4 \cos t\\\to y(t) = 4 \sin t\\\to z(t)=1\\$

$\to x^2+y^2= 16 \cos^2 t +16 \sin^2 t\\\\\to x^2+y^2= 16 (\cos^2 t +\sin^2 t)\\\\\therefore (\cos^2 t +\sin^2 t =1)\\\\\to x^2+y^2= 16 (1)\\\\\to x^2+y^2= 16\\\\\to z=1$

So, the curve $r(t) = 4 \cos (t)i + 4 sin(t)j + k$ is basically a circle, which is defined in the attachment file.

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3 years ago

What is (x^-2 y^10)^2/x^5 y^-3 using only positive exponents

Eddi Din [679]

Answer:

$\frac{(x^{ - 2}{y}^{10})^{2} }{ {x}^{5} {y}^{ - 3} } \\ \frac{ {x}^{ - 4} {y}^{20} }{ {x}^{5} {y}^{ - 3 } } \\ {x}^{ - 9} {y}^{23} \\ \frac{ {y}^{23} }{ {x}^{9} } \\ ans \: c$