Thread: some Easy Questions

Hello , Good Morning !

I have couples of Exercices where i struggled to solve them

solve the inequaliti and show the solution sets on the real Line
1. x-3 / 2 >= -4+x / 3

My resolution is x-1 /6 >= 0

x-1>= 0 --> x>=1 / Thats All ?

2.
I don't know what to do with the absolute value and solve this

3.write the Equation through (-1 ,2) and perpendicular to 1/2 x + 1/3 y = 1
My solution is y = -2x right ?

4. cos(-x) = -cosx ? or cos x?

5.y = -2+radical (1-x) i need to find the domain and range
for me domain is ]- infinity ,1] how to find the Range ?? do i need to skatch it ?

6.y= ln (x-3) +1 how to find the domain and Range ?

7.prove : 1-cos x / sin x = sin x / 1+cosx

b- 1-cosx / 1+cos x = tan^2 (x/2)

tks my friends help me , i did 70% of the homework and 30% i didn't know how so i need ur help to complete

Originally Posted by iceman1

x-3 / 2 >= -4+x / 3

My [solution] is ...x>=1

Hello iceman1:

Yes, your solution is correct.

In the future, instead of putting spaces around the slashes to denote the numerator and denominator, please use grouping symbols.

(x - 3)/2 >= (-4 + x)/3

Also, putting spaces around symbols like = + - > < helps improve readability.

Cheers,

~ Mark

Originally Posted by iceman1

I don't know what to do with the absolute value ...

Hi iceman1:

Remember that absolute value may be thought of as the distance from zero on the real number line.

|x| < 4

This means that the number x is less than four units away from zero on the real number line.

But wait. There are two different directions away from zero. X can be less than four units away from zero on either the positive side of the real number line or on the negative side. In other words, x must be inbetween -4 and 4.

Therefore, to get rid of the absolute value symbols, we write the following.

-4 < x < 4

You can get rid of the absolute value symbols on your exercise (2) in the same way.

-5 <= (2x + 7)/3 <= 5

Can you finish it now?

Cheers,

~ Mark

Thank Mark

Now i have to take each one separetly ? -5<= 2x+7/3 and solve it ? and then take 2x+7/3 <= 5 ?

Originally Posted by iceman1

3. Write the equation [of the line] through (-1 ,2) and perpendicular to 1/2 x + 1/3 y = 1

My solution is y = -2x right ?

Hi iceman1:

I get a different equation.

When you found the slope of the line from the given equation, did you get m = -3/2 ?

The perpendicular slope of this is m = 2/3.

Use the point-slope form of the line with slope 2/3 and point (-1,2), and show us what you get.

Cheers,

~ Mark

Originally Posted by iceman1

Now i have to take each one separetly ? -5<= 2x+7/3 and solve it ? and then take 2x+7/3 <= 5 ?

You do not have to do them separately.

Take the inequality that I wrote, and multiply through by 3 to get rid of the fraction.

Whatever steps you carry out to isolate x, you must do to all three parts of the inequality.

Cheers,

~ Mark

Originally Posted by mmm4444bot

Hi iceman1:

I get a different equation.

When you found the slope of the line from the given equation, did you get m = -3/2 ?

The perpendicular slope of this is m = 2/3.

Use the point-slope form of the line with slope 2/3 and point (-1,2), and show us what you get.

Cheers,

~ Mark

why the slope is m = 3/2 ? i thought its -2 ? coz perpendicular means the result must be -1 , so (1/2 )* (-2/1 )= -1

Originally Posted by mmm4444bot

You do not have to do them separately.

Take the inequality that I wrote, and multiply through by 3 to get rid of the fraction.

Whatever steps you carry out to isolate x, you must do to all three parts of the inequality.

Cheers,

~ Mark

Great so I'll have -15 <= 2x+7 <= 15 then ? sorry man but iam very weak in maths and i try to exercice so im sorry if you are angry if iam asking alot sorry my friend

Originally Posted by iceman1

4. cos(-x) = -cosx ? or cos x?

The cosine function is an even function.

Graphically speaking, an even function has symmetry about the vertical axis.

y = cos(x)

You should know that the graph of this cosine function crosses the y-axis at the point (0,1).

Because the graph to the left of the y-axis is a reflection of the graph to the right of the y-axis, the value of y will be the same whether x is positive or negative.

In other words, the following is true.

y = cos(0.1) = cos(-0.1)

y = cos(0.2) = cos(-0.2)

y = cos(0.3) = cos(-0.3)

It does not matter how many units along the x-axis that you travel. The value of y will be the same regardless of traveling the same number of units away from zero to the right or to the left.

Even functions have the following property.

f(x) = f(-x)

In other words, changing the sign on the input to an even function has no effect on the output. Both outputs are the same number.

Cheers,

~ Mark

Originally Posted by iceman1

why the slope is m = 3/2 ? i thought its -2 ? coz perpendicular means the result must be -1 , so (1/2 )* (-2/1 )= -1

You are given the following equation.

1/2 x + 1/3 y = 1

We cannot read off the slope of this line as m = 1/2 because this equation is not in slope-intercept form.

You need to put the given equation into the form y = m*x + b.

In other words, solve the given equation for y. Then you can read off the slope.

Cheers,

~ Mark

Originally Posted by iceman1

Great so I'll have -15 <= 2x+7 <= 15 then ?

You are trying to isolate the x in the middle of the inequality. The steps are the same as if you were solving an equation.

In order to isolate the x, you need to subtract 7, followed by dividing by 2.

Cheers,

~ Mark

PS: I do not get mad because nothing matters to me.

Originally Posted by mmm4444bot

You are given the following equation.

1/2 x + 1/3 y = 1

We cannot read off the slope of this line as m = 1/2 because this equation is not in slope-intercept form.

You need to put the given equation into the form y = m*x + b.

In other words, solve the given equation for y. Then you can read off the slope.

Cheers,

~ Mark

oh thank you i solve it and got y= 3/2 x so the slope m must be m =-2/3

yks u so much my friend

Originally Posted by iceman1

i solve it and got y= 3/2 x so the slope m must be m =-2/3

Double check your work, iceman1.

I got y = -(3/2)x + 3.

Cheers,

~ Mark

Originally Posted by mmm4444bot

Double check your work, iceman1.

I got y = -(3/2)x + 3.

Cheers,

~ Mark

i got y = -3/2 x + 3 not -3 Thank you so far please can you help me also with the other questions ?

Originally Posted by iceman1

5.y = -2+radical (1-x) i need to find the domain and range
for me domain is ]- infinity ,1] how to find the Range ?? do i need to skatch it ?

Hi iceman1:

I'm assuming that you typed "radical" to mean the square-root symbol. We can type square roots as follows.

y = -2 + sqrt(1 - x)

Your domain is correct, except for the square bracket next to -infinity.

When we write interval notation, a square bracket indicates that the endpoint of the interval is included.

Infinity is not a number; therefore, it can never be part of the interval.

Always use a parenthesis next to infinity or negative infinity.

Domain: (-infinity, 1]

For the range, you could draw the graph to see it. However, are you familiar with the following graph?

y = sqrt(x)

If you know the range on this one, then you can figure out the range in your exercise in your head.

It's the same, but it starts at -2 instead of zero because the addition of -2 in the function definition shifts the graph down two units.

Cheers,

~ Mark

MY EDITS: fixed the vertical shift from up to down, and the numbers 2 to -2, because I missed the negative sign originally.