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Zinaida [17]
2 years ago
6

I have to describe the angles and the sides of the triangle I don't have a clue​

Mathematics
1 answer:
babymother [125]2 years ago
7 0

Answers:

  • 0 acute angles
  • 0 obtuse angles
  • 0 right angles
  • 0 sides of equal length

===========================================================

Explanation:

The smaller sides a = 3 and b = 4 add to a+b = 3+4 = 7 which does NOT exceed the third side c = 7.

Therefore, a+b > c is not true and a triangle cannot be formed. The sum of any two sides of a triangle must exceed the third side. Refer to the triangle inequality theorem for more information.

Since a triangle cannot be constructed with those side lengths, this means we'll have 0's for each of the boxes.

---------------

If your teacher made a typo and didn't mean to use those numbers, then let me know and I'll update my solution.

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Answer:

- 4 \leqslant  - x \leqslant 3  \\  4 \geqslant x \geqslant  - 3

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Use the power series method to solve the given initial-value problem. (Format your final answer as an elementary function.)
olya-2409 [2.1K]

You're looking for a solution of the form

\displaystyle y = \sum_{n=0}^\infty a_n x^n

Differentiating twice yields

\displaystyle y' = \sum_{n=0}^\infty n a_n x^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n

\displaystyle y'' = \sum_{n=0}^\infty n(n-1) a_n x^{n-2} = \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n

Substitute these series into the DE:

\displaystyle (x-1) \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n - x \sum_{n=0}^\infty (n+1) a_{n+1} x^n + \sum_{n=0}^\infty a_n x^n = 0

\displaystyle \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^{n+1} - \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n \\\\ \ldots \ldots \ldots - \sum_{n=0}^\infty (n+1) a_{n+1} x^{n+1} + \sum_{n=0}^\infty a_n x^n = 0

\displaystyle \sum_{n=1}^\infty n(n+1) a_{n+1} x^n - \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n \\\\ \ldots \ldots \ldots - \sum_{n=1}^\infty n a_n x^n + \sum_{n=0}^\infty a_n x^n = 0

Two of these series start with a linear term, while the other two start with a constant. Remove the constant terms of the latter two series, then condense the remaining series into one:

\displaystyle a_0-2a_2 + \sum_{n=1}^\infty \bigg(n(n+1)a_{n+1}-(n+1)(n+2)a_{n+2}-na_n+a_n\bigg) x^n = 0

which indicates that the coefficients in the series solution are governed by the recurrence,

\begin{cases}y(0)=a_0 = -7\\y'(0)=a_1 = 3\\(n+1)(n+2)a_{n+2}-n(n+1)a_{n+1}+(n-1)a_n=0&\text{for }n\ge0\end{cases}

Use the recurrence to get the first few coefficients:

\{a_n\}_{n\ge0} = \left\{-7,3,-\dfrac72,-\dfrac76,-\dfrac7{24},-\dfrac7{120},\ldots\right\}

You might recognize that each coefficient in the <em>n</em>-th position of the list (starting at <em>n</em> = 0) involving a factor of -7 has a denominator resembling a factorial. Indeed,

-7 = -7/0!

-7/2 = -7/2!

-7/6 = -7/3!

and so on, with only the coefficient in the <em>n</em> = 1 position being the odd one out. So we have

\displaystyle y = \sum_{n=0}^\infty a_n x^n \\\\ y = -\frac7{0!} + 3x - \frac7{2!}x^2 - \frac7{3!}x^3 - \frac7{4!}x^4 + \cdots

which looks a lot like the power series expansion for -7<em>eˣ</em>.

Fortunately, we can rewrite the linear term as

3<em>x</em> = 10<em>x</em> - 7<em>x</em> = 10<em>x</em> - 7/1! <em>x</em>

and in doing so, we can condense this solution to

\displaystyle y = 10x -\frac7{0!} - \frac7{1!}x - \frac7{2!}x^2 - \frac7{3!}x^3 - \frac7{4!}x^4 + \cdots \\\\ \boxed{y = 10x - 7e^x}

Just to confirm this solution is valid: we have

<em>y</em> = 10<em>x</em> - 7<em>eˣ</em>   ==>   <em>y</em> (0) = 0 - 7 = -7

<em>y'</em> = 10 - 7<em>eˣ</em>   ==>   <em>y'</em> (0) = 10 - 7 = 3

<em>y''</em> = -7<em>eˣ</em>

and substituting into the DE gives

-7<em>eˣ</em> (<em>x</em> - 1) - <em>x</em> (10 - 7<em>eˣ </em>) + (10<em>x</em> - 7<em>eˣ</em> ) = 0

as required.

8 0
3 years ago
Delia is a graphic designer who charges an hourly rate. She made the graph to show her total earnings for various numbers of hou
stich3 [128]

Answer:

approximately 7 hours

see the explanation

Step-by-step explanation:

The picture of the question in the attached figure

Let

x ----> the time worked in hours

y ----> the amount earning in dollars

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form k=\frac{y}{x} or y=kx

In a proportional relationship the constant of proportionality k is equal to the slope m of the line <u><em>and the line passes through the origin</em></u>

so

In this problem we have a proportional relationship

step 1

Find approximately how many hours would it take Delia to earn $250 ?

Looking at the graph

For y=$250 The value of x is approximately 7 hours

see the attached figure N 2

step 2

Find the exact value

from the graph take the point (5,175)

Find the value of k

k=\frac{y}{x} ----> k=\frac{175}{5}=\$35\ per\ hour

The linear equation is

y=35x

For y=$250

substitute

250=35x

solve for x

x=7.14\ hours

8 0
3 years ago
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