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xenn [34]
3 years ago
12

Mary lives 6/10 mile from school. Thad lives 9/8 miles from school. Mary says Thad lives farther from school. Is she correct? Ex

plain.
Mathematics
2 answers:
vagabundo [1.1K]3 years ago
5 0
Mary is correct If you think about it maby you will get it by now
Katen [24]3 years ago
4 0
Answer: Mary is correct.

9/8 is an improper fraction, so we change it to 1 1/8.

Mary lives less than a mile from school, so we know that Thad lives farther since he lives a mile and 1/8 away, and Mary only lives 6/10 away, which is less than a mile.

Glad I could help, and good luck!

You might be interested in
7th grade math , I’ll give brainliest
Dennis_Churaev [7]

25 X 8 = 200

500 - 200 = 300

So if you have 8 weeks and spend 25 dollars a week then you would have $300 left over at the end of the 8 weeks.

Hope this helps Plz mark brainliest

5 0
3 years ago
Read 2 more answers
An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t
USPshnik [31]
The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

\displaystyle\int_a^bf(x)\,\mathrm dx

Split up the interval [a,b] into n equal subintervals,

[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]

where a=x_0 and b=x_n. Each subinterval has measure (width) \dfrac{a-b}n.

Now denote the left- and right-endpoint approximations by L and R, respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are \{x_0,x_1,\cdots,x_{n-1}\}. Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, \{x_1,x_2,\cdots,x_n\}.

So, you have

L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)
R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)

Now let T denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)

Factoring out \dfrac12 and regrouping the terms, you have

T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)

which is equivalent to

T=\dfrac12\left(L+R)

and is the average of L and R.

So the trapezoidal approximation for your problem should be \dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2
4 0
3 years ago
4.
natali 33 [55]
The answer is D. 15+18+12+5= 50 and there are 5 ramblers
3 0
3 years ago
Read 2 more answers
A man sold 50 metres of cloth for Rs.5640 at a profit of 20%.
disa [49]
I) 80% x 5640 = Rs. 4512
ii) 4512 divide by 50 = Rs. 90.24
8 0
3 years ago
Need help with both questions please ‼️
Sindrei [870]

Answer:

Option A is your answer

3 0
3 years ago
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