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2 years ago

(4/3) to the power of 2 help??

Mathematics

2 answers:

bija089 [108]2 years ago

7 0

In decimal form it is 1.777 and in fraction form it is 16/9

Lelu [443]2 years ago

5 0

(4/3) ^2
= (4/3)(4/3)
= 4 * 4 / 3 * 3
= 16/9

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NEED HELP!!!

Dahasolnce [82]

Answer:

No.

Step-by-step explanation:

The solution x = 0 means that the value 0 satisfies the equation, so there is a solution. “No solution” means that there is no value, not even 0, which would satisfy the equation.

7 0

2 years ago

How do you do this?<br> Because i don't know how to do it :>

Sindrei [870]

Answer:

£ 7.95

Step-by-step explanation:

Let the total money = £ x

Money spend on drinks and cake = 1.45 + 1.20 = £ 2.65

Money left with her = (2/3) of x

So, money spend = (1/3) of x

$\dfrac{1}{3}*x= 2.65\\\\\\x=2.65*3\\\\$

x = £ 7.95

7 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

(WHO EVER HELPS ME GETS BRAINLIEST AND EXTRA POINTSSSSS :D )

Romashka [77]

Answer:d

D. 9 and 46/180

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

If sum of the perpendicular distances of a variable point P (x, y) from the lines x + y – 5 = 0 and 3x – 2y + 7 = 0 is always 10

Ksivusya [100]

Answer:

I need points!!!!!!!!

Step-by-step explanation:

EWWWWWWWWWWWWWWWWWWWWWWWWWWWWW

3 0

2 years ago