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3 years ago

Plz help! Due tonight!

Mathematics

2 answers:

Alecsey [184]3 years ago

7 0

the answer will be

48$ dollars

quester [9]3 years ago

7 0

Answer:

20 + 6x

$44

Step-by-step explanation:

So 5 CD's for 20 dollars is $4 ea. then each one added will be +6 with the variable (x).

so you can't have the first 5 becuase x is used for every one OVER 5.

$ or price = 20 + 6x

plug 4 in because 9-5 = 4

20 + 6(4)

20 + 24

$44

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Lucy plans to spend between $50 and $60, inclusive, on packages of beads and packages of charms. If she buys 5 packages of beads

worty [1.4K]

She can buy maximum of $5$ packages while staying within her budget.

Minimum spending limit $=$ $ $50$ .

Maximum spending limit $=$ $ $60$ .

Cost of one package of beads $=$ $ $4.95$ .

Number of packages of beads bought $= 5$ .

Cost of one package of charms $=$ $ $6.55$ .

Let the number of packages of beads bought be $x$ .

$4.956 \times 5 + x \times 6.55 \geq 50$

$24.78+6.55x\geq 50$

$6.55x\geq 25.22$

$x\geq 3.85$

$4.956 \times 5 + x \times 6.55 \leq 60$

$24.78+6.55x\leq 60$

$6.55x\leq 35.22$

$x\leq 5.38$

So, she can buy maximum of $5$ packages while staying within her budget.

Learn more about solving inequalities here:

brainly.com/question/12189350?referrer=searchResults

7 0

2 years ago

Whats the answer??????

Temka [501]

4/1 x 3/2 = 12/2
6/9 x 1/7 = 6/63
5/6 x 2/1 = 10/6
8/3 x 1/6 = 8/18
8/1 x 5/8 = 40/8

7 0

3 years ago

Read 2 more answers

<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>

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