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SOVA2 [1]
3 years ago
8

A problem states: "There are 9 more children than parents in a room. There are 25 people in the room in all. How many children a

re there in the room?" What are the unknowns in this problem? Select each correct answer. 1. the total number of children in the room 2. the total number of parents in the room 3.The total number of teachers in the room 4. the total number of people in the room
Mathematics
1 answer:
Dmitrij [34]3 years ago
6 0
To work out who the unknowns in the room are, we can establish that c = p+ 9
c + p = 25. The unknowns in the room are there the total number of parents in the room, and the total number of teachers in the room.
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Katie and her brother are both writing a report. Katie write 7/8 of her report and her brother wrote 3/7 of his report. What is
skad [1K]

Answer:

Answer is 25/56

Step-by-step explanation:

Given that

Katie and her brother are both writing a report. Katie write 7/8 of her report and her brother wrote 3/7 of his report

TO compare these two fractions we have to make both of them have the same denominator.

We find that as such the two fractions have denominators as 7 and 8 and hence Least common multiple is 56

LEt us make equivalent fractions

7/8 = 49/56

3/7=24/56

Difference i.e. Katie wrote more than her brother = 49/56-24/56

=25/56

4 0
3 years ago
Please help me answer this question
ratelena [41]

By the definition of the hyperbolic function tanh x, we have proven that \frac{1-tanh \ x}{1 + tanh \ x}=e^{-2x}

<h3>Hyperbolic functions & Proof of identities </h3>

By definition

tanh \ x=\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}}

Then,

\frac{1-tanh \ x}{1 + tanh \ x}=\frac{1-\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} }{1+ \frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} }

=1-\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} \div (1+ \frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} })

=\frac{e^{x} +e^{-x}-(e^{x} -e^{-x})}{e^{x} +e^{-x}} \div (\frac{e^{x} +e^{-x}+e^{x} -e^{-x}}{e^{x} +e^{-x}} })

=\frac{e^{x} +e^{-x}-e^{x} +e^{-x}}{e^{x} +e^{-x}} \div \frac{e^{x} +e^{-x}+e^{x} -e^{-x}}{e^{x} +e^{-x}} }

=\frac{e^{-x}+e^{-x}}{e^{x} +e^{-x}} \div \frac{e^{x} +e^{x} }{e^{x} +e^{-x}} }

=\frac{2e^{-x}}{e^{x} +e^{-x}} \div \frac{2e^{x} }{e^{x} +e^{-x}} }

=\frac{2e^{-x}}{e^{x} +e^{-x}} \times \frac{e^{x} +e^{-x}}{2e^{x}}

=\frac{2e^{-x}}{1} \times \frac{1}{2e^{x}}

=\frac{2e^{-x}}{2e^{x}}

=\frac{e^{-x}}{e^{x}}
=e^{-x} \times \frac{1}{e^{x}}

= e^{-x} \times e^{-x}

= e^{-x+-x}

= e^{-x-x}

= e^{-2x}

Hence, we have proven that \frac{1-tanh \ x}{1 + tanh \ x}=e^{-2x}

Learn more on Proof of Identities here: brainly.com/question/2561079

#SPJ1

6 0
2 years ago
There are 8 people in line for the self-checkout at the grocery story. There are 6
murzikaleks [220]

Answer:

The ratio is 3;4 or 4;3

Srry if wrong

Hope this helps

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3 years ago
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