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3 years ago

What are the answers for a-c... Need the answers asap...thank you

Mathematics

1 answer:

xxMikexx [17]3 years ago

7 0

For Letter C You Need to Count The Squares in the Pond And the Half Squares are a Half. The Answer for C is 33 Because The Half and the Other Half is a Whole So I Made 3 New Wholes and Added it to the 30.

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Please help me with precalculus?? linear and angular speed

Anit [1.1K]

13)

there are 2π radians in 1 revolution, and there are 60 seconds in 1 minute, so keeping that in mind, then,

$\bf \cfrac{4\underline{\pi} }{5~\underline{s}}\cdot \cfrac{rev}{2\underline{\pi} }\cdot \cfrac{60~\underline{s}}{min}\implies \cfrac{4\cdot 60~rev}{5\cdot 2~min}\implies \cfrac{240~rev}{10~min}\implies 24\frac{rev}{min}$

14)

$\bf \textit{linear velocity}\\\\
v=rw\quad 
\begin{cases}
r=radius\\
w=angular~speed\\
----------\\
v=32\frac{m}{sec}\\
w=100\frac{rev}{min}
\end{cases}\\\\
-------------------------------\\\\
\textit{let's convert \underline{w} to }\frac{radians}{sec}$

$\bf \cfrac{100~\underline{rev}}{\underline{min}}\cdot \cfrac{2\pi }{\underline{rev}}\cdot \cfrac{\underline{min}}{60~sec}\implies \cfrac{100\cdot 2\pi }{60~sec}\implies \cfrac{10\pi }{3~sec}\implies \cfrac{10\pi }{3}\frac{radians}{sec}\\\\
-------------------------------\\\\
v=rw\implies \cfrac{v}{w}=r\implies \cfrac{\frac{30~m}{sec}}{\frac{10\pi }{3~sec}}\implies r=\cfrac{30~m}{\underline{sec}}\cdot \cfrac{3~\underline{sec}}{10\pi }
\\\\\\
r=\cfrac{90}{10\pi }m$

15)

what is the radians per seconds "w" in revolutions per minute? just another conversion like in 13)

$\bf \cfrac{\underline{\pi} }{3~\underline{sec}}\cdot \cfrac{rev}{2\underline{\pi }}\cdot \cfrac{60~\underline{sec}}{min}\implies \cfrac{60 ~rev}{3\cdot 2 ~min}\implies \cfrac{60 ~rev}{6 ~min}\implies 10\frac{rev}{min}$

4 0

3 years ago

Factor x^2+5x-50 A. (x+25)(x-2) B. (x-5)(x+10) C. (x-10)(x+5) D. (x+2)(x-25)

Drupady [299]

The answer would be B

7 0

3 years ago

Use lagrange multiplier techniques to find the local extreme values of f(x, y) = x2 − y2 − 2 subject to the constraint x2 + y2 =

dexar [7]

Given $f(x,\ y)=x^2-y^2-2$ subject to the constraint $x^2+y^2=16$

Let $g(x,\ y)=x^2+y^2$ .

The gradient vectors of $f$ and $g$ are:

$\nabla f(x,\ y)=\langle2x,-2y\rangle$ and $\nabla g(x,\ y)=\langle2x,2y\rangle$

By Lagrange's theorem, there is a number $\lambda$ , such that

$\langle2x,-2y\rangle=\lambda\langle2x,2y\rangle=\langle2\lambda x,2\lambda y\rangle$

$\lambda=\pm1$

It can be seen that $f(x,\ y)=x^2-y^2-2$ has local extreme values at the given region.

6 0

3 years ago

I need this answer!!!

Kamila [148]

Answer:

2 feet per second

Step-by-step explanation:

6/3 is 2

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3 years ago

Read 2 more answers

Im dehydrated i need water fr

Margarita [4]

Bone dry, sorry nothing I can do about that

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3 years ago