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andrezito [222]
3 years ago
13

Pheneg-Xu drinks 1 cup of milk every day.how many gallons does he drink each week.

Mathematics
1 answer:
Gwar [14]3 years ago
3 0
Let x = number of cups drank per week
let y = number of gallons drank per week

x = 1 cup x 7 days
x = 7 cups per one week

y = 7 cups per week x 0.0625 gallons
y = 0.4375 liquid gallons each week

Therefore Pheneg-Xu drinks 0.4375 liquid gallons each week
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When a teacher divided her students into groups of four, she had three students remaining. when she divided them into groups of
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39 students....39/4 = 9 remainder 3
                       39/5 = 7 remainder 4

ur answer is 39
5 0
3 years ago
What is 2,606 + 7,025?
cestrela7 [59]

Answer:

2606 + 7025 = 9631

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Evaluate the function f(x) = 6x - 5 at f(1)
Likurg_2 [28]
Remove the parentheses around the expression 1
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3 years ago
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
Rounded to the nearest thousand of 275,608
olga nikolaevna [1]
The thousandths place in 275,608 is the 5. Depending on the next number (which in this case is the 6) it will either round up or down. Since the next number is 6 and it's 5 or bigger, the 5 will round up. So your final answer is 276,000.
4 0
3 years ago
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