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PtichkaEL [24]
2 years ago
13

hey all I added my teacher to the teacher appreciation contest and would appreciate everyones vote for her. Her name is Mrs.kwak

under the steam category!​
Mathematics
1 answer:
IRISSAK [1]2 years ago
4 0

Ok I will nominate her

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If the property tax on a 300,000 Is 2,220 what is the tax on a 100,000 home
ikadub [295]

Answer:

SO first you have to divise 300,000 by 2,220 and then multiply it by 100,000.To get 13636363.6364.SO the tax on a 100k home would be about 13636363.6364

Step-by-step explanation:

6 0
3 years ago
yolanda had 47 customers on saturday from 6pm to 7pm on monday from 5 pm to 6 pm she had 20 customers what is the amount of perc
rusak2 [61]
Yolanda had 47 customers on Saturday from 6 to 7 pm. Then on Monday she only had 20 customers from 5pm to 6pm. Question: Let’s find how much percentage rate did Yolanda’s customer’s decreased over time. Solution => Saturday = 47 customers. => Monday = 20 customer. Let’s start solcing => 47 – 20 = 27 There are 27 customer that were lost over time. => 27 / 47 = 0.57 * 100 = 57% Let’s try checking our answer: => 47 * .57 = 26.79 or round off to 27, since we rounded off also our percentage.
8 0
3 years ago
A child who is 4 ½ feet tall casts a 6-foot shadow. At the same time, a nearby statue casts a 15-foot shadow. How tall is the st
stiv31 [10]

Answer:

13 1/2 tall

Step-by-step explanation:

If 4 1/2 is the childs hight and their shadow is 6 feet, the shadow adds 2 1/2 to the height.

5 0
2 years ago
Quadratic inverse for y=x^2-1
stepan [7]

let's recall that to find the inverse of any expression, we start off by doing a quick switcheroo on the variables, and then solve for "y".

\bf y = x^2-1\implies \stackrel{switcheroo}{x=y^2-1}\implies x+1=y^2\implies \sqrt{x+1}=\stackrel{f^{-1}(x)}{y}

8 0
2 years ago
Which definite integral approximation formula is this: the integral from a to b of f(x)dx ≈ (b-a)/n * [<img src="https://tex.z-d
Stella [2.4K]

The answer is most likely A.

The integration interval [<em>a</em>, <em>b</em>] is split up into <em>n</em> subintervals of equal length (so each subinterval has width (<em>b</em> - <em>a</em>)/<em>n</em>, same as the coefficient of the sum of <em>y</em> terms) and approximated by the area of <em>n</em> rectangles with base (<em>b</em> - <em>a</em>)/<em>n</em> and height <em>y</em>.

<em>n</em> subintervals require <em>n</em> + 1 points, with

<em>x</em>₀ = <em>a</em>

<em>x</em>₁ = <em>a</em> + (<em>b</em> - <em>a</em>)/<em>n</em>

<em>x</em>₂ = <em>a</em> + 2(<em>b</em> - <em>a</em>)/<em>n</em>

and so on up to the last point <em>x</em> = <em>b</em>. The right endpoints are <em>x</em>₁, <em>x</em>₂, … etc. and the height of each rectangle are the corresponding <em>y </em>'s at these endpoints. Then you get the formula as given in the photo.

• "Average rate of change" isn't really relevant here. The AROC of a function <em>G(x)</em> continuous* over an interval [<em>a</em>, <em>b</em>] is equal to the slope of the secant line through <em>x</em> = <em>a</em> and <em>x</em> = <em>b</em>, i.e. the value of the difference quotient

(<em>G(b)</em> - <em>G(a)</em> ) / (<em>b</em> - <em>a</em>)

If <em>G(x)</em> happens to be the antiderivative of a function <em>g(x)</em>, then this is the same as the average value of <em>g(x)</em> on the same interval,

g_{\rm ave}=\dfrac{G(b)-G(a)}{b-a}=\dfrac1{b-a}\displaystyle\int_a^b g(x)\,\mathrm dx

(* I'm actually not totally sure that continuity is necessary for the AROC to exist; I've asked this question before without getting a particularly satisfying answer.)

• "Trapezoidal rule" doesn't apply here. Split up [<em>a</em>, <em>b</em>] into <em>n</em> subintervals of equal width (<em>b</em> - <em>a</em>)/<em>n</em>. Over the first subinterval, the area of a trapezoid with "bases" <em>y</em>₀ and <em>y</em>₁ and "height" (<em>b</em> - <em>a</em>)/<em>n</em> is

(<em>y</em>₀ + <em>y</em>₁) (<em>b</em> - <em>a</em>)/<em>n</em>

but <em>y</em>₀ is clearly missing in the sum, and also the next term in the sum would be

(<em>y</em>₁ + <em>y</em>₂) (<em>b</em> - <em>a</em>)/<em>n</em>

the sum of these two areas would reduce to

(<em>b</em> - <em>a</em>)/<em>n</em> = (<em>y</em>₀ + <u>2</u> <em>y</em>₁ + <em>y</em>₂)

which would mean all the terms in-between would need to be doubled as well to get

\displaystyle\int_a^b f(x)\,\mathrm dx\approx\frac{b-a}n\left(y_0+2y_1+2y_2+\cdots+2y_{n-1}+y_n\right)

7 0
3 years ago
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