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Roman55 [17]
3 years ago
10

The rule T5, -0.5 oR 0, 180(x,y) is applied to triangle FGH to produce triangle F’G’H’. What are the coordinates of vertex F’ of

triangle F’G’H’?
Mathematics
2 answers:
Neporo4naja [7]3 years ago
3 0

Answer:

Step-by-step explanation:

Rule for the transformations has been given as,

T_{(5, -0.05)o\text{R}(0, 180)}(x,y) defines the rules as the translation by 5 units right and 0.5 units down followed by rotation of the triangle by 180° about the origin.

In other words,

Rule for translation:

F(x, y) → F"(x + 5, y - 0.5)

Followed by the rotation of 180° about the origin,

Rule of rotation,

F"(x', y') → F'(-x, -y)

Following these rules coordinates of the new triangle F'G'H' can be obtained.

aleksandrvk [35]3 years ago
3 0

Answer:

A

Step-by-step explanation:

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Vitek1552 [10]

If the Height of Eiffel Tower in Paris is 1063 feet tall and we took an elevator 3/4 up the tower then we would be 797.25 feet from the ground.

Given The height of the Eiffel Tower is 1063 feet tall. We have taken an elevator that is 3/4 up the tower.

Height is the measurement of someone  from head to foot or from base to top. It is the distance upward.

Height of The Eiffel tower is 1063 feet.

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2 years ago
Solve for x 2(x-3)+21=-3
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jeyben [28]
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3 0
3 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

So

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

Sin(Sin-¹u - tan-¹v) = u(1/√1+v²) + [(√1-u²)(v/√1+v²)]

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

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