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Vadim26 [7]
2 years ago
9

I WILL GUVE BRAINIEST I NEED ASAP.

Mathematics
1 answer:
Naily [24]2 years ago
8 0

Answer:

What I got for my answer for one of the cylinders is 502.65

I hope I helped :3 Let me know if your confused on it or need something done plz I will gladly help ^w^

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(x, y) ---> (cx, cy)
Nookie1986 [14]
Do you have another picture?
6 0
3 years ago
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PLZ HELP PLZ HELP brainlyist for correct answer
Amiraneli [1.4K]
So you basically plug in the x value beside the 5 so if your on number 3 the answer would be 5(3) which is 15

4 0
3 years ago
A gold mine has two elevators one for equipment and another for miners. The elevator for the miners descends 13 feet per second.
saveliy_v [14]

Answer:

The equipment elevator is positioned 481 feet closer and the miner equipment is positioned 342 feet closer relative to the surface. The equipment elevator is deeper.

Step-by-step explanation:

The rate that both elevators descend are the same according to the information given in the question.

After 18 seconds, the equipment elevator has descended 18 x 13 = 234 feet.

After another 19 seconds, the equipment elevator has descended for a total of 37 seconds which results in 481 feet and the miner elevator has descended for 19 seconds which gives us a descend of 342 feet.

So they are 481 and 342 feet closer relative to the surface and the equipment elevator is deeper since it has descended for a longer period of time.

I hope this answer helps.

8 0
3 years ago
Demetrius and Yosef are training for a race.
Andru [333]

Answer:

You could use the equation y = mx + b to solve this problem. The m would equal 7.5 and the be would be zero, so the equation would be y = 7.5x + 0. Add values for x and you would get y.

8 0
2 years ago
Question regarding logarithms.
Eddi Din [679]

5^{x-2}-7^{x-3}=7^{x-5}+11\cdot5^{x-4}\\\\5^{x-2}-7^{x-2-1}=7^{x-2-3}+11\cdot5^{x-2-2}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\5^{x-2}-\dfrac{7^{x-2}}{7^1}=\dfrac{7^{x-2}}{7^3}+11\cdot\dfrac{5^{x-2}}{5^2}\\\\5^{x-2}-\dfrac{1}{7}\cdot7^{x-2}=\dfrac{1}{343}\cdot7^{x-2}+\dfrac{11}{25}\cdot5^{x-2}\\\\-\dfrac{1}{7}\cdot7^{x-2}-\dfrac{1}{343}\cdot7^{x-2}=\dfrac{11}{25}\cdot5^{x-2}-5^{x-2}\\\\\left(-\dfrac{1}{7}-\dfrac{1}{343}\right)\cdot7^{x-2}=\left(\dfrac{11}{25}-1\right)\cdot5^{x-2}

\left(-\dfrac{49}{343}-\dfrac{1}{343}\right)\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\\\\-\dfrac{50}{343}\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\qquad\text{multiply both sides by}\ \left(-\dfrac{25}{14}\right)\\\\\dfrac{50\cdot25}{343\cdot14}\cdot7^{x-2}=5^{x-2}\qquad\text{divide both sides by}\ 7^{x-2}\\\\\dfrac{25\cdot25}{343\cdot7}=\dfrac{5^{x-2}}{7^{x-2}}\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}

\dfrac{5^2\cdot5^2}{7^3\cdot7}=\left(\dfrac{5}{7}\right)^{x-2}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\\dfrac{5^4}{7^4}=\left(\dfrac{5}{7}\right)^{x-2}\\\\\left(\dfrac{5}{7}\right)^4=\left(\dfrac{5}{7}\right)^{x-2}\iff x-2=4\qquad\text{add 2 to both sides}\\\\\boxed{x=6}

6 0
3 years ago
Read 2 more answers
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