Mathematics, 12.08.2020 06:01 jadens25

I NEED THIS ASAP In this task, you will practice finding the area under a nonlinear function by using rectangles. You will use graphing skills in addition to the knowledge gathered in this unit. Sketch the graph of the function y = 20x − x2, and approximate the area under the curve in the interval [0, 20] by dividing the area into the given numbers of rectangles. Part A Use five rectangles to approximate the area under the curve. Part B Use 10 rectangles to approximate the area under the curve. Part C Calculate the area under the curve using rectangles as their number becomes arbitrarily large (tends to infinity). You do not need to sketch the rectangles.

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Answer from: Qpaoswp5914

$\large \boxed{\text{Part A. 1280; Part B. 1320; Part C; 1333}}$

Step-by-step explanation:

Part A. Using five rectangles.

I will use the left end-point approximation.

1. Divide the interval [0,20] into five equal subintervals

Δx = (20 - 0)/5 = 4

The intervals are:

[0,4], [4,8], [8,12], [12,16], [16,20]

2. Evaluate the function at the left end-points and calculate the areas

$\begin{array}{rrr}\mathbf{x} & \mathbf{f (x)}&\textbf{Area} \\0 &0&0 \\4 & 64&256 \\8 & 96&384 \\12 & 96&384 \\16 & 64&256 \\&\textbf{TOTAL =}& \mathbf{1280}\end{array}\\\text{The area of five rectangles is $\large \boxed{\mathbf{1280}}$}$

Fig. 1 shows the graph with five rectangles.

Part B. Using 10 rectangles

1. Divide the interval [0,20] into 10 equal subintervals

Δx = (20 - 0)/10 = 2

The intervals are:

[0,2], [2,4], [4,6], [6,8], [8,10], [10,12], [12,14], [14,16], [16,18], [18,20]

2. Evaluate the function at the left end-points and calculate the areas

$\begin{array}{rrr}\mathbf{x} & \mathbf{f (x)}&\textbf{Area} \\0 &0&0 \\2 & 36&72 \\4 & 64&128 \\6 & 84&168 \\8 & 96&192 \\10 &100&200 \\12 & 96&192 \\14 &84&168 \\16 & 64&128 \\18 &36&72 \\&\textbf{TOTAL =}& \mathbf{1320}\end{array}\\\text{The area of 10 rectangles is $\large \boxed{\mathbf{1320}}$}$

Fig. 2 shows the graph with 10 rectangles.

Part C. The limit as n ⟶ ∞

$\displaystyle \lim_{n\longrightarrow \,\infty} \sum_{i = 1}^{n} f(x_{i-1})\Delta x= \int_{a}^{b}f(x)dx$

Integrate the function.

$\displaystyle \int_{0}^{20}(20x-x^{2})dx= \left[10x^{2}- \dfrac{1}{3}x^{3}\right]_{0}^{20} =4000 - \dfrac{8000}{3} = \dfrac{12000 - 8000 }{3}\\\\ = \dfrac{4000}{3} \approx\mathbf{1333}$

I NEED THIS ASAP In this task, you will practice finding the area under a nonlinear function by usin

Answer from: Quest

Answer: question no. 1 -a. yes, the sides are in the ratio 2 : 5.question no. 2 -b. 59°question no. 3 -c. angles x and y are remote interior angles.question no. 4 -d. 90°, 45° and 45°question no. 5 -b. mn and qrstep-by-step explanation: question no. 1 -rs : xu = 3 : 7.5 = 2 : 5rt : x? = 6.4 : 16 = 2 : 5st : u? = 6 : 15 = 2 : 5as the ratios of all sides are equivalent, the triangles must be congruent and the sides must be proportional.therefore, the answer is: a. yes, the sides are in the ratio 2 : 5.question no. 2 -in the second diagram, angle b is the only letter not shown. therefore, the number must be equivalent to angle b.if we substitute the value of angle b as being 22° in the first diagram, we can now use the angle sum of a triangle to obtain the value of angle a.angle sum of a triangle = 180°therefore: let angle a = aa + 99 + 22 = 180a = 180 - 99 - 22a = 59°therefore, the answer is: b. 59°question no. 3 -angles x and y don't share the vertex of the triangle with the exterior angle w. remote interior angles don't share a vertex of a triangle with an exterior angle. this means that angles x and y are remote interior angles.therefore, the answer is: c. angles x and y are remote interior angles.question no. 4 -the sum of all interior angles of a triangle must be equivalent to 180°.90°, 45° and 45° are the only set of angle measures that are equivalent to 180°.therefore, the answer is: d. 90°, 45° and 45°question no. 5 -the only pair of sides stated in the options which are corresponding are mn and qr.therefore, the answer is: b. mn and qr.

Answer from: Quest

$7.00 \cdot 10^{-7} m - 1\cdot 10^{-3} m$

explanation:

infrared radiation has wavelength in the range between 700 nm and 1 mm. lets' convert these values into scientific notation, keeping in mind that:

$1 nm = 10^{-9} m\\1 mm = 10^{-3} m$

let's start from the lower value, we have:

$700 nm = 700 \cdot 10^{-9} m = 7.00 \cdot 10^{-7} m$ (we moved the dot two places to the left, so we increased the exponent by 2 units, from -9 to -7)