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Murrr4er [49]
4 years ago
11

A 1414​-ft ladder leans against a wall at a point 55 feet above the ground. how far is the bottom of the ladder from the​ wall?

give an exact​ answer, then approximate the distance to the nearest hundredth of a foot.
Mathematics
1 answer:
viva [34]4 years ago
7 0
Assume ladder length is 14 ft and that the top end of the ladder is 5 feet above the ground.

Find the distance the bottom of the ladder is from the base of the wall.

Picture a right triangle with hypotenuse 14 feet and that the side opposite the angle is h.  Then sin theta = h / 14, or theta = arcsin 5/14.  theta is

0.365 radian.  Then the dist. of the bot. of the lad. from the base of the wall is 

14cos theta = 14cos 0.365 rad = 13.08 feet.  This does not seem reasonable; the ladder would fall if it were already that close to the ground.

Ensure that y ou have copied this problem accurately from the original.

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Which of the following expressions is equal to 5^6/5^2
Natali5045456 [20]

Answer:

\dfrac{5^6}{5^2} = 5{\cdot}5{\cdot}5{\cdot}5

Step-by-step explanation:

The given expression is :

\dfrac{5^6}{5^2}

We need to find this expression is equal to what.

\dfrac{5^6}{5^2}=\dfrac{5^4\times 5^2}{5^2}\\\\=5^4\\\\=5\times 5\times 5\times 5\\\\\text{or}\\\\=5{\cdot}5{\cdot}5{\cdot}5

Hence, \dfrac{5^6}{5^2} is equal to 5{\cdot}5{\cdot}5{\cdot}5. Hence, the correct option is (c).

5 0
3 years ago
Evaluate the expression a•b for a =24 and b=8
Butoxors [25]

\huge\text{Hey there!}


\mathsf{a\times b}}

\mathsf{= 24\times8}

\mathsf{= 192}


\huge\textbf{Therefore, your answer should be:}

\huge\boxed{\mathsf{192}}\huge\checkmark


\huge\text{Good luck on your assignment \& enjoy your day!}


<h3>~\frak{Amphitrite1040:)}</h3>
4 0
2 years ago
Using the distributive property what is 4(2x -1) in simplest form?
Marina86 [1]

Answer:

8x -4

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
For f(x)=4sin(x2) between x=0 and x=3, find the coordinates of all intercepts, critical points, and inflection points to two dec
telo118 [61]

Answer:

Intercepts:

x = 0, y = 0

x = 1.77, y = 0

x = 2.51, y = 0

Critical points:

x = 1.25, y = 4

x = 2.17 , y = -4

x = 2.8, y = 4

Inflection points:

x = 0.81, y = 2.44

x = 1.81, y = -0.54

x = 2.52, y = 0.27

Step-by-step explanation:

We can find the intercept by setting f(x) = 0

4sin(x^2) = 0

sin(x^2) = 0

x^2 = n\pi where n = 0, 1, 2,3, 4, 5,...

x = \sqrt(n\pi)

Since we are restricting x between 0 and 3 we can stop at n = 2

So the function f(x) intercepts at y = 0 and x:

x = 0

x = 1.77

x = 2.51

The critical points occur at the first derivative = 0

f^{'}(x) = 4cos(x^2)2x = 8xcos(x^2) = 0

xcos(x^2) = 0

x = 0 or

cos(x^2) = 0

x^2 = \frac{\pi}{2} + n\pi where n = 0, 1, 2, 3

x = \sqrt{\pi(n+1/2)}

Since we are restricting x between 0 and 3 we can stop at n =  2

So our critical points are at

x = 1.25, y = f(1.25) = 4sin(1.25^2) = 4

x = 2.17 , y = f(2.17) = 4sin(2.17^2) = -4

x = 2.8, y = f(2.8) = 4sin(2.8^2) = 4

For the inflection point, we can take the 2nd derivative and set it to 0

f^[''}(x) = 8(cos(x^2) - xsin(x^2)2x) = 8cos(x^2) - 16x^2sin(x^2) = 0

cos(x^2) = 2x^2sin(x^2)

tan(x^2) = \frac{1}{2x^2}

We can solve this numerically to get the inflection points are at

x = 0.81, y = f(0.81) = 4sin(0.81^2) = 2.44

x = 1.81, y = f(1.81) = 4sin(1.81^2) = -0.54

x = 2.52, y = f(2.52) = 4sin(2.52^2) = 0.27

3 0
3 years ago
Several years​ ago, 50​% of parents who had children in grades​ K-12 were satisfied with the quality of education the students r
galina1969 [7]

Answer:

The 95% confidence interval for the proportion of parents that are satisfied with their children's education is (0.4118, 0.4618). 0.5 is not part of the confidence interval, so this represents evidence that​ parents' attitudes toward the quality of education have changed.

Step-by-step explanation:

We have to see if 50% = 0.5 is part of the confidence interval.

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 1095, \pi = \frac{478}{1095} = 0.4365

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4365 - 1.96\sqrt{\frac{0.4365*0.5635}{1095}} = 0.4118

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4365 + 1.96\sqrt{\frac{0.4365*0.5635}{1095}} = 0.4612

The 95% confidence interval for the proportion of parents that are satisfied with their children's education is (0.4118, 0.4618). 0.5 is not part of the confidence interval, so this represents evidence that​ parents' attitudes toward the quality of education have changed.

5 0
3 years ago
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