Determine the variables

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Cloud [144]

Answer:

a - b2

Step-by-step explanation:

STEP 1 :

Trying to factor as a Difference of Squares:

1.1 Factoring: a-b2

Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)

Proof : (A+B) • (A-B) =

A2 - AB + BA - B2 =

A2 - AB + AB - B2 =

A2 - B2

Note : AB = BA is the commutative property of multiplication.

Note : - AB + AB equals zero and is therefore eliminated from the expression.

Check : a1 is not a square !!

Ruling : Binomial can not be factored as the difference of two perfect squares

Final result :

a - b2

HOPE THIS HELPS!

PLEASE MARK BRAINLIEST! :)

7 0

3 years ago

What’s 5y/42-5/6? Solve

Vedmedyk [2.9K]

First try to isolate the variable, in this case "y". the first step to this is add 5/6 to both sides to get 5y/42=5/6. next multiply each side by 42 to get 5y=35. next divide by 5 to get y=7. so y=7 is the answer.

7 0

3 years ago

1 What is the slope of a line? A Its length. B Its width. C Its location in space. D Its steepness. PLEASEEEE HELP

Nadusha1986 [10]

Answer:

its steepness

Step-by-step explanation:

3 0

3 years ago

Choose SSS, SAS, or neither to compare

blondinia [14]

Answer:

SAS

Step-by-step explanation:

The triangles are compared with 2 sides and 1 angle. So, it is SAS

8 0

3 years ago

Read 2 more answers

The coordinates of the vertices of ∆PQR are P(-2,5), Q(-1,1), and R(7,3). Determine whether ∆PQR is a right triangle. Show your

Mashutka [201]

Given

∆PQR points are P(-2,5), Q(-1,1), and R(7,3)

Determine whether ∆PQR is a right triangle

To proof

As given ∆PQR points are P(-2,5), Q(-1,1), and R(7,3)

First find out the sides of triangle

FORMULA

Distance formula between two points

$D^{2}= (x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}$

Distance between two points P(-2,5) and Q(-1,1)

$PR = \sqrt{(-1+2)^{2}+(1-5)^{2} }$

$PR = \sqrt{17}$

Distance between two points Q(-1,1)and R(7,3)

$QR = \sqrt{(7+1)^{2} +(3-1)^{2} }$

$QR =\sqrt{68}$

Distance between two points R(7,3) and P(-2,5)

$RP =\sqrt{(-2-7)^{2} + (5-3)^{2} }$

$RP=\sqrt{85}$

now show that ∆PQR is a right triangle

$RP^{2} = PQ^{2} +QR^{2}$

Putting the value given above

$(\sqrt{85}) ^{2} = \sqrt{17} ^{2} +\sqrt{68} ^{2}$

85 = 17 +68

85 =85

In the right triangle

HYPOTENUSE² = BASE² + PERPENDICULAR²

This is prove above

Hence ∆PQR is a right triangle

Hence proved

7 0

4 years ago