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3 years ago

How do you change 1/16 into a decimal?

Mathematics

2 answers:

Lynna [10]3 years ago

5 0

Convert the fraction to a decimal by dividing the numerator by the denominator.

1/16= 0.0625

Lady bird [3.3K]3 years ago

4 0

Basically divide the top number with the bottom number.... like...
$\frac{1}{16}$
like for that... u divide 1 by 16 so the formula for that would be 1÷16. i dont have a calc with me but u can use a calc to find the answer

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Please tell explain what the right answer is

kupik [55]

Angle 3 and angle 4 are supplementary angles and when added together need to equal 180 degrees.

To find angle 4 subtract angle 3 from 180.

Angle 4 = 180 - 101 = 79 degrees.

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3 years ago

Read 2 more answers

What are the missing angles? What is the relationship between the measures of supplementary angles?

goldenfox [79]

Answer:

6=100°

8=80°

7=100°

9=80°

Step-by-step explanation:

If 6 is 100°

in a straight line theres 180 degrees

180-100=80°

8 is 80°

opposite 6 to 7 there are parallel corresponding angles meaning the angle opposite it will be the same. 8 with 9.

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3 years ago

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

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

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

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

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

Then the coefficients in the power series solution are governed by the recurrence relation,

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

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

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2 years ago

I told you people that i would be sending out a series of problems todayy!!

qwelly [4]

Each student would need to raise $27.56

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2 years ago

If a polygon has an interior angle of 144, what kind of polygon is it? (find the # of sides first

ikadub [295]

Each Interior Angle = ((Number of Sides -2) • 180 degrees) ÷ (Number of Sides)

Solving for number of sides

Each Interior Angle / 180 = (N -2) / (N)

144 / 180 = (n-2) / n

144 * n = 180n -360
360 = 36 n

n = 10 the number of sides

Source:
http://www.1728.org/polygon.htm

5 0

3 years ago