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3 years ago

What is 0.06 divided by 8 working out

Mathematics

2 answers:

Feliz [49]3 years ago

6 0

It’s an answer for your question

sladkih [1.3K]3 years ago

4 0

Answer:

0.0075

Step-by-step explanation:

0. 0 0 7 5

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8 ÷ 0. 0 6 0 0

− 0

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0 0

− 0

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0 6

− 0

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6 0

− 5 6

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4 0

− 4 0

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50% of the cake Jenny bike last week we're party cakes 1/5 with fruit cakes and the remainder of sponge cakes what percentage of

Alik [6]

100%-1/5(20%)=80%
80%-50%=30%

7 0

3 years ago

Find the value of z.

AVprozaik [17]

Answer:

z = 110°

Step-by-step explanation:

Angles z and the one marked 70° form an linear pair, hence are supplementary.

z = 180° -70° = 110°

<h3>Angles where chords cross</h3>

The angle made by two chords is half the sum of the arcs intercepted by those chords. Here, that means ...

70° = 1/2(60° +x) ⇒ x = 140° -60° = 80°

The arc w completes the circle of 360°, so we have ...

w +x +79° +60° = 360° ⇒ w = 360° -219° = 141°

Finally, z is the average of w and 79°:

z = (w +79°)/2 = (141° +79°)/2 = 110° . . . as above

8 0

2 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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3 years ago

The question and choices are in the picture below NO LINKS

satela [25.4K]

i think its a if its not let me know oki

AND BTW I HATE LINKS

MARK BRAINLYEST PLZ

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3 years ago

Please show x < 80 on a number line

nevsk [136]

Answer:

on the number line make an open circle at 80 (am open circle is this: ○) and connect that to an arrow that points back to 0 and the negative numbers. It doesn't matter how far back as long as you make it an arrow

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3 years ago