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artcher [175]
3 years ago
11

What is the annual bonus for a salesperson who sells 55 cars?

Mathematics
2 answers:
Keith_Richards [23]3 years ago
8 0

Answer:

$825 dollars

Step-by-step explanation:

is the answer i belive hope this helps

katovenus [111]3 years ago
7 0
825
1,800/120 = 15
55 x 15 = 825
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Problem 4: Solve the initial value problem
pishuonlain [190]

Separate the variables:

y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx

Separate the left side into partial fractions. We want coefficients a and b such that

\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}

\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}

\implies 1 = a(y-2)+b(y+1)

\implies 1 = (a+b)y - 2a+b

\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13

So we have

\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx

Integrating both sides yields

\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx

\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C

\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C

\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C

\dfrac{y-2}{y+1} = e^{3x + C}

\dfrac{y-2}{y+1} = Ce^{3x}

With the initial condition y(0) = 1, we find

\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12

so that the particular solution is

\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}

It's not too hard to solve explicitly for y; notice that

\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}

Then

1 - \dfrac3{y+1} = -\dfrac12 e^{3x}

\dfrac3{y+1} = 1 + \dfrac12 e^{3x}

\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}

y+1 = \dfrac6{2+e^{3x}}

y = \dfrac6{2+e^{3x}} - 1

\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}

7 0
2 years ago
Need urgent help in mathematics!! Will mark brainliest!!!!
sesenic [268]

Hey there!!

How do we solve this problem :

We will use the combinations formula to solve this :

c ( n , r ) where n = 11 and r = 2

c ( n , r ) =  n ! / r ! ( n - r ) !

... 11 ! / 2 ! ( 11 - 2 ) !

... 11! / 2! × 9!

... 11! / 2 × 9!

... 11×10×9×8×7×6×5×4×3×2 / 2×9×8×7×6×5×4×3×2

... 11×10 / 2

... 11 × 5

... 55 combinations.

Hence, the required answer = 55 , option ( d )

Hope my answer helps!

8 0
3 years ago
the figure shown consists of a triangle resting on a square of side 5 cm. the triangle has a height of 4 cm. what is the total a
aniked [119]
Because its a square all its sides are 5cm
So area of square is 5x5=25cm^2
Triangle is 5x4= 20
20/2 is 10cm^2
25+10=35cm^2
4 0
2 years ago
The local library had a sale to get rid of the books that were slightly damaged it’s on paper back books for two dollars and har
kondor19780726 [428]

Answer:

The Local Library sold <u>31</u> hardcover books.

Step-by-step explanation:

Let number of paper back books be 'x'.

Let number of hardcover books be 'y'.

Given:

Total number of books sold = 89

Now Total number of books sold is equal to sum of number of paper back books and number of hardcover books.

framing in equation form we get;

x+y=89 \ \ \ \ \ equation \ 1

Now Given:

Cost of each paper back books = $2

Cost of each hardcover books = $5

Total Amount raised by library = $271

Now We know that Total Amount raised by library is equal to Cost of each paper back books multiplied by number of paper back books plus Cost of each hardcover books multiplied number of hardcover books.

framing in equation form we get;

2x+5y=271 \ \ \ \ equation \ 2

Now Multiplying equation 1 by 2 we get;

2(x+y)=89\times2\\\\2x+2y = 178 \ \ \ \ equation \ 3

Now Subtracting equation 3 from equation 2 we get;

(2x+5y)-(2x+2y)=271-178\\\\2x+5y-2x-2y=93\\\\3y = 93\\\\y=\frac{93}{3}=31

Substituting the value of y in equation 1 we get;

x+y =89\\\\x+31=89\\\\x =89-31 = 58

Hence The Local Library sold 58 paper back books and <u>31</u> hardcover books.

8 0
3 years ago
Pls help im an idiot and cant solve this first gets brainliest
Pani-rosa [81]

Answer:

1. 6 x 7/2 = 21

2. 6 x 10/3 = 20

3. 6 x 25/6 = 25

Step-by-step explanation:

7 0
3 years ago
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