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Natasha2012 [34]
3 years ago
8

Karen planted a flower that was 14 inches tall. It grew 2 1/4 inches each week. If it is now 23 inches tall, how

Mathematics
1 answer:
NARA [144]3 years ago
8 0

Answer:

4 weeks

Step-by-step explanation:

2 1/4 = 9/4 = 2.25

if it grows 2.25 inches in one week then in two weeks it would have grown 4.5 inches

keep adding the 2.25 inches to 14 until u get 23

:) hope this was heplful

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(7 + 41) ÷ 2 – 15 =? <br><br> Find the solution for this question.
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precedence is:

1 Parentheses (simplify inside 'em)

2 Exponents

3 Multiplication and Division (from left to right)

4 Addition and Subtraction (from left to right)

(7+41) = 48

48 ÷ 2 = 24

24 -15 = 9

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2 years ago
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First answer get best marks ​
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The three expressions are the first option, the third option, and the forth option

Step-by-step explanation:

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Find the area of quadrilateral ABCD. A. 27.28 units² B. 33.08 units² C. 28.53 units² D. 26.47 units²
Ludmilka [50]

Answer:

<h3>\boxed{ \boxed{ \bold{ \purple{ \sf{28.53 \:  {units}^{2} }}}}}</h3>

Option C is the correct option.

Step-by-step explanation:

For ABD

\sf{ =  \frac{2.89 + 8.59 + 8.6}{2} }

\sf{ =  \frac{20.08}{2} }

\sf{ = 10.04}

∆ ABD = \sf{ =  \sqrt{10.04(10.04 - 2.89)(10.04 - 8.59)(10.04 - 8.6)} }

\sf{  = \sqrt{10.04 \times 7.15 \times 1.45 \times 1.44 }}

\sf{  =  \sqrt{149.8891} }

\sf{ = 12.2429}

For ∆ ACD ,

\sf{s =  \frac{8.6 + 4.3 + 7.58}{2} }

\sf{ =   \frac{20.48}{2} }

\sf{ = 10.24}

∆ ACD = \sf{ =  \sqrt{10.24(10.24 - 8.6)(10.24 - 4.3)(10.24 - 7.58} }

\sf{ \sqrt{10.24 \times 1.64 \times 5.94 \times 2.66} }

\sf{ =  \sqrt{265.3456} }

\sf{ = 16.2894}

Area of quadrilateral ABCD = \sf{12.2429 + 16.2894}

\sf{ = 28.5323}

\sf{ = 28.53} units ²

Hope I helped!

Best regards!

8 0
3 years ago
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