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valkas [14]
3 years ago
15

Yvonne is a salesperson who earns a fixed amount of $1,850 per month. She also earns a commission of 4% on the amount of goods t

hat she sells. If she wants to earn more than $2,300 in one month, how many dollars (x) in goods must she sell?
A. x > 4,150
B. x > 11,250
C. x > 10,050
D. x < 11,250
Mathematics
2 answers:
Blizzard [7]3 years ago
7 0
Let the amount of dollars in goods sold be x, then her earning is 1850 + (4/100)x = 1850 + 0.04x.
For her to earn more than $2300, we have:
1850 + 0.04x > 2300
0.04x > 2300 - 1850
0.04x > 450
x > 450 / 0.04
x > $11,250
Slav-nsk [51]3 years ago
3 0

Let's see the answer is d which is x<11,250


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<h2>¡Hola! </h2>

<h2>Explicación paso a paso Espero que te ayude</h2>

2x(x - 1) \\  \\  {2x}^{2}  - 2x

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kozerog [31]

Answer:

The possible values of x is given by x\leq  -2 and they include: -2, -3, -4, -5, -6 and so on

The greatest possible value of 7x is -14

Step-by-step explanation:

We are given the equation: x + 8 <= 6

Let collect the like terms:

x + 8 \leq  6\\x\leq  6 -8\\x \leq  -2

Therefore, x <= -2 and the possible values include: -2, -3, -4, -5, -6 and so on

To find the greatest possible value of 7x is when x = -2.

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6 0
3 years ago
<img src="https://tex.z-dn.net/?f=prove%20that%5C%20%20%5Ctextless%20%5C%20br%20%2F%5C%20%20%5Ctextgreater%20%5C%20%5Cfrac%20%7B
inysia [295]

\large \bigstar \frak{ } \large\underline{\sf{Solution-}}

Consider, LHS

\begin{gathered}\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } \\ \end{gathered}

We know,

\begin{gathered}\boxed{\sf{  \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }} \\ \end{gathered}  \\  \\  \text{So, using this identity, we get} \\  \\ \begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - ( {sec}^{2}\theta - {tan}^{2}\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}

We know,

\begin{gathered}\boxed{\sf{  \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: \: }} \\ \end{gathered}  \\

So, using this identity, we get

\begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - (sec\theta + tan\theta )(sec\theta - tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}

can be rewritten as

\begin{gathered}\rm\:=\:\dfrac {(\sec \theta + tan\theta ) - (sec\theta + tan\theta )(sec\theta -tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered} \\  \\  \\\begin{gathered}\rm \: = \:\dfrac {(\sec \theta + tan\theta ) \: \cancel{(1 - sec\theta + tan\theta )}} { \cancel{ \tan \theta - \sec \theta + 1} } \\ \end{gathered} \\  \\  \\\begin{gathered}\rm \: = \:sec\theta + tan\theta \\\end{gathered} \\  \\  \\\begin{gathered}\rm \: = \:\dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \\ \end{gathered} \\  \\  \\\begin{gathered}\rm \: = \:\dfrac{1 + sin\theta }{cos\theta } \\ \end{gathered}

<h2>Hence,</h2>

\begin{gathered} \\ \rm\implies \:\boxed{\sf{  \:\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \:\dfrac{1 + sin\theta }{cos\theta } \: \: }} \\ \\ \end{gathered}

\rule{190pt}{2pt}

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