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horsena [70]
3 years ago
6

cindy says she found a shortcut for doing multipication problems .when she multiplies 3 times 24 ,she says, 3 times 4 is 12 ones

,or 1ten and 2 ones . then there's just 2 tens left in 24, so add it up and you get 3 tens and 2 0nes ." do you think cindy's shortcut works ? explain your thinking in words to justify your response using a model or partial prouducts.
Mathematics
2 answers:
agasfer [191]3 years ago
6 0
Yes she would be right and that is a smart short cut
Xelga [282]3 years ago
4 0
No because 3 times 24 actually equals 72
she forgot to multiply the 2 tens by 3
if she would've done that the answer would be 7 tens and 2 ones
:-)
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Let's simplify step-by-step. 
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3 years ago
What is the range of the function f(x) = 4x + 9, given the domain D = {-4, -2, 0, 2}?
anyanavicka [17]

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f( - 4) = 4( - 4) + 9 =  - 7

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f( - 2) = 4( - 2) + 9 = 1

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f(0) = 4(0) + 9 = 9

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f(2) = 4(2) + 9 = 17

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Thus ;

R = { - 7 , 1 , 9 , 17 }

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The correct answer is (( D )) .

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3 years ago
Mary Hernandez had a policy with a $500 deductible which paid 80% of her covered charges less deductible. She had medical expens
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4 0
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galina1969 [7]

Answer:

The key is to prove that the triangles are congruent, then their corresponding parts are congruent.

Step-by-step explanation:

#5)

It it given that:

Angle CBD is congruent to angle CDB.

Angle BAE is congruent to angle DEA.

Triangle CBD is isosceles because it's base angles are equal. Therefore, segment CB is congruent to segment CD.

Triangle CAE is isosceles because it's base angles are equal. Therefore, segment CA is congruent to segment CE.

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CE-CD=DE

segment BD is congruent to segment DB.

Angle ABD is supplementary to angle CBD (180-angle CBD= angle ABD).

Angle EDB is supplementary to angle CDB (1980-angle CDB=angle EDB).

therefore, angle ABD=angle EDB.

Triangles ABD and EDB are congruent because of the Side-angle-side theorem.

Triangles ABD and EDB are congruent, segment AD is congruent to segment EB because corresponding legs of congruent triangles are congruent.

#6)

It is given that:

Angle EBC is congruent to angle ECB

Segment AE is congruent to segment DE.

Triangle BEC is iscsceles because it's base angles are equal. Therefore, segment EB is congruent to segment EC.

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segment BC is congruent to segment CB.

Triangle BAC is congruent to triangle CDB because of the side-angle-side theorem.

Segment AD is congruent to segment DC because corresponding legs of congruent triangles are congruent.

8 0
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