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3 years ago

13

e greater roadrunner bird can run 14 miles per hour. at’s 7 times faster than an ostrich can walk. How fast does an ostrich walk

?

1 answer:

ExtremeBDS [4]3 years ago

3 0

14 / 7 = 2

An ostrich walks 2 miles an hour.

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What is the minimum number of degrees in order for the pentagon to map onto itself

amid [387]

Answer:

We can divide the pentagon (5 sides) into five inscribed triangles with central vertex angles of 72 degrees. 360 degrees ÷ 5 = 72 degrees....complete rotation is 360 degrees So each rotation of the pentagon of 72 degrees will be identical

Step-by-step explanation:

hope it helps :))

3 0

3 years ago

Which is the inverse of the function a(d)=5d-3? And use the definition of inverse functions to prove a(d) and a-1(d) are inverse

Drupady [299]

Answer:

$a'(d) = \frac{d}{5} + \frac{3}{5}$

$a(a'(d)) = a'(a(d)) = d$

Step-by-step explanation:

Given

$a(d) = 5d - 3$

Solving (a): Write as inverse function

$a(d) = 5d - 3$

Represent a(d) as y

$y = 5d - 3$

Swap positions of d and y

$d = 5y - 3$

Make y the subject

$5y = d + 3$

$y = \frac{d}{5} + \frac{3}{5}$

Replace y with a'(d)

$a'(d) = \frac{d}{5} + \frac{3}{5}$

Prove that a(d) and a'(d) are inverse functions

$a'(d) = \frac{d}{5} + \frac{3}{5}$ and $a(d) = 5d - 3$

To do this, we prove that:

$a(a'(d)) = a'(a(d)) = d$

Solving for $a(a'(d))$

$a(a'(d)) = a(\frac{d}{5} + \frac{3}{5})$

Substitute $\frac{d}{5} + \frac{3}{5}$ for d in $a(d) = 5d - 3$

$a(a'(d)) = 5(\frac{d}{5} + \frac{3}{5}) - 3$

$a(a'(d)) = \frac{5d}{5} + \frac{15}{5} - 3$

$a(a'(d)) = d + 3 - 3$

$a(a'(d)) = d$

Solving for: $a'(a(d))$

$a'(a(d)) = a'(5d - 3)$

Substitute 5d - 3 for d in $a'(d) = \frac{d}{5} + \frac{3}{5}$

$a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}$

Add fractions

$a'(a(d)) = \frac{5d - 3+3}{5}$

$a'(a(d)) = \frac{5d}{5}$

$a'(a(d)) = d$

Hence:

$a(a'(d)) = a'(a(d)) = d$

7 0

2 years ago

Help no time plzzzz plz plz

Sav [38]

Answer:

FALSE

Step-by-step explanation:

<E in ∆AED ≅ <E in ∆CEB.

Both are 90°.

Side ED ≅ Side EB

Side AD ≅ Side CB.

Thus, two sides (ED and AD) and a non-included angle (<E) of ∆AED are congruent to corresponding two sides (EB and CB) and a non-included angle (<E) of ∆CEB. Therefore, by A-S-S Congruence Theorem, both triangles are congruent to each other not by SSS.

8 0

2 years ago

Where should the point P be chosen on line segment AB so as to maximize the angle θ? (Assume a = 4 units, b = 5 units, and c = 9

taurus [48]

From the figure, let the distance of point P from point A on line segment AB be x and let the angle opposite side a be M and the angle opposite side c be N.

Using pythagoras theorem,
$\tan M= \frac{a}{b-x} \\ \\ M=\tan^{-1}\left(\frac{a}{b-x}\right)$
and
$\tan N= \frac{c}{x} \\ \\ N=\tan^{-1}\left(\frac{c}{x}\right)$

Angle θ is given by
$\theta=180-M-N \\ \\ =180-\tan^{-1}\left(\frac{a}{b-x}\right)-\tan^{-1}\left(\frac{c}{x}\right)$

Given that a = 4 units, b = 5 units, and c = 9 units, thus
$\theta=180-\tan^{-1}\left(\frac{4}{5-x}\right)-\tan^{-1}\left(\frac{9}{x}\right)$

To maximixe angle θ, the differentiation of <span>θ with respect to x must be equal to zero.
i.e.
$\frac{d\theta}{dx} = -\frac{4}{x^2-10x+41} + \frac{9}{x^2+81} =0 \\ \\ -4(x^2+81)+9(x^2-10x+41)=0 \\ \\ -4x^2-324+9x^2-90x+369=0 \\ \\ 5x^2-90x+45=0 \\ \\ x^2-18x+9=0 \\ \\ x=9\pm6 \sqrt{2}$

Given that x is a point on line segment AB, this means that x is a positive number less than 5.

Thus
$x=9-6 \sqrt{2}=0.5147$

Therefore, The distance from A of point P, so that </span>angle θ is maximum is 0.51 to two decimal places.

6 0

3 years ago

The heights of adult males in the United States are approximately normally distributed. The mean height is 70 inches (5 feet 10

steposvetlana [31]

Using the normal distribution, it is found that the probability is 0.16.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem, the mean and the standard deviation are given by, respectively, $\mu = 70, \sigma = 3$ .

The proportion of students between 45 and 67 inches is the p-value of Z when <u>X = 67 subtracted by the p-value of Z when X = 45</u>, hence:

X = 67:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{67 - 70}{3}$

Z = -1

Z = -1 has a p-value of 0.16.

X = 45:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{45 - 70}{3}$

Z = -8.3

Z = -8.3 has a p-value of 0.

0.16 - 0 = 0.16

The probability is 0.16.

More can be learned about the normal distribution at brainly.com/question/24663213

3 0

2 years ago