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3 years ago

14

Kevin ate 4 pies in 132 seconds what is the unit rate ?

2 answers:

Alex Ar [27]3 years ago

6 0

The answer is 33 your answer

solong [7]3 years ago

3 0

132/4=33

you divde 132 into 4

and then get Answer-33

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9/20 divided by 4/15 is what

jenyasd209 [6]

9/20 / 4/15

9/20 * 15/4

9/5 * 3/4

27/20
or
1 7/20

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3 years ago

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Determine whether each set of data can be modeled by an exponential function. If so, find the exponential

Sergeeva-Olga [200]

It can be modeled by an exponential function, f(x) is divided by 3 each time x increases by 1.

This means that the base of the exponential function is 1/3, and since the value of f(x) when x=0 is 81, $f(x)=81\left(\frac{1}{3} \right)^{x}$

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2 years ago

Which of the following equations is the result after the first step in solving 3x + 6 = 12?

Verizon [17]

3x=6 is the answer after solving the first step in the equation

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3 years ago

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Y= -27(x+4)³+1 the real zero(s) is (are)

rjkz [21]

$\bf y=-27(x+4)^3+1\implies 0=-27(x+4)^3+1
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\\\\\\
\sqrt[3]{\cfrac{1}{27}}=x+4\implies \cfrac{\sqrt[3]{1}}{\sqrt[3]{27}}=x+4\implies \cfrac{1}{3}=x+4\implies \cfrac{1}{3}-4=x
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-\cfrac{11}{3}=x$

4 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