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2 years ago

9

Please tell the truth please help me please help me with the work please please help me with the work please help me with the wo

rk I do I need a little more work done

2 answers:

katovenus [111]2 years ago

6 0

1.

a. -10/7

b. 5/3

c. 3.02

d. 6 1/2 (it is a bit blury, hard to tell)

explanation- so the opposite of a positive is a negative, the opposite of a negative is a positive. so it is the same for this. for example 10/7 is positive so you make it a negative for it be the opposite.

i would help with the rest, but it is hard when you have to label things on a diagram when you are doing it on a computer like this. sorry!

vagabundo [1.1K]2 years ago

6 0

Answer:

1a. -10/7

1b. 5/3

1c. 3.02

1d. 6 1/2

Step-by-step explanation:

i can see everything because blury

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The sequence 2, 3, 5, 7, 11 are the first five prime numbers. A prime number is a positive integer which has exactly 2 positive integer factore, one factor being 1 and the other factor is the positive number itself. So the next prime numbers after 11 are 13, then 17, 19, 23, 29, 31, 37, 41, 43

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3 years ago

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Sonbull [250]

Answer:

-3.5

Step-by-step explanation:

To find the slope we use

m = (y2-y1)/(x2-x1)

= (-15-20)/(-6 - -16)

=(-15-20)/(-6+16)

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2 years ago

Use Pythagorean Theorem to solve #1

denpristay [2]

The answer is 6the answer is 63

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2 years ago

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the volume of hemisphere is 1527.4cm^3.find the radius of the hemisphere to the nearest whole whole number.(take π to be 22/7)

skelet666 [1.2K]

Answer:

radius of the hemisphere = 9 cm approx

Step-by-step explanation:

Volume of hemisphere = $\frac{2}{3} \pi r^{3}$

1527.4 = $\frac{2}{3} * \frac{22}{7} * r^{3}$

$\frac{7*15274*3}{2*22*10}$ = $r^{3}$

$\frac{320754}{440}$ = $r^{3}$

$r^{3}$ = 729

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

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2 years ago