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mart [117]
3 years ago
9

Is 8:2 greater than 5:1? I need someone to help me clarify if this is correct. Thanks.

Mathematics
2 answers:
kaheart [24]3 years ago
3 0

Answer:

No, 8:2 = 4:1 which is less than 5:1

Step-by-step explanation:

svet-max [94.6K]3 years ago
3 0

Answer:

\huge \fbox \pink {A}\huge \fbox \green {n}\huge \fbox \blue {s}\huge \fbox \red {w}\huge \fbox \purple {e}\huge \fbox \orange {r}

8 : 2   \\  =  \frac{8}{2}  \\  = 4

5 : 1 \\  =  \frac{5}{1}  \\  = 5

8 : 2 = 4, while 5 : 1 = 5.

4 < 5.

•°• 8 : 2 is <u>not </u><u>greater </u><u>than </u>5 : 1.

ʰᵒᵖᵉ ⁱᵗ ʰᵉˡᵖˢ

\huge\blue{ \mid{ \underline{ \overline{ \tt ꧁❣ ʀᴀɪɴʙᴏᴡˢᵃˡᵗ2²2² ࿐ }} \mid}}

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lets see some examples to prove this conclusion:

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another example : 5 and 7 ... in this ... if we do it .. .we get the LCM as 37 ... which is greater than both the numbers

another case where both the numbers are same ....

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lianna [129]

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6 0
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Read 2 more answers
Pls Need Answer Now!!
iris [78.8K]

Answer:

Option B, 14.6 units

Step-by-step explanation:

<u>Step 1:  Find the distance of AB</u>

<u />AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

AB = \sqrt{(3 - 1)^2 + (5 - 3)^2}

AB = \sqrt{(2)^2 + (2)^2}

AB = 2\sqrt{2} or 2.83...

<u>Step 2:  Find the distance of BC</u>

BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

<u />BC = \sqrt{(3 -1)^2 + (-1 - 3)^2}

BC = \sqrt{(2)^2 + (-4)^2}

BC = \sqrt{4 + 16}

BC = \sqrt{20}

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<u>Step 3:  Find the perimeter</u>

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5.66 + 8.94

14.6 units

Answer: Option B, 14.6 units

5 0
3 years ago
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