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3 years ago

Lana works as an accountant and ears $78,832 per year. What is Lana’s approximate weekly earnings

Mathematics

2 answers:

Nadusha1986 [10]3 years ago

7 0

Answer:

$1516

Step-by-step explanation:

$78,832 / 52 weeks = $1516 per week

-If Lana worked every week of the year with no vacation with only weekends as a break.

ad-work [718]3 years ago

7 0

Answer:

Lana earns $1,511.85

Step-by-step explanation:

There are 52.1429 weeks in a year, you would divide the amount of money ($78,832) by the amount of weeks in a year (52.1429).

$78,832÷52.1429=1511.845333

Which rounded equals $1,511.85

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The first three terms of an arithmetic series are 6p+2, 4p²-10 and 4p+3 respectively. Find the possible values of p. Calculate t

Doss [256]

Answer:

First Case:

$\displaystyle p=\frac{5}{2}\text{ and } d=-2$

Second Case:

$\displaystyle p=-\frac{5}{4}\text{ and } d=\frac{7}{4}$

Step-by-step explanation:

We know that the first three terms of an arithmetic series are:

$6p+2, 4p^2-10, \text{ and } 4p+3$

Since this is an arithmetic sequence, each subsequent term is d more than the previous term, where d is our common difference.

Therefore, we can write the second term as;

$4p^2-10=(6p+2)+d$

And, likewise, for the third term:

$4p+3=(6p+2)+2d$

Let's solve for d for each of the equations.

Subtracting in the first equation yields:

$d=4p^2-6p-12$

And for the second equation:

$2d=-2p+1$

To avoid fractions, let's multiply the first equation by 2. Hence:

$2d=8p^2-12p-24$

Therefore:

$8p^2-12p-24=-2p+1$

Simplifying yields:

$8p^2-10p-25=0$

Solve for p. We can factor:

$8p^2+10p-20p-25=0$

Factor:

$2p(4p+5)-5(4p+5)=0$

Grouping:

$(2p-5)(4p+5)=0$

Zero Product Property:

$\displaystyle p_1=\frac{5}{2} \text{ or } p_2=-\frac{5}{4}$

Then, we can use the second equation to solve for d. So:

$2d_1=-2p_1+1$

Substituting:

$\begin{aligned} 2d_1&=-2(\frac{5}{2})+1 \\ 2d_1&=-5+1 \\ 2d_1&=-4 \\ d_1&=-2\end{aligned}$

So, for the first case, p is 5/2 and d is -2.

Likewise, for the second case:

$\begin{aligned} 2d_2&=-2(-\frac{5}{4})+1 \\ 2d_2&=\frac{5}{2}+1 \\ 2d_2&=\frac{7}{2} \\ d_2&=\frac{7}{4}\end{aligned}$

So, for the second case, p is -5/4, and d is 7/4.

By using the values, we can determine our series.

For Case 1, we will have:

17, 15, 13.

For Case 2, we will have:

-11/2, -15/4, -2.

8 0

3 years ago

2b + 8 − 5b + 3 = -13 + 8b − 5 b = __

Bumek [7]

Answer:

b= 29/11

Step-by-step explanation:

2b+8−5b+3=−13+8b−5

Step 1: Simplify both sides of the equation.

2b+8−5b+3=−13+8b−5

2b+8+−5b+3=−13+8b+−5

(2b+−5b)+(8+3)=(8b)+(−13+−5)(Combine Like Terms)

−3b+11=8b+−18

−3b+11=8b−18

Step 2: Subtract 8b from both sides.

−3b+11−8b=8b−18−8b

−11b+11=−18

Step 3: Subtract 11 from both sides.

−11b+11−11=−18−11

−11b=−29

Step 4: Divide both sides by -11.

−11b/−11 = −29/−11

b= 29/11

4 0

3 years ago

100 person had a food provision for 24 days if 20 person left the place the provision will last for

stepladder [879]

100 person had a food provision for 24 days if 20 person left the place the provision will last for 30 days

4 0

3 years ago

Tania measured the growth of her plant each week. The firstt week, the plant's height measured 2.65 Decimeters. During the secon

Evgesh-ka [11]

Answer:

3.35 decimeter.

Step-by-step explanation:

We have been given that the first week, the plant's height measured 2.65 Decimeters. During the second week, Tania''s plant grew 0.7 Decimeter.

To find the height of Tania's plant at the end of second we will add 0.7 to 2.65.

$\text{The length of Tania's plant at the end of 2nd week}=2.65+0.7$

$\text{The length of Tania's plant at the end of 2nd week}=3.35$

Therefore, Tania's plant was 3.35 decimeter tall at the end of 2nd week

3 0

3 years ago

Find area of the cross section of the dam pictured below

kondaur [170]

Answer:

(w + h cotθ)h

Step-by-step explanation:

Cross section is the isosceles trapezoid.

Area of trapezoid:

A = 1/2(a + b)h

We have one of the bases and the height.

Let the difference of the base lengths is x.

We have:

(x/2)/h = cot θ ⇒ x = 2h cot θ

The area is:

A = 1/2(w + w + 2h cot θ)h = (w + h cotθ)h

4 0

3 years ago