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3 years ago

How do I simplify this?

Mathematics

2 answers:

kakasveta [241]3 years ago

4 0

I believe this would be the solution.

Mrac [35]3 years ago

3 0

First of all, turn the entire problem (except for the exponent) upside down. This will change the sign of your exponent (-2) to positive (2).

(x)^2=x^2

(y^4)^2 = y^8

The new numerator is thus (x^2)(y*8). What's the denominator going to look like?

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How are different ideas about geometry connected?

Margarita [4]

They areall related to geoemtry

8 0

3 years ago

State of the given functions are inverses. Part 3. NO LINKS!!!!

Effectus [21]

Answer:

h(x)=-x-1

let y =-x-1

interchanging role of x & y

x=-y-1

y=-x-1

h-¹(x)=-x-1

again

f(x)=(2-3x)/2

let

y=(2-3x)/2

interchanging role of x & y

x=(2-3y)/2

2x-2=-3y

y=(2-2x)/3

f-¹(x)=(2-2x)/3

Given function are not inverse of each other.

10.

f(x)=-x+3

let y=-x+3

interchanging role of x & y

x=-y+3

y=3-x

f-¹(x)=-x+3

equal to g(x)=-x+3

Given function are inverse of each other.

8 0

2 years ago

Read 2 more answers

I need help with 25.

noname [10]

Sum of linear pair of angles = 180°

<EFG + <GFH = 180°

2n + 21 + 4n + 15 = 180°

6n + 36 = 180°

6n = 144

n = 24

<EFG = 2(24) + 21 = 69°

<GFH = 4n + 15 = 4(24) + 15 = 111°

6 0

3 years ago

I need the answer I will give brainliest answer !!!!!

Blizzard [7]

Answer:

Step-by-step explanation:

been struggling and crying I try my best and I still can’t I give up and please help me this is due today

6 0

3 years ago

There is a parallelogram ABCD with diagonals AC and BD. The diagonals AC and BD intersects each other at point E. Side AB is con

grandymaker [24]

Answer:

SAS theorem

Step-by-step explanation:

Given

$\square ABCD$

$\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD$

$\angle BAC = \angle DCA$

Required

Which theorem shows △ABE ≅ △CDE.

From the question, we understand that:

AC and BD intersects at E.

This implies that:

$\[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC$

and

$\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED$

So, the congruent sides and angles of △ABE and △CDE are:

$\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD$ ---- S

$\angle BAC = \angle DCA$ ---- A

$\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED$ or $\[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC$ --- S

Hence, the theorem that compares both triangles is the SAS theorem

4 0

2 years ago