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ryzh [129]
3 years ago
15

Highest common factor of 72 and 96

Mathematics
1 answer:
Likurg_2 [28]3 years ago
4 0

Answer:

24

Step-by-step explanation:

Prime Factorise 72:

72=2*36\\72=2*2*18\\72=2*2*2*9\\72=2*2*2*3*3\\\\72=2^3*3^2

Prime factorise 96:

96=2*48\\96=2*2*24\\96=2*2*2*12\\96=2*2*2*2*6\\96=2*2*2*2*2*3\\96=2^5*3

Multiply the common factors of 2^3*3^2 and 2^5*3:

2^3*3=24

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Step-by-step explanation:

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Given: P is a point on the perpendicular bisector, I of MN.<br> Prove: PM = PN
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Step-by-step explanation:

I can't make specific statements about the proof because the midpoint is missing.

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There are two right angles created by where the perpendicular bisector meats MN. Both are 90 degrees.

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Therefore the two triangles are congruent by SAS

PM = PN                Parts contained in Congruent triangles are congruent.

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3 years ago
Write the following polynomial in standard<br> form: <br><br> 4x+3-4x2-10x3
Ne4ueva [31]

Answer:

− 10 x ^3 − 4 x ^2 + 4 x + 3

Step-by-step explanation:

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3 years ago
I don’t understand this and I did take notes on this on the computer but I still don’t understand it.
Serggg [28]

Answer:

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Step-by-step explanation:

The line of reflection is at <em>y = −</em><em>2</em>,<em> </em>so looking at the line and where the coordinates of the shape are, you will have this:

<em>I</em>[−3, −2] → <em>I</em>'[−3, −2] (on the line, so it remains the same)

<em>Q</em>[2, −4] → <em>Q</em>'[2, 0]

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<em>E</em>[−3, −3] → <em>E</em>'[−3, −1]

I am joyous to assist you anytime.

3 0
3 years ago
Write the following expression with as few symbols as possible, using the notational conventions:
My name is Ann [436]

Answer:

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Step-by-step explanation:

The expressions provided are:

(i)\ x+((y\times (z')) +t)\\\\(ii)\ x\times((x+(y'))+z)

(i)

Simplify the first expression with as few symbols as possible:

x+((y\times (z')) +t)=x+(yz'+t)

                              =x+yz'+t

(ii)

Simplify the second expression with as few symbols as possible:

x\times((x+(y'))+z)=x\times ((x+y')+z)

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Thus, the simplified expressions are (<em>x</em> + <em>y·</em>z' + <em>t</em>) and <em>x·</em>(<em>x</em> + <em>y</em>' + <em>z</em>) respectively.

6 0
3 years ago
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