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frutty [35]
3 years ago
11

Megan sent 12 postcards last week. This week she sent 4 postcards. How many more postcards didn't she sent last week?

Mathematics
2 answers:
choli [55]3 years ago
7 0

Answer:

She sent 8 more postcards last week.

Step-by-step explanation:

Consider the provided information.

Megan sent 12 postcards last week. This week she sent 4 postcards.

We need to find How many more postcards did she sent last week?

To find this simply subtract the numbers as shown:

12-4 = 8

Hence, she sent 8 more postcards last week.

Gala2k [10]3 years ago
5 0
8 postcards she didn't sent last week.
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the length of the sides of a square are initially 0 cm and increase at a constant rate of 11 cm per second. suppose the function
Likurg_2 [28]

The function of the area of the square is A(t)=121t^{2}

Given that The length of a square's sides begins at 0 cm and increases at a constant rate of 11 cm per second. Assume the function f determines the area of the square (in cm2) given several seconds, t since the square began growing and asked to find the function of the area

Lets assume the length of side of square is x

\frac{dx}{dt} = 11 \frac{cm}{sec}

⇒x=11t

Area of square=(length of side)^{2}

Area of square=(11t)^{2}{as the length of side is 11t}{varies by time}

Area of square=121t^{2}

Therefore,The function of the area of the square is A(t)=121t^{2}

Learn more about The function of the area of the square is A(t)=121t^{2}

Given that The length of a square's sides begins at 0 cm and increases at a constant rate of 11 cm per second. Assume the function f determines the area of the square (in cm2) given several seconds, t since the square began growing and asked to find the function of the area

Lets assume the length of side of square is x

\frac{dx}{dt} = 11 \frac{cm}{sec}

⇒x=11t

Area of square=(length of side)^{2}

Area of square=(11t)^{2}{as the length of side is 11t}{varies by time}

Area of square=121t^{2}

Therefore,The function of the area of the square is A(t)=121t^{2}

Learn more about area here:

brainly.com/question/27683633

#SPJ4

3 0
2 years ago
Plz answer correctly both questions if do will give brainliest!!!!
Marina86 [1]
54 square inches
Basically you just multiply
6 0
3 years ago
Read 2 more answers
The amount of time it takes a swimmer to swim a race varies inversely as the average speed of the swimmer. A swimmer finishes a
yarga [219]

Answer:

5 average speed

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
A bag contains 2 white marbles and 7 purple marbles. Two marbles are drawn at random. One marble is drawn and not replaced. Then
Masja [62]
A.
\frac{7purple}{9total} *  \frac{2white}{8total}  \\  \frac{14}{72}  \\ =0.194
The denominator of the second number goes down one because you aren't replacing the initial marble pulled, making the amount of marbles in the bag go down, but not affecting the white count because you pulled a purple marble.
b.
\frac{2white}{9total} *  \frac{1white}{8total}  \\  \frac{2}{72}  \\ =0.278
The same can be said about the denominator for this problem, but not the numerator. The numerator must go down one number as well, because you already chose one white marble, taking one of the white marbles out of the bag, leaving you with one solitary, white marble.
c.
Chance of selecting two purple marbles:
<u />\frac{7purple}{9total} *  \frac{6purple}{8total}  \\  \frac{42}{72}  \\ =0.583
Chance of selecting two white marbles:
\frac{2white}{9total} * \frac{1white}{8total} \\ \frac{2}{72} \\ =0.278
<u>Chance of selecting two purple</u> vs. <em>Chance of Selecting 2 white:</em>
<u>0.583</u>><em>0.278</em>.
It is more likely to select 2 purple marbles.

Hope this helps!
6 0
3 years ago
Read 2 more answers
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
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