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2 years ago

Add or Subtract to Simplify: (x5 + x3) - (6x - x3 + 6x5)

Mathematics

1 answer:

ziro4ka [17]2 years ago

3 0

Answer:

$\boxed{ \bold{ \boxed{ \sf{ - 5 {x}^{5} + 2 {x}^{3} - 6x}}}}$

Step-by-step explanation:

$\sf{( {x}^{5} + {x}^{3} ) - (6x - {x}^{3} + 6 {x}^{5}) }$

When there is a ( - ) in front of an expression in parentheses , change the sign of each term.

Also, remove the parentheses

⇒ $\sf{ {x}^{5} + {x}^{3} - 6x + {x}^{3} - 6 {x}^{5} }$

Collect like terms

⇒ $\sf{ {x}^{5} - 6 {x}^{5} + {x}^{3} + {x}^{3} - 6x}$

⇒ $\sf{ - 5 {x}^{5} + 2 {x}^{3} - 6x}$

Hope I helped!

Best regards!!

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In 3-4 sentences, describe how you would find a line parallel to the line 2x + 5y = 15 that goes through the point (-10, 1). Be

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the parallel line is 2x+5y+15=0.

Step-by-step explanation:

ok I hope it will work

soo,

Solution

given,

given parallel line 2x+5y=15

which goes through the point (-10,1)

now,

let 2x+5y=15 be equation no.1

then the line which is parallel to the equation 1st

2 x+5y+k = 0 let it be equation no.2

now the equation no.2 passes through the point (-10,1)

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2 years ago

Need this really quick plz

Oduvanchick [21]

Answer:

Let

x=sin-¹u

Sinx=u

let y=tan-¹v

tany=v

Substituting

Sin[x + y]

Applying the sine expansion

Sinxcosy + CosxSiny

Recall x =Sin-¹u

y=tan-¹v

Sin(Sin-¹u)Cos(tan-¹v) +Cos(sin-¹u)Sin(tan-¹v)

Now at this point

Here's what you do

For the first expression

Sin(Sin-¹u)

Let's simplify this

Let P = Sin-¹u

Taking sine of both sides

SinP=u

Draw a Right angled angle for this

Since Sine from SOHCAHTOA is OPP/HYP

Where P is the angle and u is the opposite and 1 is the hypotenuse since u is the same as u/1

substituting Sin-¹u = P

You have

Sin(Sin-¹u) = SinP

and from the triangle you drew

SinP = u

Taking the second express

Cos(tan-¹v)

Let Q=Tan-¹v

taking tan of both sides

tanQ=v

Draw a right angled triangle for this too

Since Tan from SOHCAHTOA is OPP/ADJ

Find the Hypotenuse cos you'll need it

Now Let's do the substitution again

We first said tan-¹v = Q

When we substitute it in Cos(tan-¹v)

We have CosQ

Cos Q from the second right angle triangle you drew is 1/√1+v²

Because CAH is adj/Hyp

the first part of the original Express

Which is

Sin(Sin-¹u)Cos(tan-¹v) is now simplified to

u(1/√1+v²).

Let's Move to the second part of the Original Expression

Cos(Sin-¹u)Sin(tan-¹v)

From our first solution

We said Sin-¹u= P

So replacing it here

we have Cos(sin-¹u) = CosP

let's leave the second one for now which is sin(tan-1v) We'll deal with this after the first

so Cos(Sin-¹u) = CosP

we can still use our first Right angle triangle for this because the angle was P.

so Cos P from that triangle will be

CosP= √1-u²

Now onto the next

Sin(tan-¹v)

From the Second solution of the first we did

we said let Tan-¹v =Q

Substituting this

we have

Sin(tan-¹v) = SinQ

using the second Right angle triangle because its angle is Q

We have

SinQ= v/√1+v²

Answer for second phase Which is

Cos(sin-¹u)Sin(tan-¹v) = √1-u²(v/√1+v²)

We're done

compiling our answers

The answer to

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You can still choose to factor out 1/√1+v² since it appears on both sides

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