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3 years ago

If you start work at 2:45 and worked 2 hours how many hours did you work?

Mathematics

2 answers:

Goryan [66]3 years ago

8 0

You work for 2 hours

:/

musickatia [10]3 years ago

6 0

California Workers Get Double Time Pay In Two Cases1. Hours exceeding 12 in a day

When the hours worked in one day exceed 12, employees are paid double time for every hour worked thereafter.

2. Seventh consecutive day

If an employee works seven consecutive days, they are entitled to double time pay after the first 8 hours on that seventh workday.

From the California DIR:

For all hours worked in excess of 12 hours in any workday and for all hours worked in excess of eight on the seventh consecutive day of work in a workweek.

Why employees might not get double time

If the 7th day falls outside of the workweek, double time would not apply. A workweek starts over after 7 days so if an employee works 5 days in one workweek and 2 in another, the 7th consecutive day rule would not apply. The same is true of workdays. Read more about it here.

We’ll Do the Calculations For You!

Timesheets.com is an online time tracking software that takes the headaches out of time calculations. Our software can be configured for complicated California rules and you’ll never have to think about it again!

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Help ASAP solve: -3/4n = -6

anyanavicka [17]

Answer:

$\huge\boxed{n=8}$

Step-by-step explanation:

$-\dfrac{3}{4}n=-6\qquad|\text{multiply both sides by (-4)}\\\\(-4\!\!\!\!\diagup)\left(-\dfrac{3}{4\!\!\!\!\diagup}n\right)=(-4)(-6)\\\\3n=24\qquad|\text{divide both sides by 3}\\\\\dfrac{3\!\!\!\!\diagup n}{3\!\!\!\!\diagup}=\dfrac{24\!\!\!\!\!\diagup}{3\!\!\!\!\diagup}\\\\n=8$

3 0

3 years ago

Read 2 more answers

Plz help Factor the expression using the GCF 16m + 20n

lara [203]

Answer:

4 (4m+5n)

Step-by-step explanation:

hope this helps

4 0

3 years ago

F(x)=2x^2-5;find when f(x)=-7

Daniel [21]

Answer:

x=i or no solution exists depending on the user's grade

Step-by-step explanation:

f(x) = 2x^2-5

-7 = 2x^2-5

-2 = 2x^2, x^2=-1, x=i

6 0

2 years ago

Read 2 more answers

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

estimate by rounding each number to the same place value. please help with these i did these but i wanna make sure they are corr

valkas [14]

Im not too sure with these, but this is what i got

11. 39.90
12.16
13.182.8
14.220
15.499
16.19.5

4 0

3 years ago