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likoan [24]
3 years ago
11

Can you guys help me out with this? Thanks! <3

Mathematics
1 answer:
Tamiku [17]3 years ago
6 0

Answer:

Step-by-step explanation:

1.Jenna's interest was $480

2. 7 years Mike has saved & eared interest of $280.

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Please help?
Mazyrski [523]

Answer:

23\sqrt{3}\ un^2

Step-by-step explanation:

Connect points I and K, K and M, M and I.

1. Find the area of triangles IJK, KLM and MNI:

A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\

2. Note that

A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

7 0
3 years ago
If 45% of a number g is 225 what is 74% of g
Gelneren [198K]
370
You first have to divide 225 by 45 and then multiply by 74
5 0
3 years ago
Help Please Mathhh!!!!!!!!!!!!!!!
irga5000 [103]

Answer: the Nswer for this is 30 people

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Convert 5/11 to a percent​
atroni [7]

Pretty sure it's 45.45%

5 0
3 years ago
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What is another way to write 2 7/5 as a fraction with the same denominator
lukranit [14]

Answer:

\frac{17}{5}

Step-by-step explanation:

As an improper fraction

4 0
3 years ago
Read 2 more answers
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