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3 years ago

Find decimal notation. 0.8%

Mathematics

2 answers:

professor190 [17]3 years ago

6 0

To convert 0.8% to decimal divide by 100

0.8 ÷ 100 = 0.008

ipn [44]3 years ago

3 0

To convert a percent to a decimal, divide the percent by 100.

0.8/100 = 0.008 <<<<<< answer.

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Janet is trying to identity a number. She knows the following properties of the unknown number. The opposite of the unknown numb

cluponka [151]

We know that the number's opposite is 5. In this context, the opposite means the negative of the given number. Since the number's opposite is 5, the number itself must be -5. This is consistent with the second piece of information, since the absolute value of a number is its distance to 0 and the absolute value of -5 is 5.

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3 years ago

Lena is making two dishes for an event. Each batch of her mac-n-cheese recipe calls for 6 ounces of cheese and 2 tablespoons of

kvasek [131]

No she needs four more ounces of cheese to make 2 batches of mac n cheese and 3 pizzas

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4 years ago

Jay is paid £2000 each month. He saves 6% of the £2000 each month. How many months will it take Jay to save £480?

BARSIC [14]

Answer:

4 months

Step-by-step explanation:

(2000 *6)/100 = 120

120x = 480

X = 480/120

X = 4 months

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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

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

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

4 0

3 years ago

-the equations x-y=-14 and -x-y=14 what is the number of solutions? A.) no solutions B.) infinitely many solutions C.) one solut

Yuki888 [10]

Answer:

The answer is C one solution

Step-by-step explanation:

x-y=-14

-y=-x-14 bring x to the other side

y=x+14 divide everything by -1

-x-y=14

-y=x+14 bring x to the other side

y=-x-14 divide everything by -1

6 0

3 years ago

Find decimal notation. 0.8​%

Find decimal notation. 0.8%