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3 years ago

What is a formula for the nth term of the given sequence? 180, -240,320...

Mathematics

1 answer:

Semmy [17]3 years ago

8 0

Answer:

Tn=T1+(n-1)*d

Step-by-step explanation:

Tn is the nth term

T1 is the first term of the sequence

n is the nth term as well

d is the common difference

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Candice spent 5 1/4 hours doing her homework. Her brother, Ronald, spent 1/2 that number of hours doing his homework. How many h

Elden [556K]

Let

x-----> the number of hours that Candice spent doing her homework

y-----> the number of hours that Ronald spent doing his homework

we know that

$x=5\frac{1}{4}\ hours$

convert mixed number to an improper fraction

$5\frac{1}{4}\ hours=\frac{5*4+1}{4}= \frac{21}{4}\ hours$

$x=\frac{21}{4}\ hours$ -------> equation A

$y=\frac{1}{2}x$ ------> equation B

Substitute equation A in equation B

$y=\frac{1}{2}*\frac{21}{4}=\frac{21}{8}\ hours$

convert an improper fraction to a mixed number

$\frac{21}{8}\ hours=2.625=2+0.625=2+\frac{5}{8}=2\frac{5}{8}\ hours$

therefore

the answer is the option B

$2\frac{5}{8}\ hours$

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 $\dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2$