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IgorLugansk [536]

3 years ago

5

Help me answer the question.

1 answer:

Studentka2010 [4]3 years ago

7 0

1/8 = 12.5 and 3/4 = 75 so he can wash 6 times before running out.

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i think of a number, multiply it by 2 subract 6 and then divide by 4. If the answer is 8 what is the original number

ivanzaharov [21]

Answer:

19

Step-by-step explanation:

8 x 4 = 32

32 + 6 = 38

38 / 2 = 19

6 0

3 years ago

Ebola has an Ro of 2. How<br> many third-wave cases can<br> doctors expect?

stellarik [79]

Answer:

4 I guess

Step-by-step explanation:

Because

3-1=2

2^2=2*2=4

4 0

3 years ago

Type the correct answer in each box. Spell all words correctly, and use numerals instead of words for numbers.

Natasha2012 [34]

<span>The missing angle measure in triangle ABC is 55°.

The measure of angle BAC in triangle ABC is equal to the measure of angle
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5 0

2 years ago

The product of two numbers is 50 and their difference is 5. What are the two numbers?

lyudmila [28]

27.5 and 22.5 are the two numbers

6 0

2 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

2 years ago