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hammer [34]
2 years ago
14

A girl weight increased by 12% between her tenth and four teen birthdays . If she weighed 45kg on her tenth birthday. What did s

he weigh on her fourteenth birthday
Mathematics
1 answer:
jeka57 [31]2 years ago
6 0
Ok so the first thing you are going to do is chang the percentage into a decimal. 12 percent would change into 0.12, because in order to change a percentage into a decimal you move the decimal two places to the left. Next you would multiply 0.12 by 45 which gives you 5.4. But we are not done yet. By multiplying 0.12 x 45 we find how much weight she has gained. So our next and final step should be add 5.4 to 45 which gives us 50.4. Therefore your answer should be 50.4kg
Hope this helps!!!
Please tell me if I have made a mistake, I enjoy learning from mistakes!!
Have a great rest of your day!!
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HELP ASAPPPP NOWWW PLEASEEE
pickupchik [31]
X=0 because when you have something to the power of zero your answer is 1.
x=0 since 1 to the power of zero equals 1.


hope im not too late
5 0
3 years ago
Read 2 more answers
Carly is a catcher on the school softball team. By the end of the season she played 45% of the 40 games the team played. In how
klemol [59]
She played 18 games, so subtract 18 from 40 to get your answer
She did NOT play 22 games
Hope I helped!

4 0
3 years ago
Read 2 more answers
Can u help me with this
NISA [10]
9.7 x 4.658 = 45.1826

hope it helps
...................
7 0
3 years ago
Read 2 more answers
Solve A = a+b/2 for b
jeka94

Answer:

Step-by-step explanation:

I'm not sure if the right side is \frac{a + b}{2} or a + \frac{b}{2}, so I'll provide an answer for both:

A = \frac{a + b}{2}

2A = a + b

2A - a = b

A = a + \frac{b}{2}

A - a = \frac{b}{2}

2(A - a) = b

2A - 2a = b

6 0
3 years ago
Read 2 more answers
A dealer sells a certain type of chair and a table for $40. He also sells the same sort of table and a desk for $83 or a chair a
vesna_86 [32]
Chair=x

Table=y

Desk=z

\begin{Bmatrix}x+y&=&40\\y+z&=&83\\x+z&=&77\end{matrix}

keep the first row as normal, then in the other ones, we can isolate Y and X

\begin{Bmatrix}x+y&=&40\\y&=&83-z\\x&=&77-z\end{matrix}

now we can replace at first row...

\begin{Bmatrix}(77-z)+(83-z)&=&40\\y&=&83-z\\x&=&77-z\end{matrix}

\begin{Bmatrix}160-2z&=&40\\y&=&83-z\\x&=&77-z\end{matrix}

\begin{Bmatrix}-2z&=&40-160\\y&=&83-z\\x&=&77-z\end{matrix}

\begin{Bmatrix}-2z&=&-120\\y&=&83-z\\x&=&77-z\end{matrix}

\begin{Bmatrix}z&=&60\\y&=&83-z\\x&=&77-z\end{matrix}

now we can replace the Z to discovery the other value

\begin{Bmatrix}z&=&60\\y&=&83-60\\x&=&77-60\end{matrix}

\boxed{\boxed{\boxed{\begin{Bmatrix}Chair&=&17\\Table&=&23\\Desk&=&60\end{matrix}}}}
6 0
3 years ago
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