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xeze [42]
3 years ago
11

A grain silo is show below ,168 length and 6 width what is the volume of the green in the completely fill the cello rounded to t

he nearest whole number use 22/7
Mathematics
1 answer:
Gnom [1K]3 years ago
7 0

Answer:

around the globe

Step-by-step explanation:

the reason why is not important

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Find the value of g(-1)
Anna11 [10]

Answer:

g(-1) = -3

Step-by-step explanation:

g(-1) is the y value of the graph when x = -1

when x=-1 y=-3

g(-1) = -3

6 0
3 years ago
A circle has a circumference of 9π what is the diameter?
mixas84 [53]
C=πd  where C=circumference, π=the constant pi, and d=diamter

d=C/π  and since we are told that C=9π

d=9π/π

d=9
7 0
3 years ago
Subtract. (4x2+8x−2)−(2x2−4x+3) Enter your answer, in standard form, in the box.
Crazy boy [7]

the answer is 2x2+12z-5 (its also in the picture)

3 0
3 years ago
If f(x)=5x^2-8, what is the value of F(-3)?
mr Goodwill [35]

Answer:

F(-3)=5(-3)^2-8

=37

Step-by-step explanation:

Get it?

8 0
3 years ago
<img src="https://tex.z-dn.net/?f=prove%20that%5C%20%20%5Ctextless%20%5C%20br%20%2F%5C%20%20%5Ctextgreater%20%5C%20%5Cfrac%20%7B
inysia [295]

\large \bigstar \frak{ } \large\underline{\sf{Solution-}}

Consider, LHS

\begin{gathered}\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } \\ \end{gathered}

We know,

\begin{gathered}\boxed{\sf{  \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }} \\ \end{gathered}  \\  \\  \text{So, using this identity, we get} \\  \\ \begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - ( {sec}^{2}\theta - {tan}^{2}\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}

We know,

\begin{gathered}\boxed{\sf{  \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: \: }} \\ \end{gathered}  \\

So, using this identity, we get

\begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - (sec\theta + tan\theta )(sec\theta - tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}

can be rewritten as

\begin{gathered}\rm\:=\:\dfrac {(\sec \theta + tan\theta ) - (sec\theta + tan\theta )(sec\theta -tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered} \\  \\  \\\begin{gathered}\rm \: = \:\dfrac {(\sec \theta + tan\theta ) \: \cancel{(1 - sec\theta + tan\theta )}} { \cancel{ \tan \theta - \sec \theta + 1} } \\ \end{gathered} \\  \\  \\\begin{gathered}\rm \: = \:sec\theta + tan\theta \\\end{gathered} \\  \\  \\\begin{gathered}\rm \: = \:\dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \\ \end{gathered} \\  \\  \\\begin{gathered}\rm \: = \:\dfrac{1 + sin\theta }{cos\theta } \\ \end{gathered}

<h2>Hence,</h2>

\begin{gathered} \\ \rm\implies \:\boxed{\sf{  \:\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \:\dfrac{1 + sin\theta }{cos\theta } \: \: }} \\ \\ \end{gathered}

\rule{190pt}{2pt}

5 0
2 years ago
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