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VladimirAG [237]
3 years ago
15

Please help me with this question!

Mathematics
1 answer:
gavmur [86]3 years ago
6 0
First off, let's convert the percentages to decimal format, so our 77% turns to 77/100 or 0.77, and our 55% turns to 55/100 or 0.55 and so on

now, the sum of both salines, must add up to the 77% mixture, let's say is "y"
so, 11 + 4 = y, and whatever the concentration level is, must also sum up to the mixture's concentration of 77%

anyway   thus

\bf \begin{array}{lccclll}
&amount&concentration&
\begin{array}{llll}
concentrated\\
amount
\end{array}\\
&-----&-------&-------\\
\textit{first sol'n}&11&x&11x\\
\textit{second sol'n}&4&0.55&2.20\\
------&-----&-------&-------\\
mixture&y&0.77&0.77y
\end{array}\\\\
-------------------------------\\\\
\begin{cases}
11+4=y\implies 15=\boxed{y}\\
11x+2.2=0.77y\\
----------\\
11x+2.2=0.77\cdot \boxed{15}
\end{cases}

solve for "x"
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Perform the indicated operations. Express the result in scientific notation. (5.7 × 105) + (4.6 × 106) – (2.9 × 105)
alex41 [277]

Answer:

1.3905 × 10^4

Step-by-step explanation:

First you should multiply all the two numbers  inside the parentheses

5.7 × 105=598.5

4.6 × 106=487.6

2.9 × 105=304.5

Now add the three numbers

598.5+487.5+304.5= 1390.5

scientific notation: Put the decimal after the first digit. To find the exponent count the number of places from the decimal to the end of the number.

the answer in scientific notation: 1.3905 × 10^4

8 0
3 years ago
Suppose that 5 t shirts cost $55 what's the unit rate?
Setler [38]
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answer
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3 years ago
If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar
sleet_krkn [62]

This question is incomplete, the complete question is;

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple are written in increasing order but are not necessarily distinct.

In other words, how many 5-tuples of integers  ( h, i , j , m ), are there with  n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1 ?

Answer:

the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Step-by-step explanation:

Given the data in the question;

Any quintuple ( h, i , j , m ), with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1

this can be represented as a string of ( n-1 ) vertical bars and 5 crosses.

So the positions of the crosses will indicate which 5 integers from 1 to n are indicated in the n-tuple'

Hence, the number of such quintuple is the same as the number of strings of ( n-1 ) vertical bars and 5 crosses such as;

\left[\begin{array}{ccccc}5&+&n&-&1\\&&5\\\end{array}\right] = \left[\begin{array}{ccc}n&+&4\\&5&\\\end{array}\right]

= [( n + 4 )! ] / [ 5!( n + 4 - 5 )! ]

= [( n + 4 )!] / [ 5!( n-1 )! ]

= [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Therefore, the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

4 0
3 years ago
Find the indicated products.<br> (3x-5)(3x+5)
frutty [35]
9x²- 25


3x(3x) = 9x²
3x(5) = 15x
-5(3x) = -15x
-5(5) = -25
9x² + 15x - 15x - 25
9x² - 25 is your answer

hope this helps
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