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3 years ago

What is the perimeter of this figure

Mathematics

1 answer:

IceJOKER [234]3 years ago

3 0

Perimeter? Add up the sides along the outer boxes.
I see you've already marked them.
6+1+2+2+2+1+6+1+2+1=24

Perimeter: 24

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Please help me with this question.

Sav [38]

Answer:

-2; -7

Step-by-step explanation:

Hello, the graph is split in two lines, right?

Go to the first line at the left and move to the right until you reach the end of the line, which is for x = -1,

*** This is the meaning of x tending to

$(-1)^{-}$ ***

which value can you see for the y-axis? $\boxed{-2}$

Now, go to the second line at the right and move to the left until you reach the end of the line, which is for x = -1,

*** This is the meaning of x tending to

$(-1)^{+}$ ***

which value can you see for the y-axis?

$\boxed{-7}$

Thanks

7 0

3 years ago

The large sample 98% confidence interval for the proportion of hotel reservations that are canceled on the intended arrival day

Digiron [165]

Answer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Please help me

Whitepunk [10]

First solve your inequality step by step.

y+15<3

= y<-12

Answer is A. y<-12

Answer will help you.

Brainliest answer and Verified answers.

4 0

3 years ago

What is the slope of the line through (6,9) and (7,1)

pychu [463]

Answer:

-8

Step-by-step explanation:

m= y₂- y₁/ x₂-x₁

m= 1-9/7-6

m= -8/1

m= -8

8 0

3 years ago

Read 2 more answers

D/d{cosec^-1(1+x²/2x)} is equal to

SIZIF [17.4K]

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 + {x}^{2} }{2x} \bigg)$

Let assume that

$\rm :\longmapsto\:y = {cosec}^{ - 1} \bigg( \dfrac{1 + {x}^{2} }{2x} \bigg)$

We know,

$\boxed{\tt{ {cosec}^{ - 1}x = {sin}^{ - 1}\bigg( \dfrac{1}{x} \bigg)}}$

So, using this, we get

$\rm :\longmapsto\:y = sin^{ - 1} \bigg( \dfrac{2x}{1 + {x}^{2} } \bigg)$

Now, we use Method of Substitution, So we substitute

$\red{\rm :\longmapsto\:x = tanz \: \rm\implies \:z = {tan}^{ - 1}x}$

So, above expression can be rewritten as

$\rm :\longmapsto\:y = sin^{ - 1} \bigg( \dfrac{2tanz}{1 + {tan}^{2} z} \bigg)$

$\rm :\longmapsto\:y = sin^{ - 1} \bigg( sin2z \bigg)$

$\rm\implies \:y = 2z$

$\bf\implies \:y = 2 {tan}^{ - 1}x$

So,

$\bf\implies \: {cosec}^{ - 1}\bigg( \dfrac{1 + {x}^{2} }{2x} \bigg) = 2 {tan}^{ - 1}x$

Thus,

$\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 + {x}^{2} }{2x} \bigg)$

$\rm \: = \: \dfrac{d}{dx}(2 {tan}^{ - 1}x)$

$\rm \: = \: 2 \: \dfrac{d}{dx}( {tan}^{ - 1}x)$

$\rm \: = \: 2 \times \dfrac{1}{1 + {x}^{2} }$

$\rm \: = \: \dfrac{2}{1 + {x}^{2} }$

Hence, 

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 + {x}^{2} }{2x} \bigg) = \frac{2}{1 + {x}^{2} }}}}$

Hence, Option (d) is correct.

6 0

2 years ago