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IgorC [24]
2 years ago
11

How do you do inequalitys?

Mathematics
1 answer:
nata0808 [166]2 years ago
4 0
I think this quisthionbchk shjvsjjwv hsjvsbkzbxbs hhsvsvwjkwvs hskvsvdod
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If beyonce had 5 single ladies and she multiplied the single ladies by 13 how many single ladies would there be.
mars1129 [50]

Answer:

63

Step-by-step explanation:

5 x 13 = 63

is this what you asked for?

6 0
3 years ago
Read 2 more answers
Gianni says the lateral area of the square pyramid is 624 in2 Do u agree or disagree with gianni?
kirza4 [7]

Answer: agree

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
What is the slope of the line that passes through the points (-4, -4)(−4,−4) and (-4, -9) ?(−4,−9)? Write your answer in simples
Akimi4 [234]

Answer:

undefined slope

Step-by-step explanation:

The points contain the same x-value. This means that the line is vertical.

All vertical lines have an undefined slope.

Hope this helps.

5 0
2 years ago
40 POINTS! HELP ME. QUICKLY.
dybincka [34]

Answer:

Part A: <em>D,E</em>    

Part B:

            m°= <em>90°</em>

            x°= <em>70°</em>

            y°=  <em>70°0</em>

            z°=  <em>110°</em>

7 0
2 years ago
Find maclaurin series
Mumz [18]

Recall the Maclaurin expansion for cos(x), valid for all real x :

\displaystyle \cos(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}

Then replacing x with √5 x (I'm assuming you mean √5 times x, and not √(5x)) gives

\displaystyle \cos\left(\sqrt 5\,x\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt5\,x\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^{2n}}{(2n)!}

The first 3 terms of the series are

\cos\left(\sqrt5\,x\right) \approx 1 - \dfrac{5x^2}2 + \dfrac{25x^4}{24}

and the general n-th term is as shown in the series.

In case you did mean cos(√(5x)), we would instead end up with

\displaystyle \cos\left(\sqrt{5x}\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt{5x}\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^n}{(2n)!}

which amounts to replacing the x with √x in the expansion of cos(√5 x) :

\cos\left(\sqrt{5x}\right) \approx 1 - \dfrac{5x}2 + \dfrac{25x^2}{24}

7 0
2 years ago
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