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vovangra [49]
3 years ago
6

Nichole ran 11 3/4 miles on Monday, 10 1/5 miles on Tuesday and only 7 7/8 miles on Wednesday. Estimate how many miles she ran i

n those 3 days to the nearest whole mile
Mathematics
1 answer:
nikklg [1K]3 years ago
7 0

Answer: 30 miles


Step-by-step explanation:

1st day = 11 3/4 = 11.75 miles

2nd day = 10 1/5 = 10.2 miles

3rd day = 7 7/8 = 7.875 miles

If we add these three distances

11.75 + 10.2 + 7.875 = 29.825 miles

That is nearest to 30 whole miles

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QUESTION IN THE ATTACHMENT
eimsori [14]

Answer:

A. The sum of the first 10th term is 100.

B. The sum of the nth term is n²

Step-by-step explanation:

Data obtained from the question include:

Sum of 20th term (S20) = 400

Sum of 40th term (S40) = 1600

Sum of 10th term (S10) =..?

Sum of nth term (Sn) =..?

Recall:

Sn = n/2[2a + (n – 1)d]

Sn is the sum of the nth term.

n is the number of term.

a is the first term.

d is the common difference

We'll begin by calculating the first term and the common difference. This is illustrated below:

Sn = n/2 [2a + (n – 1)d]

S20 = 20/2 [2a + (20 – 1)d]

S20= 10 [2a + 19d]

S20 = 20a + 190d

But:

S20 = 400

400 = 20a + 190d .......(1)

S40 = 40/2 [2a + (40 – 1)d]

S40 = 20 [2a + 39d]

S40 = 40a + 780d

But

S40 = 1600

1600 = 40a + 780d....... (2)

400 = 20a + 190d .......(1)

1600 = 40a + 780d....... (2)

Solve by elimination method

Multiply equation 1 by 40 and multiply equation 2 by 20 as shown below:

40 x equation 1:

40 x (400 = 20a + 190d)

16000 = 800a + 7600. ........ (3)

20 x equation 2:

20 x (1600 = 40a + 780d)

32000 = 800a + 15600d......... (4)

Subtract equation 3 from equation 4

Equation 4 – Equation 3

32000 = 800a + 15600d

– 16000 = 800a + 7600d

16000 = 8000d

Divide both side by 8000

d = 16000/8000

d = 2

Substituting the value of d into equation 1

400 = 20a + 190d

d = 2

400 = 20a + (190 x 2)

400 = 20a + 380

Collect like terms

400 – 380 = 20a

20 = 20a

Divide both side by 20

a = 20/20

a = 1

Therefore,

First term (a) = 1.

Common difference (d) = 2.

A. Determination of the sum of the 10th term.

First term (a) = 1.

Common difference (d) = 2

Number of term (n) = 10

Sum of 10th term (S10) =..?

Sn = n/2 [2a + (n – 1)d]

S10 = 10/2 [2x1 + (10 – 1)2]

S10 = 5 [2 + 9x2]

S10 = 5 [2 + 18]

S10 = 5 x 20

S10 = 100

Therefore, the sum of the first 10th term is 100.

B. Determination of the sum of the nth term.

First term (a) = 1.

Common difference (d) = 2

Sum of nth term (Sn) =..?

Sn = n/2 [2a + (n – 1)d]

Sn = n/2 [2x1 + (n – 1)2]

Sn = n/2 [2 + 2n – 2]

Sn = n/2 [2 – 2 + 2n ]

Sn = n/2 [ 2n ]

Sn = n²

Therefore, the sum of the nth term is n²

6 0
3 years ago
What is 2/3 times 1/4
barxatty [35]
(2/3)(1/4)=
2/12=
1/6
multiply the numerators by the numerators, (top numbers)
and the denominators by the denominators, (bottom numbers)
6 0
3 years ago
I need help with d and I need to show my work.
cupoosta [38]
Ok
So when you multiply by a fraction just multiply top by top and bottom by bottom
So top by top is 2 by 4 so 8
And bottom by bottom is 3 by 5 which is 15
So answer is 8/15
Hope this helps and don’t forget to mark as brainliest if you thought it was most helpful :)
7 0
3 years ago
The law of sines states that if a triangle with sides a,b, and c, then the following is true.
Andrej [43]
C, because that’s the way to prove the law of sines.
7 0
3 years ago
Read 2 more answers
how many different plates are possible if letters and digits can be repeated with 3 letters follwed by 2 digits
WARRIOR [948]
We know that there are 26 letters in the alphabet, and 10 different digit possibilities (0-9). We also know that the plates each have 3 letters and 2 digits. and they can be repeated. To solve this, we multiply 26 X 26 X 26 X 10 X 10 to get the total amount of possibilities 1,757,600. Hope this helps.
5 0
3 years ago
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