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Valentin [98]
3 years ago
9

What is the area of the school crossing sign represented below?

Mathematics
1 answer:
vovangra [49]3 years ago
4 0
C. 675 square inches
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What is the greatest common factor of 42, 56, and 112
VashaNatasha [74]

Answer: gcf, hcf, gcd (42; 112) = 14 = 2 × 7: greatest (highest) common factor (divisor), calculated.

Step-by-step explanation:

8 0
2 years ago
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Do you agree or disagree
lakkis [162]
Disagree she needs 7 cups of bananas because 1.75×4=7  
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HURRY I WILL GIVE BRAINLIEST!!
nika2105 [10]
Hi there! 

The formula for the volume of a cone is V = 1/3(pi x r^2 x h). Using this formula, we can solve for the volume of the cone.

WORK:
V = 1/3(3.14 x 6.5^2 x 14)
V = 1857.31 / 3
V = 619.10 cm^3

ANSWER:
A) 619.10cm^3

Hope this helps!! :)
If there's anything else that I can help you with, please let me know!
6 0
3 years ago
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chris has 3 1/2 boxes of tomatoes at his restaurant he uses 1 3/4 boxes to make tomato sauce for tonight's dinner. How many boxe
Sladkaya [172]

Answer:

1 3/4

Step-by-step explanation:

3 1/2 boxes of tomatoes currently. uses 1 3/4, taking it away basically.

3 1/2 - 1 3/4

3 2/4 - 1 3/4 ;; i did this with finding a common denominator for it to be easier for me to solve.

3 - 1 = 2

2/4 - 3/4 = -1/4

2 + -1/4 = 1 3/4

correct me if wrong.

5 0
2 years ago
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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
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