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Lynna [10]
3 years ago
5

Write in descending order 360t + 1013 - 12012

Mathematics
1 answer:
liubo4ka [24]3 years ago
5 0
I’m pretty sure it’s the way you have it. 360t+1013-12012
if not...someone else will probably tell you the right way
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A rental car company charges $15/day for a rental car plus 20¢ for every mile driven that day. Which equation models the relatio
nikdorinn [45]
<span>R=15d+0.20R=15d+0.20 is the answer</span>
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3 years ago
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Find the surface area of 6 in. 4 in. And 8 in.
hammer [34]

Answer:

208

Step-by-step explanation:

6×4=24

24×2=48

8×4=32

32×2=64

8×6=48

48×2=96

48+64+96=208

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Slope of 1/10 passes through the points (0,1) and (c,2) what is the value of c<br> ten pts?
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c=10

Step-by-step explanation:

3 0
3 years ago
Which of these numbers is in the solution set of x-15=5?
BlackZzzverrR [31]

Answer:

x=20

Step-by-step explanation:

x-15=5

x=5+15

x=20

3 0
2 years ago
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
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