All categories

3 years ago

Roy weighs 150 pounds and plans to lose 1.5 pounds a week. Luis weighs 200 pounds and plans to lose 2.5 pounds a week.

Mathematics

2 answers:

solniwko [45]3 years ago

4 0

Answer: when Luis is on his his 10th week of lossing weigh and when roy is on his 9th week they will weigh the same

Step-by-step

Elden [556K]3 years ago

4 0

Answer:

150-1.5x = 200-2.5x

Step-by-step explanation:

Read more about inequality at

brainly.com/question/24372553

#SPJ1

7 0

2 years ago

Can some one please help<br><br> Simplify 4(11-6)+2

Nata [24]

Answer: 22

Step-by-step explanation:

11 - 6 = 5

4 x 5 = 20

20 + 2 = 22

' pls can i have brainliest? i need it for my goal'

4 0

2 years ago

Read 2 more answers

Help me please!!!!! its so hard

Murljashka [212]

Answer: 384 in

Step-by-step explanation: 12 in*10 in=384 in

6 0

3 years ago

Read 2 more answers

13 copies of the small rectangle shown below have been used to make the larger rectangle.

Nikolay [14]

Answer:

4800

Step-by-step explanation:

length = 20mm X 3 = 60mm

width = 20mm X 4 = 80mm

area = length X width = 60 X 80 = 4800 m $m^{2}$

I hope it helps you!

8 0

2 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