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Korolek [52]
3 years ago
12

The monthly cost of driving a car depends on the number of miles driven. Lynn found that in May it cost her $375 to drive 380 mi

and in June it cost her $455 to drive 700 mi. (a) Express the monthly cost C as a function of the distance driven d, assuming that a linear relationship gives a suitable model. C(d)
Mathematics
1 answer:
attashe74 [19]3 years ago
8 0

Answer:

idk

Step-by-step explanation:

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olga nikolaevna [1]

Answer:

24

Step-by-step explanation:

5 0
3 years ago
Write an equation in slope intercept form given the following m = 5/4 and passing through (- 4, - 3)
raketka [301]
Y = 5/4x + 2


y - (-3) = 5/4 ( x - (-4))

1. Distribute 5/4

y + 3 = 5/4x + 5

2. Subtract three to collect like terms

y = 5/4x + 2
8 0
3 years ago
Find the simple interest of $2300 invested at 7.5%for 10years
san4es73 [151]

Use the formula I = Prt. Plug $2300 into P.

Plug 7.5% (decimal form is 0.075) into r.

Plug 10 into t.

Your simple interest after 10 years is $1725

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}

5 0
2 years ago
Please help timed test
beks73 [17]

Answer:

e

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
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