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2 years ago

NEED HELP FAST PLEASE

1 answer:

5 0

Hey!

To solve this equation, we'd first have to apply the distributing rule to this equation.

Original Equation :
$( 3a - 2 )^{2}$

New Equation {Changed by Applying the Distribution Rule} :
$( 3a )^{2} - 2$ · 3a · 2 + $2 ^{2}$

And now we simplify.

Old Equation :
$( 3a ) ^{2} -2$ · 3a · 2 + $2 ^{2}$

New Equation {Old Equation Simplified} :
$9a ^{2}$ - 12a + 4

So, the equation $(3a - 2) ^{2}$ simplified is $9a ^{2}$ - 12a + 4.

Hope this helps!

- Lindsey Frazier ♥

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What is the value of st÷(6r) if r=5, s=32, and t=45

Lana71 [14]

S = 32 and t = 45

You replace the letters into numbers...

Since (st) is the same thing as (s × t), it would be 32 × 45, which equals to 1,440.

Then replace the r with 5, so now it's (6 × 5), instead of (6r).

st ÷ (6r) → 1,440 ÷ 30 = 48

Value of st ÷ (6r) = 48

4 0

2 years ago

Segments

Kazeer [188]

Given

AB and CD intersect

AC, CB, BD and AD are congruent.

Prove that AB is the bisector of ∠CAD and ray CD is the bisector of ∠ACB.

and AB and CD are perpendicular.

To proof

Bisector

A bisector is that which cut an angle in two equal parts.

In ΔACB and ΔADB

AD = AC ( Given )

AB = AB ( common )

BC = DB ( Given )

by SSS congurence property

we have

ΔACB ≅ΔADB

∠CAB =∠ DAB

∠CBA = ∠DBA

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Thus AB is the bisector of the ∠CAD.

InΔ DAC and ΔDBC

AD = DB (Given)

AC = CB ( Given )

CD = CD (common)

By SSS congurence property

ΔDAC≅ Δ DBC

∠ ACD =∠ BCD

∠ADC =∠BDC

( By corresponding sides of the congurent triangle )

Therefore CD is the bisector of the CAD.

In ΔBOC andΔ BOD

BO = BO ( Common )

∠BCO = ∠BDO

( As prove above ΔACB ≅ΔADB

Thus ∠ACB = ∠ADB by corresponding sides of the congurent triangle , CD is a bisector

∠BCO = ∠BDO )

CB = DB ( given )

by SAS congurence property

ΔBOC ≅ ΔBOD

∠BOC =∠ BOD

∠BOC +∠ BOD = 180 °( Linear pair )

2∠ BOC = 180°

∠BOC = 90°

∠BOC =∠ BOD = 90°

also

In ΔCOA and ΔAOD

AO = AO ( Common )

∠ACO =∠ ADO

( As prove above ΔACB ≅ΔADB Thus ACB = ADB by corresponding sides of congurent triangle ,CD is a bisector

thus ∠ACO = ∠ADO )

AC =AD ( given )

by SAS congurence property

Δ COA ≅ ΔAOD

∠AOC = ∠AOD

( By corresponding angle of corresponding sides )

∠AOC + ∠AOD = 180°

2∠ AOC = 180° ( Linear pair )

∠AOC = 90°

∠AOC = ∠AOD = 90 °

Thus AB and CD are perpendicular.

Hence proved

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