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2 years ago

What is y-intercept for the line y=8x+21

Mathematics

1 answer:

Anna71 [15]2 years ago

7 0

Change Y to 0.

y=8x+21

0=8x+21

-8x=21

$x=\frac{21}{-8\\}$

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A car rental company charges a $50 flat fee and an additional $20 per day. A second company also

aivan3 [116]

The difference between the flat fees of the two companies is $20

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2 years ago

Daniella has $210 in the bank, and her balance is growing at a rate of $3 each month. Lori has $187 in the bank,but her balance

just olya [345]

Answer: 15 1/3 months

Step-by-step explanation:

Daniella has $210 in the bank, and her balance grows at a rate of $3 each month. This can be written as:

b = 3m + 210 ...... equation i

Lori has $187 in the bank and her balance grows at a rate of $4.50 each month. This can be written as:

b = 4.5m + 187 ....... equation ii

Combine both equations

b = 3m + 210

b = 4.5m + 187

3m + 210 = 4.5m + 187

4.5m - 3m = 210 - 187

1.5m = 23

m = 23/1.5

m = 15 1/3 months

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2 years ago

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jek_recluse [69]

Answer: I believe the answer would be about 1

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3 years ago

It takes Benjamin 28 minutes to mow two lawns. Assuming along with the same size and Benjamin works at the same speed about how

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Step-by-step explanation:

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2 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

2 years ago