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zmey [24]
3 years ago
10

John is making 4 cylindrical wax candles. If he plans to make candles with a diameter of 7 cm and a height of 12 cm, approximate

ly how many cubic centimeters of wax will John need to make the candles?
Mathematics
1 answer:
Stells [14]3 years ago
5 0
He will need 527.52 cubic centimeters of wax. You divide the diameter by 2 ten multiply that number by itself to get the circumference. From there you multiply it by 12 then by 4 and there's your answer
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Please help me with NUMBER 7 !! Thank you so much! 20 points!! :))
mart [117]

Answer:

x<-3 or x>-1

Step by step explanation:

Start with -5x>15. To get x by itself, we have to divide each side by -5. When we do that we get x<-3. (Don't forget to flip the sign when dealing with negative  · and ÷.) Now answer the second equation. x-5>-6 To get x by itself, add 5 to both sides to get x>-1. Now we have x<3 or x>-1.

8 0
3 years ago
Dylan has a bank that sorts coins as they are dropped into it. A panel on the front displays the total number of coins inside as
vesna_86 [32]

Answer:

1). System of equations- D + Q = 90 and D + 2.5Q = 175.50

2). There are 33 dimes and 57 quarters in the bank.

Step-by-step explanation:

Dylan has a bank that sorts coins as they are dropped in it.

A panel shows the total number of coins inside as well as the total value of these coins.

Let the number of dimes in the bank = D

and the number of quarters in the bank = Q

If the panel shows total number of coins = 90

Then the equation will be

D + Q = 90 -------(1)

And the panel displays the amount of the coins = $17.55

Then equation will be

0.10D + 0.25Q = 17.55 [1 Dime = $0.10 and 1 quarter = $0.25]

D + 2.5Q = 175.50 ------------(2)

Now we subtract equation (1) form equation (2)

D + 2.5Q - (D + Q) = 175.50 - 90

D + 2.5Q - D - Q = 85.5

1.5Q = 85.5

Q = \frac{85.5}{1.5}

   = 57

By putting Q = 57 in the equation (1)

D + 57 = 90

D = 90 - 57 = 33

Therefore, there are 33 dimes and 57 quarters in the bank.

4 0
3 years ago
Which is not equivalent to the others please help
PtichkaEL [24]

Answer:0.801

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
How do you find a vector that is orthogonal to 5i + 12j ?
Rashid [163]
\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{a}{b}\\\\&#10;slope=\cfrac{a}{{{ b}}}\qquad negative\implies  -\cfrac{a}{{{ b}}}\qquad reciprocal\implies - \cfrac{{{ b}}}{a}\\\\&#10;-------------------------------\\\\

\bf \boxed{5i+12j}\implies &#10;\begin{array}{rllll}&#10;\ \textless \ 5&,&12\ \textgreater \ \\&#10;x&&y&#10;\end{array}\quad slope=\cfrac{y}{x}\implies \cfrac{12}{5}&#10;\\\\\\&#10;slope=\cfrac{12}{{{ 5}}}\qquad negative\implies  -\cfrac{12}{{{ 5}}}\qquad reciprocal\implies - \cfrac{{{ 5}}}{12}&#10;\\\\\\&#10;\ \textless \ 12, -5\ \textgreater \ \ or\ \ \textless \ -12,5\ \textgreater \ \implies \boxed{12i-5j\ or\ -12i+5j}

if we were to place <5, 12> in standard position, so it'd be originating from 0,0, then the rise is 12 and the run is 5.

so any other vector that has a negative reciprocal slope to it, will then be perpendicular or "orthogonal" to it.

so... for example a parallel to <-12, 5> is say hmmm < -144, 60>, if you simplify that fraction, you'd end up with <-12, 5>, since all we did was multiply both coordinates by 12.

or using a unit vector for those above, then

\bf \textit{unit vector}\qquad \cfrac{\ \textless \ a,b\ \textgreater \ }{||\ \textless \ a,b\ \textgreater \ ||}\implies \cfrac{\ \textless \ a,b\ \textgreater \ }{\sqrt{a^2+b^2}}\implies \cfrac{a}{\sqrt{a^2+b^2}},\cfrac{b}{\sqrt{a^2+b^2}}&#10;\\\\\\&#10;\cfrac{12,-5}{\sqrt{12^2+5^2}}\implies \cfrac{12,-5}{13}\implies \boxed{\cfrac{12}{13}\ ,\ \cfrac{-5}{13}}&#10;\\\\\\&#10;\cfrac{-12,5}{\sqrt{12^2+5^2}}\implies \cfrac{-12,5}{13}\implies \boxed{\cfrac{-12}{13}\ ,\ \cfrac{5}{13}}
4 0
3 years ago
You have two exponential functions. One function has the formula g(x) = 5 x . The other function has the formula h(x) = 5-x . Wh
AVprozaik [17]
Given the exponential functions g(x)= 5^{x} and h(x)= 5^{-x}

k(x)=(g-h)(x)
k(x)=g(x)-h(x)&#10;
k(x)= 5^{x} - 5^{-x}
k(x)= 5^{x} - \frac{1}{ 5^{x} }
k(x)= \frac{ 5^{x} 5^{x}  }{ 5^{x} } - \frac{1}{ 5^{x} }
k(x)= \frac{ 5^{2x} }{ 5^{x} } - \frac{1}{ 5^{x} }
k(x)= \frac{ 5^{2x}-1 }{ 5^{x} }
4 0
2 years ago
Read 2 more answers
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