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4 years ago

What is the converse of the following conditional statement? If two segments have the same length, then they are congruent

Mathematics

1 answer:

Nana76 [90]4 years ago

5 0

If two segments are congruent, then they have the same length.

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The table below shows randy breuers height in inches leading into his teen years. The graph below displays a scatter plot of the

-Dominant- [34]

The least-square regression line has a slope of:

m=(nΣxy-ΣyΣx)/(nΣx²-ΣxΣx)

and a y-intercept of:

b=(Σy-mΣx)/n

In this case: n=7, Σxy=4899, Σy=391, Σx=85, Σx²=1153 so

m=(7(4899)-391*85)/(7(1153)-85*85))=1058/846

b=(391*846-85*1058)/(7*846)=34408/846

So the line of best fit is:

y=(1058x+34408)/846 and if we approximated this as your answers see to have done....

y=1.25x+40.67

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4 years ago

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damaskus [11]

Your answer will be -1/7

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3 years ago

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Fill in the table! Please help! My teacher taught this the day I was absent!

Goshia [24]

The parabola is going downwards.

The equation for the axis of symmetry is x = -1.

The vertex is at -1,9

The x-intercepts are at -4, 2

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Whatis 4b to the -6 power in its simplest form

Olenka [21]

4b^-6

When u have a negative exponent, u simply take the reciprocal. So the answer is

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Be sure to answer all parts. List the evaluation points corresponding to the midpoint of each subinterval to three decimal place

gayaneshka [121]

Answer:

The Riemann Sum for $\int\limits^5_4 {x^2+4} \, dx$ with n = 4 using midpoints is about 24.328125.

Step-by-step explanation:

We want to find the Riemann Sum for $\int\limits^5_4 {x^2+4} \, dx$ with n = 4 using midpoints.

The Midpoint Sum uses the midpoints of a sub-interval:

$\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f\left(\frac{x_0+x_1}{2}\right)+f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_2+x_3}{2}\right)+...+f\left(\frac{x_{n-2}+x_{n-1}}{2}\right)+f\left(\frac{x_{n-1}+x_{n}}{2}\right)\right)$

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

We know that a = 4, b = 5, n = 4.

Therefore, $\Delta{x}=\frac{5-4}{4}=\frac{1}{4}$

Divide the interval [4, 5] into n = 4 sub-intervals of length $\Delta{x}=\frac{1}{4}$

$\left[4, \frac{17}{4}\right], \left[\frac{17}{4}, \frac{9}{2}\right], \left[\frac{9}{2}, \frac{19}{4}\right], \left[\frac{19}{4}, 5\right]$

Now, we just evaluate the function at the midpoints:

$f\left(\frac{x_{0}+x_{1}}{2}\right)=f\left(\frac{\left(4\right)+\left(\frac{17}{4}\right)}{2}\right)=f\left(\frac{33}{8}\right)=\frac{1345}{64}=21.015625$

$f\left(\frac{x_{1}+x_{2}}{2}\right)=f\left(\frac{\left(\frac{17}{4}\right)+\left(\frac{9}{2}\right)}{2}\right)=f\left(\frac{35}{8}\right)=\frac{1481}{64}=23.140625$

$f\left(\frac{x_{2}+x_{3}}{2}\right)=f\left(\frac{\left(\frac{9}{2}\right)+\left(\frac{19}{4}\right)}{2}\right)=f\left(\frac{37}{8}\right)=\frac{1625}{64}=25.390625$

$f\left(\frac{x_{3}+x_{4}}{2}\right)=f\left(\frac{\left(\frac{19}{4}\right)+\left(5\right)}{2}\right)=f\left(\frac{39}{8}\right)=\frac{1777}{64}=27.765625$

Finally, use the Midpoint Sum formula

$\frac{1}{4}(21.015625+23.140625+25.390625+27.765625)=24.328125$

This is the sketch of the function and the approximating rectangles.

5 0

4 years ago