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zhannawk [14.2K]
3 years ago
14

I need help converting 25 oz to g

Mathematics
2 answers:
GenaCL600 [577]3 years ago
8 0

Answer: 708.738 because 1 oz = 28.35 grams

25 x 28.35 = 708.738

Step-by-step explanation:

s2008m [1.1K]3 years ago
3 0
The answer is 708.74 grams
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How do I answer this?​
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If you are trying to find the area the answer would be 48
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Dennis decided he wanted to go to the carnival. The carnival charges $6.00 for entry and $1.50 for each ride.
myrzilka [38]

We want to create a linear equation to model the given situation.

A) c(r) = $6.00 + $1.50*r

B) 19 rides.

We know that the carnival charges $6.00 for entry plus $1.50 for each ride.

A) With the given information we can see that if you ride for r rides, then the cost equation will be:

c(r) = $6.00 + $1.50*r

Where c(r) is the cost for going to the carnival and doing r rides.

B) If you have $35.00, then we can solve:

c(r) = $35.00 =  $6.00 + $1.50*r

Now we can solve the equation for r.

$35.00 =  $6.00 + $1.50*r

$35.00 -  $6.00 = $1.50*r

$29.00 = $1.50*r

$29.00/$1.50 = r = 19.33

Rounding to the next whole number we get: r = 19

This means that with $35.00, Dennis could go to 19 rides.

If you want to learn more, you can read:

brainly.com/question/13738061

7 0
3 years ago
Find the smallest 4 digit number such that when divided by 35, 42 or 63 remainder is always 5
alex41 [277]

The smallest such number is 1055.

We want to find x such that

\begin{cases}x\equiv5\pmod{35}\\x\equiv5\pmod{42}\\x\equiv5\pmod{63}\end{cases}

The moduli are not coprime, so we expand the system as follows in preparation for using the Chinese remainder theorem.

x\equiv5\pmod{35}\implies\begin{cases}x\equiv5\equiv0\pmod5\\x\equiv5\pmod7\end{cases}

x\equiv5\pmod{42}\implies\begin{cases}x\equiv5\equiv1\pmod2\\x\equiv5\equiv2\pmod3\\x\equiv5\pmod7\end{cases}

x\equiv5\pmod{63}\implies\begin{cases}x\equiv5\equiv2\pmod 3\\x\equiv5\pmod7\end{cases}

Taking everything together, we end up with the system

\begin{cases}x\equiv1\pmod2\\x\equiv2\pmod3\\x\equiv0\pmod5\\x\equiv5\pmod7\end{cases}

Now the moduli are coprime and we can apply the CRT.

We start with

x=3\cdot5\cdot7+2\cdot5\cdot7+2\cdot3\cdot7+2\cdot3\cdot5

Then taken modulo 2, 3, 5, and 7, all but the first, second, third, or last (respectively) terms will vanish.

Taken modulo 2, we end up with

x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod2

which means the first term is fine and doesn't require adjustment.

Taken modulo 3, we have

x\equiv2\cdot5\cdot7\equiv70\equiv1\pmod3

We want a remainder of 2, so we just need to multiply the second term by 2.

Taken modulo 5, we have

x\equiv2\cdot3\cdot7\equiv42\equiv2\pmod5

We want a remainder of 0, so we can just multiply this term by 0.

Taken modulo 7, we have

x\equiv2\cdot3\cdot5\equiv30\equiv2\pmod7

We want a remainder of 5, so we multiply by the inverse of 2 modulo 7, then by 5. Since 2\cdot4\equiv8\equiv1\pmod7, the inverse of 2 is 4.

So, we have to adjust x to

x=3\cdot5\cdot7+2^2\cdot5\cdot7+0+2^3\cdot3\cdot5^2=845

and from the CRT we find

x\equiv845\pmod2\cdot3\cdot5\cdot7\implies x\equiv5\pmod{210}

so that the general solution x=210n+5 for all integers n.

We want a 4 digit solution, so we want

210n+5\ge1000\implies210n\ge995\implies n\ge\dfrac{995}{210}\approx4.7\implies n=5

which gives x=210\cdot5+5=1055.

5 0
3 years ago
CAN SOMEONE HELP ME WITH THIS PLEASE AND THANK YOU.
Fantom [35]

Answer:

4 11/12

Step-by-step explanation:

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