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kotykmax [81]
3 years ago
7

Can i also know the working out for this question? thanks in advance!​

Mathematics
1 answer:
shepuryov [24]3 years ago
7 0
4x-1=18
4x=18+1
4x=19
x=19/4
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(-3) + (+1) =<br> steps of solving
viktelen [127]

Answer:

-2

Step-by-step explanation:

Opening brackets,

-3+1=-2

3 0
3 years ago
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4. Find the solution for each equation:<br> x + 4 = 16
Alex777 [14]

Answer:

x=12

Step-by-step explanation:

when solving and equation with a variable, you must work backwards and do the opposite of the equation

sooo… you must subtract 4 (opposite of +4 is -4)

16-4=2

x=12

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2 years ago
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PLEASE HELP!!!!! HAVING TROUBLE <br><br> (see attached images)
qaws [65]

Answer:

The Proof for

Part C , Qs 9 and Qs 10  is below.

Step-by-step explanation:

PART C .

Given:

AD || BC ,

AE ≅ EC

To Prove:

ΔAED ≅ ΔCEB

Proof:

Statement                             Reason

1. AD || BC                           1. Given

2. ∠A ≅ ∠C                        2. Alternate Angles Theorem as AD || BC

3. ∠AED ≅ ∠CEB               3. Vertical Opposite Angle Theorem.

4. AE ≅ EC                        4. Given

5. ΔAED ≅ ΔCEB              5. By A-S-A congruence test....Proved

Qs 9)

Given:

AB ≅ BC ,

∠ABD ≅ ∠CBD

To Prove:

∠A ≅ ∠C

Proof:

Statement                             Reason

1. AB ≅ BC                        1. Given

2. ∠ABD ≅ ∠CBD            2. Given      

3. BD ≅ BD                       3. Reflexive Property

4. ΔABD ≅ ΔCBD             4. By S-A-S congruence test

5. ∠A ≅ ∠C                       5. Corresponding parts of congruent Triangles Proved.

Qs 10)

Given:

∠MCI ≅ ∠AIC

MC ≅ AI

To Prove:

ΔMCI ≅ ΔAIC

Proof:

Statement                             Reason

1. ∠MCI ≅ ∠AIC       1. Given

2. MC ≅ AI              2. Given

3. CI ≅ CI                3. Reflexive Property

4. ΔMCI ≅ ΔAIC     4. By S-A-S congruence test

4 0
3 years ago
What is 3 tenths as a percentage
Valentin [98]
It is thirty percent (30%). there are ten tenths in one hundred. One tenth is equal to ten percent.

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Every year, a teacher surveys his students about the number of hours a week they watch television. In 2002, his students watched
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58.3 percent rounded to the nearest tenth
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