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3 years ago

543,982 rounded to the nearest hundred thousand

Mathematics

2 answers:

Aleksandr-060686 [28]3 years ago

6 0

Answer:

500,000

Step-by-step explanation:

Bezzdna [24]3 years ago

3 0

Hello! :)

543,982 rounded to the nearest hundred thousand would be 500,000. Because the digit 4 is less than 5, it would be 600,000. If the digit where the 4 is was 5 or more, it would round to 600,000,

543,982 to the nearest hundred is 500,000.

Hope this helped you!

THEDIPER

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PLEASE HELP ASAP In this task, you will practice finding the area under a nonlinear function by using rectangles. You will use g

mrs_skeptik [129]

Answer:

a) $1280 u^{2}$

b) $1320 u^{2}$

c) $\frac{4000}{3} u^{2}$

Step-by-step explanation:

In order to solve this problem we must start by sketching the graph of the function. This will help us visualize the problem better. (See attached picture)

You can sketch the graph of the function by plotting as many points as you can from x=0 to x=20 or by finding the vertex form of the quadratic equation by completing the square. You can also do so by using a graphing device, you decide which method suits better for you.

So we are interested in finding the area under the curve, so we divide it into 5 rectangles taking a right hand approximation. This is, the right upper corner of each rectangle will touch the graph. (see attached picture).

In order to figure the width of each rectangle we can use the following formula:

$\Delta x=\frac{b-a}{n}$

in this case a=0, b=20 and n=5 so we get:

$\Delta x=\frac{20-0}{5}=\frac{20}{5}=4$

so each rectangle must have a width of 4 units.

We can now calculate the hight of each rectangle. So we figure the y-value of each corner of the rectangles. We get the following heights:

h1=64

h2=96

h3=96

h4= 64

h5=0

so now we can use the following formula to find the area under the graph. Basically what the formula does is add the areas of the rectangles:

$A=\sum^{n}_{i=1} f(x_{i}) \Delta x$

which can be rewritten as:

$A=\Delta x \sum^{n}_{i=1} f(x_{i})$

So we go ahead and solve it:

$A=(4)(64+96+96+64+0)$

so:

$A= 1280 u^{2}$

B) The same procedure is used to solve part B, just that this time we divide the area in 10 rectangles.

In order to figure the width of each rectangle we can use the following formula:

$\Delta x=\frac{b-a}{n}$

in this case a=0, b=20 and n=10 so we get:

$\Delta x=\frac{20-0}{10}=\frac{20}{10}=2$

so each rectangle must have a width of 2 units.

We can now calculate the hight of each rectangle. So we figure the y-value of each corner of the rectangles. We get the following heights:

h1=36

h2=64

h3=84

h4= 96

h5=100

h6=96

h7=84

h8=64

h9=36

h10=0

so now we can use the following formula to find the area under the graph. Basically what the formula does is add the areas of the rectangles:

$A=\sum^{n}_{i=1} f(x_{i}) \Delta x$

which can be rewritten as:

$A=\Delta x \sum^{n}_{i=1} f(x_{i})$

So we go ahead and solve it:

$A=(2)(36+64+84+96+100+96+84+64+36+0)$

so:

$A= 1320 u^{2}$

In order to find part c, we calculate the area by using limits, the limit will look like this:

$\lim_{n \to \infty} \sum^{n}_{i=1} f(x^{*}_{i}) \Delta x$

so we start by finding the change of x so we get:

$\Delta x =\frac{b-a}{n}$

$\Delta x =\frac{20-0}{n}$

$\Delta x =\frac{20}{n}$

next we find $x^{*}_{i}$

$x^{*}_{i}=a+\Delta x i$

so:

$x^{*}_{i}=0+\frac{20}{n} i=\frac{20}{n} i$

and we find $f(x^{*}_{i})$

$f(x^{*}_{i})=f(\frac{20}{n} i)=-(\frac{20}{n} i)^{2}+20(\frac{20}{n} i)$

cand we do some algebra to simplify it.

$f(x^{*}_{i})=-\frac{400}{n^{2}}i^{2}+\frac{400}{n}i$

we do some factorization:

$f(x^{*}_{i})=-\frac{400}{n}(\frac{i^{2}}{n}-i)$

and plug it into our formula:

$\lim_{n \to \infty} \sum^{n}_{i=1}-\frac{400}{n}(\frac{i^{2}}{n}-i) (\frac{20}{n})$

And simplify:

$\lim_{n \to \infty} \sum^{n}_{i=1}-\frac{8000}{n^{2}}(\frac{i^{2}}{n}-i)$

$\lim_{n \to \infty} -\frac{8000}{n^{2}} \sum^{n}_{i=1}(\frac{i^{2}}{n}-i)$

And now we use summation formulas:

$\lim_{n \to \infty} -\frac{8000}{n^{2}} (\frac{n(n+1)(2n+1)}{6n}-\frac{n(n+1)}{2})$

$\lim_{n \to \infty} -\frac{8000}{n^{2}} (\frac{2n^{2}+3n+1}{6}-\frac{n^{2}}{2}-\frac{n}{2})$

and simplify:

$\lim_{n \to \infty} -\frac{8000}{n^{2}} (-\frac{n^{2}}{6}+\frac{1}{6})$

$\lim_{n \to \infty} \frac{4000}{3}+\frac{4000}{3n^{2}}$

and solve the limit

$\frac{4000}{3}u^{2}$

4 0

3 years ago

Twice the sum of a number and 3 times a second number is 4. The difference of ten times the seco

34kurt

Two numbers that gives twice the sum of a number and 3 times a second number is 4. The difference of ten times the second number and five times the first is 90 are -10 and 4

Given :

Twice the sum of a number and 3 times a second number is 4. The difference of ten times the second number and five times the first is 90.

Let a and b be the two unknown numbers

Lets frame equation using the given statements

Twice the sum of a number and 3 times a second number is 4.

$2(a+3b)=4\\a+3b=2$

the difference of ten times the second number and five times the first is 90

$10b -5a=90\\Divide \; by \;5 \\2b-a=18$

Now use these two equations to solve a and b

$a+3b=2\\-a+2b=18$

Add both the equations

$a+3b=2\\-a+2b=18\\--------------------------\\5b=20\\b=4$

Now find out 'a'

$a+3b=2\\a+3(4)=2\\a+12=2\\a=2-12\\a=-10$

The two numbers are -10 and 4

Learn more : brainly.com/question/13856304

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2 years ago

To find the angle of rotation θ between the previous x and y axes and the new x′ and y′ axes, we use the formula: cotangent of 2

Tamiku [17]

The equation is the given one:

$\cot 2\theta=\frac{A-C}{B}$

Which is the same as:

$\tan 2\theta=\frac{B}{A-C}$

And the interval of the angle θ must be (0°,90°) so, 2θ must be (0°,180°).

So the statement is True.

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1 year ago

Describe the relationship between the number of at bats and the number of hits using the data in the scatter plot.

Fynjy0 [20]

Answer:

The number of at bats increases, the number of hits also increases

Step-by-step explanation:

Given

See attachment for graph

Required

Relationship between the plots

From the attached graph, we can observe that there is an upward trend in the graph; in other words, when the number of at bats increases, the number of hits also increases

Hence, the relationship between them is that: an increment in at bats causes an increment in hits.

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andre [41]

Answer:

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Step-by-step explanation:

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