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laiz [17]
1 year ago
10

A building is 5 feet tall. the base of the ladder is 8 feet from the building. how tall must a ladder be to reach the top of the

building? explain your reasoning.show your work. round to the nearest tenth if necessary.
Mathematics
1 answer:
iris [78.8K]1 year ago
5 0

The ladder must be 9.4 ft to reach the top of the building

Here, we want to get the length of the ladder that will reach the top of the building

Firstly, we need a diagrammatic representation

We have this as;

As we can see, we have a right triangle with the hypotenuse being the length of the ladder

We simply will make use of Pythagoras' theorem which states that the square of the hypotenuse is equal to the sum of the squares of the two other sides

Thus, we have;

\begin{gathered} x^2=5^2+8^2 \\ x^2=\text{ 25 + 64} \\ x^2\text{ = 89} \\ x=\text{ }\sqrt[]{89} \\ x\text{ = 9.4 ft} \end{gathered}

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3 years ago
Find the smallest solution to the equation 2/3 X² equals 24
iren [92.7K]

Answer:

The smallest solution is -6

Step-by-step explanation:

2/3 x^2 = 24

Multiply each side by 3/2

3/2 *2/3 x^2 = 24*3/2

x^2 = 36

Take the square root of each side

sqrt(x^2) = sqrt(36)

x = ±6

The smallest solution is -6

The largest solution is 6

4 0
3 years ago
(8.166×10^8)×(1.515×10^9)
loris [4]
The answer is ....1.237149 x 10^18
4 0
3 years ago
What is the approximate length of a pendulum that takes 2.4 pi seconds to swing back and forth?
LuckyWell [14K]

Answer:

The length of the pendulum is 14.13 meters.

Step-by-step explanation:

The length of a pendulum is given by the formula, T = 2\pi \sqrt{\frac{L}{g} }, where T is the times period and L is the length of the pendulum.

Now, given that T = 2.4π seconds.

So, from the above equation, we get,

2.4\pi  = 2\pi  \sqrt{\frac{L}{g} }

⇒ \sqrt{\frac{L}{g}} = 1.2

⇒ \frac{L}{g} = 1.44 {We have the value of g as 9.81 m/sec²}

⇒ L = 1.44 × 9.81 = 14.13 meters. (Approx.)

Therefore, the length of the pendulum is 14.13 meters. (Answer)

3 0
3 years ago
a) What is an alternating series? An alternating series is a whose terms are__________ . (b) Under what conditions does an alter
andriy [413]

Answer:

a) An alternating series is a whose terms are alternately positive and negative

b) An alternating series \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} (-1)^{n-1} b_n where bn = |an|, converges if 0< b_{n+1} \leq b_n for all n, and \lim_{n \to \infty} b_n = 0

c) The error involved in using the partial sum sn as an approximation to the total sum s is the remainder Rn = s − sn and the size of the error is bn + 1

Step-by-step explanation:

<em>Part a</em>

An Alternating series is an infinite series given on these three possible general forms given by:

\sum_{n=0}^{\infty} (-1)^{n} b_n

\sum_{n=0}^{\infty} (-1)^{n+1} b_n

\sum_{n=0}^{\infty} (-1)^{n-1} b_n

For all a_n >0, \forall n

The initial counter can be n=0 or n =1. Based on the pattern of the series the signs of the general terms alternately positive and negative.

<em>Part b</em>

An alternating series \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} (-1)^{n-1} b_n where bn = |an|  converges if 0< b_{n+1} \leq b_n for all n and \lim_{n \to \infty} b_n =0

Is necessary that limit when n tends to infinity for the nth term of bn converges to 0, because this is one of two conditions in order to an alternate series converges, the two conditions are given by the following theorem:

<em>Theorem (Alternating series test)</em>

If a sequence of positive terms {bn} is monotonically decreasing and

<em>\lim_{n \to \infty} b_n = 0<em>, then the alternating series \sum (-1)^{n-1} b_n converges if:</em></em>

<em>i) 0 \leq b_{n+1} \leq b_n \forall n</em>

<em>ii) \lim_{n \to \infty} b_n = 0</em>

then <em>\sum_{n=1}^{\infty}(-1)^{n-1} b_n  converges</em>

<em>Proof</em>

For this proof we just need to consider the sum for a subsequence of even partial sums. We will see that the subsequence is monotonically increasing. And by the monotonic sequence theorem the limit for this subsquence when we approach to infinity is a defined term, let's say, s. So then the we have a bound and then

|s_n -s| < \epsilon for all n, and that implies that the series converges to a value, s.

And this complete the proof.

<em>Part c</em>

An important term is the partial sum of a series and that is defined as the sum of the first n terms in the series

By definition the Remainder of a Series is The difference between the nth partial sum and the sum of a series, on this form:

Rn = s - sn

Where s_n represent the partial sum for the series and s the total for the sum.

Is important to notice that the size of the error is at most b_{n+1} by the following theorem:

<em>Theorem (Alternating series sum estimation)</em>

<em>If  \sum (-1)^{n-1} b_n  is the sum of an alternating series that satisfies</em>

<em>i) 0 \leq b_{n+1} \leq b_n \forall n</em>

<em>ii) \lim_{n \to \infty} b_n = 0</em>

Then then \mid s - s_n \mid \leq b_{n+1}

<em>Proof</em>

In the proof of the alternating series test, and we analyze the subsequence, s we will notice that are monotonically decreasing. So then based on this the sequence of partial sums sn oscillates around s so that the sum s always lies between any  two consecutive partial sums sn and sn+1.

\mid{s -s_n} \mid \leq \mid{s_{n+1} -s_n}\mid = b_{n+1}

And this complete the proof.

5 0
3 years ago
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