Ask question

Ask question

All categories

3 years ago

15

Choose the system of equations which matches the following graph: a line includes points 0 comma 4 and 2 comma 3. A line include

s points 0 comma 3 and 4 comma 1

2 answers:

Lena [83]3 years ago

4 0

Answer:

17

Step-by-step explanation:

patriot [66]3 years ago

3 0

Step-by-step explanation:

1838584848484884727738383

You might be interested in

157/14 written as a Mixed number

Korvikt [17]

Answer:

11 3/14

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

A dilation changes the size AND the shape of a figure

dolphi86 [110]

False, it does not change the shape of the figure, it only changes the size

7 0

3 years ago

Choose the triangle that seems to be congruent to △AEC.<br><br><br><br> △BFC<br> △ABD<br> △EFD

Mashutka [201]

Answer. Triangle ABD

Explanation: There’s no thinking in this, because you can use your eyes to look for the answer. Not all questions will have this type of question. But they do seem pretty similar to me.

6 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

Dhhdhdhdhhddhdhdhdhdhdhi

wel

Answer:

dddasssssssssssssssssssssssssssssssssssssss

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers