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ozzi
3 years ago
10

3/17=k/10 what is the missing value

Mathematics
1 answer:
Tcecarenko [31]3 years ago
3 0
I think it is about 1.8 if you round it to the nearest tenth and cross multiply
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200 = 16(6t - 13)<br><br> t = ? <br> Help!
Marrrta [24]

Step 1. Divide both side by 16

100/16 = 6t - 13

Step 2. Dimplify 200/16 to 25/2

25/2 = 6t - 13

Step 3. Add 13 to both sides

25/2 + 13 = 6t

Step 4. Simplify 25/2 + 13 to 51/2

51/2 = 6t

Step 5. Divide both sides by 6

51/2/6 = t

Step 6. Simplify 51/2/6 to 51/2 * 6

51/2 * 6 = t

Step 7. Simplify 2 * 6 to 12

51/12 = t

Step 8. Simplify 51/12 to 17/4

17/4 = t

Step 9. Switch sides

t = 17/4

4 0
3 years ago
Read 2 more answers
What is the process of comparing data with a set of rules or values to determine if the data meets certain criteria
Airida [17]

Answer:

Validation

Step-by-step explanation: Validation is a term used to describe the processes involved when we compare a set of values and observations against a set standard or rules to ensure that they meet certain expectations or criteria.

Validation is meant to prove that something, a data set etc are acceptable based on known rules, the rules or standards which is used to evaluate what can be described as valid.

3 0
3 years ago
6. Find the slope. Make sure to reduce.
zaharov [31]

Answer:

the slope is -1/7 hope this helps

5 0
3 years ago
A teacher places n seats to form the back row of a classroom layout. Each successive row contains two fewer seats than the prece
Alex_Xolod [135]

Answer:

The number of seat when n is odd S_n=\frac{n^2+2n+1}{4}

The number of seat when n is even S_n=\frac{n^2+2n}{4}

Step-by-step explanation:

Given that, each successive row contains two fewer seats than the preceding row.

Formula:

The sum n terms of an A.P series is

S_n=\frac{n}{2}[2a+(n-1)d]

    =\frac{n}{2}[a+l]

a = first term of the series.

d= common difference.

n= number of term

l= last term

n^{th} term of a A.P series is

T_n=a+(n-1)d

n is odd:

n,n-2,n-4,........,5,3,1

Or we can write 1,3,5,.....,n-4,n-2,n

Here a= 1 and d = second term- first term = 3-1=2

Let t^{th} of the series is n.

T_n=a+(n-1)d

Here T_n=n, n=t, a=1 and d=2

n=1+(t-1)2

⇒(t-1)2=n-1

\Rightarrow t-1=\frac{n-1}{2}

\Rightarrow t = \frac{n-1}{2}+1

\Rightarrow t = \frac{n-1+2}{2}

\Rightarrow t = \frac{n+1}{2}

Last term l= n,, the number of term =\frac{ n+1}2, First term = 1

Total number of seat

S_n=\frac{\frac{n+1}{2}}{2}[1+n}]

    =\frac{{n+1}}{4}[1+n}]

     =\frac{(1+n)^2}{4}

    =\frac{n^2+2n+1}{4}

n is even:

n,n-2,n-4,.......,4,2

Or we can write

2,4,.......,n-4,n-2,n

Here a= 2 and d = second term- first term = 4-2=2

Let t^{th} of the series is n.

T_n=a+(n-1)d

Here T_n=n, n=t, a=2 and d=2

n=2+(t-1)2

⇒(t-1)2=n-2

\Rightarrow t-1=\frac{n-2}{2}

\Rightarrow t = \frac{n-2}{2}+1

\Rightarrow t = \frac{n-2+2}{2}

\Rightarrow t = \frac{n}{2}

Last term l= n, the number of term =\frac n2, First term = 2

Total number of seat

S_n=\frac{\frac{n}{2}}{2}[2+n}]

    =\frac{{n}}{4}[2+n}]

     =\frac{n(2+n)}{4}

    =\frac{n^2+2n}{4}  

4 0
3 years ago
<img src="https://tex.z-dn.net/?f=1%20%5Cdiv%20%20%5Cfrac%7B9%7D%7B5%7D%20" id="TexFormula1" title="1 \div \frac{9}{5} " alt="1
jasenka [17]
The answer to your question is 5/9.
7 0
4 years ago
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