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3 years ago

Help me plot please.

Mathematics

2 answers:

Veronika [31]3 years ago

4 0

One at 100,18 one at 200,14 one at 150,15 one at 125,20 and one at 225,12

vampirchik [111]3 years ago

4 0

The graph below given with answers

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Find The Difference [Will Give Brainliest If Correct]

RUDIKE [14]

Answer:

30, 10, -10

Step-by-step explanation:

Subtract each number below from the number above.

60 - 30 = 30

50 - 40 = 10

40 - 50 = -10

7 0

3 years ago

A bird releases waste from a height of

shepuryov [24]

Answer:

5secs

Step-by-step explanation:

Given the equation of the height expressed ad;

h(t) = - 16t^2 + initial height

Given that initial height = 400feet

h(t) = - 16t^2 + 400

The waste will hit the ground at when h(t) = 0

substitute

0 = - 16t^2 + 400

16t^2 = 400

t² = 400/16

t² = 25

t = √25

t = 5secs

Hence it will take the easte 5secs to hit the ground

7 0

2 years ago

How many 5-millimeter-long spiders would it take to make a line of spiders 1 meter long?

Bogdan [553]

Answer:

1000

Step-by-step explanation:

10milimeter=1 centimeter

100 centimeters=1 meter

10x100=1000

Let me know if im right :)

3 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

Select all expressions that are equivalent to 64.

ikadub [295]

For 64 it’s A. , C. , D. and for 3 it’s D. and F

5 0

2 years ago