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3 years ago

8

First arc (centered at B)

1 answer:

marysya [2.9K]3 years ago

7 0

Fr j K rubbghhhghbvvv it’s is B

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The Jennings family paid $371.40 for the year for their cable service. If their payments were the same each month, how much was

liberstina [14]

Answer:

$30.95

Step-by-step explanation:

There are 12 months in a year. $371.40 divided by 12 gives us $30.95.

3 0

3 years ago

Read 2 more answers

Find the domain and range of the function graphed below.

geniusboy [140]

I don’t see any graph attached to the question

7 0

3 years ago

What is the factorization of the trinomial below -x^2 + 2x + 48

OlgaM077 [116]

For this case we must factor the following expression:

$-x ^ 2 + 2x + 48$

We take the sign "-" as a common factor of the expression, taking into account that:

$- * - = +\\- (x ^ 2-2x-48)$

To factor, we must find two numbers that, when multiplied, result in -48 and when added, result in -2. These numbers are -8 and +6.

$-8 + 6 = -2\\(-8) * (+ 6) = - 48$

Then, we factor the expression within the parenthesis as:

$- ((x-8) (x + 6))$

Answer:

$- (x-8) (x + 6)$

4 0

3 years ago

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,

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

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

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

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

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

<u>Hence, </u>

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

<u>Hence, Option (d) is </u><u>correct.</u>

6 0

2 years ago

Points E, F, and G are collinear. If EF = 8 and EG = 12, which point cannot lie between the other two?

Dima020 [189]

Point G can not be between

3 0

3 years ago