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Anna11 [10]
3 years ago
9

What is the standard form of this number? nine hundred fifty thousand, four hundred forty-five

Mathematics
2 answers:
gladu [14]3 years ago
6 0

Answer:

950,445

Step-by-step explanation:

ELEN [110]3 years ago
3 0
9.50445 * 10 to the power of 5
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Violet is baking a mixed berry pie that contains blueberries, cherries, blackberries, and raspberries. She uses three times as m
erastova [34]

Answer:

Blackberries= 30

Rasberries=30

Blueberries=60

Step-by-step explanation:

10 x 3 = 30 blackberries

Black berries is the same amount as rasberries

30 x 2= 60 blueberries

7 0
3 years ago
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there are 425 people in the auditorium they are sitting in rows of 9 how many rows have 9 people sitting down?​
mars1129 [50]
425/9 = 47.2

round to 47 because they want to know how many people are in rows with ALL 9 people

47 rows have 9 people sitting down
3 0
3 years ago
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The number of fish in a lake is decreasing by 400 every year, as described by the function below.
Jlenok [28]

Using the recursive function given, it is found that f(5) = 6600.

The function given is:

f(n + 1) = f(n) - 400

f(1) = 8200

To find f(5), we keep applying the function until n + 1 = 5, hence:

f(2) is f(1) subtracted by 400

f(2) = f(1) - 400 = 8200 - 400 = 7800

f(3) is f(2) subtracted by 400

f(3) = f(2) - 400 = 7800 - 400 = 7400

f(4) is f(3) subtracted by 400

f(4) = f(3) - 400 = 7400 - 400 = 7000

f(5) is f(4) subtracted by 400

f(5) = f(4) - 400 = 7000 - 400 = 6600

Hence, the result is f(5) = 6600.

A similar problem is given at brainly.com/question/21245344

4 0
3 years ago
If the square root of n = 3/2, what is the negative square root of n ?
Advocard [28]

A por k si estudie guey

Step-by-step explanation:

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3 0
3 years ago
Question (c)! How do I know that t^5-10t^3+5t=0?<br> Thanks!
astra-53 [7]
(a) By DeMoivre's theorem, we have

(\cos\theta+i\sin\theta)^5=\cos5\theta+i\sin5\theta

On the LHS, expanding yields

\cos^5\theta+5i\cos^4\theta\sin\theta-10\cos^3\theta\sin^2\theta-10i\cos^2\theta\sin^3\theta+5\cos\theta\sin^4\theta+i\sin^4\theta

Matching up real and imaginary parts, we have for (i) and (ii),


\cos5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta
\sin5\theta=5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta

(b) By the definition of the tangent function,

\tan5\theta=\dfrac{\sin5\theta}{\cos5\theta}
=\dfrac{5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta}{\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta}

=\dfrac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}
=\dfrac{t^5-10t^3+5t}{5t^4-10t^2+1}


(c) Setting \theta=\dfrac\pi5, we have t=\tan\dfrac\pi5 and \tan5\left(\dfrac\pi5\right)=\tan\pi=0. So

0=\dfrac{t^5-10t^3+5t}{5t^4-10t^2+1}

At the given value of t, the denominator is a non-zero number, so only the numerator can contribute to this reducing to 0.


0=t^5-10t^3+5t\implies0=t^4-10t^2+5

Remember, this is saying that

0=\tan^4\dfrac\pi5-10\tan^2\dfrac\pi5+5

If we replace \tan^2\dfrac\pi5 with a variable x, then the above means \tan^2\dfrac\pi5 is a root to the quadratic equation,

x^2-10x+5=0

Also, if \theta=\dfrac{2\pi}5, then t=\tan\dfrac{2\pi}5 and \tan5\left(\dfrac{2\pi}5\right)=\tan2\pi=0. So by a similar argument as above, we deduce that \tan^2\dfrac{2\pi}5 is also a root to the quadratic equation above.

(d) We know both roots to the quadratic above. The fundamental theorem of algebra lets us write

x^2-10x+5=\left(x-\tan^2\dfrac\pi5\right)\left(x-\tan^2\dfrac{2\pi}5\right)

Expand the RHS and match up terms of the same power. In particular, the constant terms satisfy

5=\tan^2\dfrac\pi5\tan^2\dfrac{2\pi}5\implies\tan\dfrac\pi5\tan\dfrac{2\pi}5=\pm\sqrt5

But \tanx>0 for all 0, as is the case for x=\dfrac\pi5 and x=\dfrac{2\pi}5, so we choose the positive root.
3 0
3 years ago
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