I think it's -35. The steps to get that number is subtracting the 3 to the right side. Then you have -2 - 3 which equals to -5. Then you still have x/7= -5. Get rid of 7 from the x and do the same thing to the other side but you multiply 7 and -5. Last you your would be x= -35.

7 0

2 years ago

A dependent means t-test determines if the differences between scores on the pretest and posttest measures differ significantly

lara31 [8.8K]

A dependent means t-test determines if the differences between scores on the pretest and post-test measures differ significantly from Zero(0).

Benefits of pre-test measurements

This design is superior to the post-test-only control design because it adds pre-tests. Addition of pre-test: Improve your design ability to detect effects. You can investigate the impact of interventions on various sublevels of pretesting.

The design of the pre-test and post-test is similar to the in-subject experiment. In this experiment, each participant is first tested in control conditions and then in therapeutic conditions.

Learn more about pretesting. here: brainly.com/question/968894

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6 0

1 year ago

Which function has a domain of all real numbers except x=π/2 ±nπ?

boyakko [2]

<span>A. y=secx This problem deals with the various trig functions and is looking for those points where they are undefined. Since the only math operations involved is division, that will happen with the associated trig function attempts to divide by zero. So let's look at the functions that are a composite of sin and cos. sin and cos are defined for all real numbers and range in value from -1 to 1. sin is zero for all integral multiples of pi, and cos is zero for all integral multiples of pi plus pi over 2. So the functions that are undefined will be those that divide by cos. tan = sin/cos, which will be undefined for x = π/2 ±nπ cot = cos/sin, which will be undefined for x = ±nπ sec = 1/cos, which will be undefined for x = π/2 ±nπ csc = 1/sin, which will be undefined for x = ±nπ Now let's look at the options and pick the correct one. A. y=secx * There's a division by cos, so this is the correct choice. B. y=cosx * cos is defined over the entire domain, so this is a bad choice. C. y=1/sinx * The division is by sin, not cos. So this is a bad choice. D. y=cotx, * The division is by sin, not cos. So this is a bad choice.</span>

4 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