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Komok [63]
2 years ago
8

Bathroom tiles cost £1.65 each.Work out the total cost of 90 tiles.​

Mathematics
2 answers:
MatroZZZ [7]2 years ago
8 0

Answer:

£144.00

Step-by-step explanation:

1.60 x 90 = 144

Hope this Helps (✿◡‿◡)

xenn [34]2 years ago
8 0

Answer:

£148.50

Step-by-step explanation:

you first have to change £1.65 into a whole number

165×90=14850

14850 × 100 = 148.50

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2.25k - 20 = 6.5k + 14<br><br><br> plsssssssssssss
Pani-rosa [81]

Answer:

k=-8

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

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Find the value of x.
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Answer:

x=16

Step-by-step explanation:

50+40=90

40-8=32

32/2=16

x=16

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A childs pool contains 17500 millileters. how many liters are in the pool
Aleks04 [339]

Answer:

17.5 litres

Step-by-step explanation:

1000 mililitres = 1 litre

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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}  

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3 years ago
7-70. Use your preferred method to solve
kirill [66]

Answer:

You would want to divide 6 by 2, getting x=3.

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2 years ago
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