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Nataly [62]
3 years ago
13

A company builds metal stands for bicycle wheels. a new design calls for a v-shaped stand that will hold wheels with a 13-inch r

adius. the sides of the stand form a 70° angle to the nearest tenth of an inch, what should the length of the side of the v-shaped stand be so that it is tangent to the wheel?
Mathematics
1 answer:
aleksandrvk [35]3 years ago
4 0
The length of the side = <span>The length of the arc intercepted by the central
<span>angle

</span></span>∴ length = Θ r
where Θ = <span>central<span> angle in radians
and     r  = radius

∴  </span></span><span>Θ = 70° = 70 * π /180
     </span><span>r  = 13 in.

∴ </span>length = <span>(70 * π /180) * 13 ≈ 15.88 in.
</span>
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Likurg_2 [28]

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Read 2 more answers
How many terms of the arithmetic sequence {1,22,43,64,85,…} will give a sum of 2332? Show all steps including the formulas used
MA_775_DIABLO [31]

There's a slight problem with your question, but we'll get to that...

Consecutive terms of the sequence are separated by a fixed difference of 21 (22 = 1 + 21, 43 = 22 + 21, 64 = 43 + 21, and so on), so the <em>n</em>-th term of the sequence, <em>a</em> (<em>n</em>), is given recursively by

• <em>a</em> (1) = 1

• <em>a</em> (<em>n</em>) = <em>a</em> (<em>n</em> - 1) + 21 … … … for <em>n</em> > 1

We can find the explicit rule for the sequence by iterative substitution:

<em>a</em> (2) = <em>a</em> (1) + 21

<em>a</em> (3) = <em>a</em> (2) + 21 = (<em>a</em> (1) + 21) + 21 = <em>a</em> (1) + 2×21

<em>a</em> (4) = <em>a</em> (3) + 21 = (<em>a</em> (1) + 2×21) + 21 = <em>a</em> (1) + 3×21

and so on, with the general pattern

<em>a</em> (<em>n</em>) = <em>a</em> (1) + 21 (<em>n</em> - 1) = 21<em>n</em> - 20

Now, we're told that the sum of some number <em>N</em> of terms in this sequence is 2332. In other words, the <em>N</em>-th partial sum of the sequence is

<em>a</em> (1) + <em>a</em> (2) + <em>a</em> (3) + … + <em>a</em> (<em>N</em> - 1) + <em>a</em> (<em>N</em>) = 2332

or more compactly,

\displaystyle\sum_{n=1}^N a(n) = 2332

It's important to note that <em>N</em> must be some positive integer.

Replace <em>a</em> (<em>n</em>) by the explicit rule:

\displaystyle\sum_{n=1}^N (21n-20) = 2332

Expand the sum on the left as

\displaystyle 21 \sum_{n=1}^N n-20\sum_{n=1}^N1 = 2332

and recall the formulas,

\displaystyle\sum_{k=1}^n1=\underbrace{1+1+\cdots+1}_{n\text{ times}}=n

\displaystyle\sum_{k=1}^nk=1+2+3+\cdots+n=\frac{n(n+1)}2

So the sum of the first <em>N</em> terms of <em>a</em> (<em>n</em>) is such that

21 × <em>N</em> (<em>N</em> + 1)/2 - 20<em>N</em> = 2332

Solve for <em>N</em> :

21 (<em>N</em> ² + <em>N</em>) - 40<em>N</em> = 4664

21 <em>N</em> ² - 19 <em>N</em> - 4664 = 0

Now for the problem I mentioned at the start: this polynomial has no rational roots, and instead

<em>N</em> = (19 ± √392,137)/42 ≈ -14.45 or 15.36

so there is no positive integer <em>N</em> for which the first <em>N</em> terms of the sum add up to 2332.

4 0
3 years ago
Solve the below: <br>4x+3y=7 3x-2y=9<br><br>​
Ahat [919]
4x+3y=7
3x-2y=9

12x+9y=21
12x-8y=36
^^Multiply top equation by 3 and multiply bottom equation by 4

12x+9y=36
- 12x-8y=36

17y=-15
^^Then subtract those two new equations

17y/17= -15/17

Y= -0.88
^^Next divide both sides by 17

4x+3* -0.88=7

4x + (-2.64) =7
-(-2.64) -(-2.64)

4x = 9.64

4x/4=9.64/4
X=2.41

^^then choose on equation and put your new y in the y spot and solve.


X=2.41

Y=-0.88

Your point/coordinate is (2.41,-.88)
7 0
3 years ago
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