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Remember that 'x' is just a box, waiting for something to be put into it. Don't let odd-looking problems scare you. For instance:

• Given f (x) = x² + 2x – 1, evaluate f (§).

Well, evaluating a function means plugging whatever they gave me in for the argument in the formula. This means that I have to plug this character '§' in for every instance of x. Here goes:

f (§) = (§)² + 2(§) – 1

This is definitely weird-looking, but it follows all the rules they gave me, so:

f (§) = §² + 2§ – 1

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The above example is kinda pointless, I'll grant you, but it clearly illustrates how the notation works. You plug the given value in for the given variable, and chug your way to the answer. Itrash 4 1 4 download free. Hence, these exercises are often referred to as 'plug-n-chug'. Your best bet is to try not to over-think them.

This brings us to to our next topic: evaluating functions at variable expressions.

• Given g(x) = 4 – x, evaluate at x = t.

To evaluate this function at x = t, I'll need to plug t into every instance of x in the formula for the function g.

g(3) = 4 – (t)

There's nothing more I can do with this, and I can't find a fully numerical value because I don't have a number to plug in for the t. So my answer is:

g(3) = 4 – t

• Given that f (x) = 3x² + 2x, find f (h).

Everywhere that my formula has an 'x', I now plug in an 'h'. I start with the formula they gave me:

If I want to be excruciatingly clear, I can start by writing the formula over again, this time with empty spaces where I'll be putting the new argument in place of the original variable:

f ( ) = 3( )² + 2( )

Now I'll fill in those blanks with the new argument:

There's nothing I can simplify, other than removing the extra parentheses. So my answer is:

f (h) = 3h² + 2h

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Not every 'evaluate at a variable expression' exercise is going to involve only variables. The variable expression can contain numbers, too.

• For h(w) = w² – 3, find h(2d + 1).

For every instance of the variable w, I'll need to plug in the expression 2d + 1. I'll use parentheses to make the replacements clear for my next step.

I need to multiply out the squared binomial next, and then simplify:

(2d + 1)² – 3

= (4d ² + 4d + 1) – 3

= 4d ² + 4d + 1 – 3

= 4d ² + 4d – 2

I've simplified as much as I can. My answer is:

Notice that, while doing the squaring halfway through my work above, I didn't ignore (and then maybe forget) the 'minus three' that was along for the ride. In whatever manner you do your work, make sure that you're doing it so you don't get so busy with one part of an exercise that you find yourself in danger of losing track of other parts.

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Not every 'evaluate at a variable expression' exercise is going to involve only variables. The variable expression can contain numbers, too.

• For h(w) = w² – 3, find h(2d + 1).

For every instance of the variable w, I'll need to plug in the expression 2d + 1. I'll use parentheses to make the replacements clear for my next step.

I need to multiply out the squared binomial next, and then simplify:

(2d + 1)² – 3

= (4d ² + 4d + 1) – 3

= 4d ² + 4d + 1 – 3

= 4d ² + 4d – 2

I've simplified as much as I can. My answer is:

Notice that, while doing the squaring halfway through my work above, I didn't ignore (and then maybe forget) the 'minus three' that was along for the ride. In whatever manner you do your work, make sure that you're doing it so you don't get so busy with one part of an exercise that you find yourself in danger of losing track of other parts.

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Itubedownloader 6 5 9 X 6 What Is It In Simplest Form

• Given that f (x) = 3x² + 2x, find f (x + h).

This one feels wrong, because it's asking me to plug something that involvesx in for the original x. But this evaluation works exactly like all the others; namely, everywhere that the original formula has an 'x', I will now plug in an 'x + h'.

f (x + h) = 3(x + h)² + 2(x + h)

= 3(x² + 2xh + h²) + 2x + 2h

= 3x² + 6xh + 3h² + 2x + 2h

If you're not sure how I got the stuff inside the parentheses (the set that the 3 was multiplied through), then you'll want to review how to simplify with parentheses and how to do polynomial multiplication.

• Given that f (x) = 3x² + 2x, find f (x + h) – f (x).

I should not try to do this all at once. Instead, I'll break this into smaller, more manageable pieces.

