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vlabodo [156]
2 years ago
15

CAN SOMEONE HELP ME WITH THIS PLEASE!!!!!! I NEED IT AS SOON AS POSSIBLE!!!

Mathematics
2 answers:
yulyashka [42]2 years ago
7 0
1. 7 units
2. 11 units
3. (5, -3)
4. (-4, 5)
Aleksandr-060686 [28]2 years ago
7 0

Answer:

Step-by-step explanation:

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? Don't Send links, don't Write random things and get the points for free.
alekssr [168]

Answer:

V = 1847.26\ ft^3

Step-by-step explanation:

Step 1:  Determine the volume

V=\pi r^2h

V = \pi *(7\ ft)^2 * (12\ ft)

V = \pi *49\ ft^2 * 12\ ft

V = \pi  * 588\ ft^3

V = 1847.26\ ft^3

Answer:   V = 1847.26\ ft^3

6 0
2 years ago
EVALUATE !!!! ILL GIVE YOU BRAINLIST HAVE TO GET IT RIGHT !!!!
elena-s [515]

Answer:

<h2>12</h2>

Step-by-step explanation:

nxy\div m\\\\n =3/4\\x =8\\y =6\\m=3\\\\\frac{3}{4} \times 8 \times 6 \div 3\\\\\mathrm{Divide\:the\:numbers:}\:6\div \:3=2\\=\frac{3}{4}\times \:8\times \:2\\\\\mathrm{Convert\:element\:to\:fraction}:\quad \:8=\frac{8}{1}\\=2\times \frac{3}{4}\times \frac{8}{1}\\\\\mathrm{Convert\:element\:to\:fraction}:\quad \:2=\frac{2}{1}\\=\frac{3}{4}\times \frac{8}{1}\times \frac{2}{1}\\\\Multiply\:fractions\\=\frac{3\times \:8\times \:2}{4\times \:1\times \:1}\\\\=\frac{48}{4}\\\\Simplify\\=12

8 0
3 years ago
Explain why there can be no infinite geometric series with a first term of 12 and a sum of 5.
Iteru [2.4K]

Answer:

Step-by-step explanation:

When you find the sum of a number you are adding two or more numbers together. therefore the only answer that you could use to get a sum of 5 when your first term is 12 would be -7

3 0
3 years ago
Simplify 27^2.<br><br> please (thxx)
adoni [48]

Answer:

729

Step-by-step explanation:

Raise 27 to the power of 2.

729

4 0
3 years ago
Use IDEAL to solve the following problems (WORTH 50 POINTS)
Dima020 [189]
So pythagrorean theorem is the ideal way and easiest way to solve this

a^2+b^2=c^2

8.5^2+11^2=x^2
72.25+121=193.25
square root of 193.25=13.9014 or 13.9 in


9.5^2+15^2=x^2
90.25+225=315.25
square root of 315.25=17.7553 or 17.8 ft
8 0
3 years ago
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