(I also note that this exercise uses the same function as the previous exercise, and one of the substitutions is the same, too. So I'm gonna cheat a bit and copy that exercise's result for f (x + h).)

This function difference is the original function subtracted from the result of the previous exercise, so:

f (x + h) – f (x)

= [3x² + 6xh + 3h² + 2x + 2h] – [3x² + 2x]

= 3x² + 6xh + 3h² + 2x + 2h – 3x² – 2x

= 3x² – 3x² + 6xh + 3h² + 2x – 2x+ 2h

= 6xh + 3h² + 2h

In the result above, notice that f (x + h) – f (x) does not equal f (x + h – x) = f (h). You cannot 'simplify' the different functions' arguments in this manner. Addition or subtraction of functions is not the same as addition or subtraction of the functions' arguments. Again, the parentheses in function notation do not indicate multiplication.

• Given that f (x) = 3x² + 2x, find [f (x + h) – f(x)] / h.

(This type of functional expression is called a 'difference quotient', and is actually something you will see again in calculus. I guess the reason this sort of exercise crops up so commonly in algebra is that they're trying to 'prep' you.

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Download mathematica 11 2 keygen for mac os x. (But, to be fair, it's not like anybody remembers these by the time they get to calculus, so it's really a lot of work for no real purpose, in my opinion. However, this type of problem is quite popular, so you should expect to need to know how to do it, and should expect to see one on the next test.)

My best course of action is to break this up into pieces. Helpfully, the previous two exercises already had me do the first two pieces, setting me up with the final expression for f (x + h) – f(x). All I really need to do here is the final division. My work looks like this:

[f (x + h) – f (x)] / h

= [6xh + 3h² + 2h] / h

= (h) [6x + 3h + 2] / h

= 6x + 3h + 2

When working on complicated exercises like the last example above, use caution and start by splitting the exercises into the smaller, simpler steps demonstrated across the last three exercises above, so that you can complete these problems successfully. It's not that these are 'hard' really, so much as being 'very prone to the making of silly mistakes'. Help yourself by taking it slow and doing one small chunk at a time.

You can use the Mathway widget below to practice evaluating functions at variable expressions. Try the entered exercise, or type in your own exercise. Then click the button and select 'Evaluate' to compare your answer to Mathway's.

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(Click on 'Tap to view steps' to be taken directly to the Mathway site for a paid upgrade.)

URL: https://www.purplemath.com/modules/fcnnot2.htm

Although named after the emerald, which often receives this rectangular cut with trimmed corners, the popular emerald cut design can be applied to many gemstones. It has a range of standard gem sizes for mass-produced jewelry settings. 'Emerald Cut Tourmaline,' 4.52 cts, slight color shift. © All That Glitters. Used with permission.

Choosing a Standard Setting or a Custom Setting

Mass-produced jewelry settings come in standard gem sizes. If a gem won'�t fit in a standard setting, you should have a setting custom made. This involves much more labor and, therefore, expense. When purchasing a gem or selecting a cut design, bear this in mind. Under some circumstances, the stone's value makes cutting for maximum yield worthwhile. You can then pay for a custom setting later. However, in other circumstances, the gem's value doesn't justify the cost of a custom setting.

Do Gems Need to Match Standard Settings Precisely?

No, a gem doesn't need to be the precise size of the jewelry setting. A skilled metalsmith can modify a standard setting to accommodate a variety of gem sizes. For example, any round gem can be put in a standard setting. For other shapes, metalsmiths must consider a variety of factors. As a general rule, gems under one carat can fall within 0.1 mm of the setting size. Over one carat, a 0.2 mm variance can usually be accommodated, sometimes more.

Estimating Weights for Gemstones

The carat weights given for standard gem sizes in the following chart apply to diamonds cut to ideal proportions. Typically, gem cutters cut colored gemstones with greater volume than diamonds. Thus, they'll weigh more than the sizes on the chart.

The density or specific gravity (SG) of the material will also affect the weight. For example, a 6.5 mm round diamond, sapphire, and opal, all cut in the same proportions, will all weigh different. The diamond, with an SG of 3.52, will weigh 1 carat. The sapphire, with an SG of 4, will weigh 1.14 carats. The opal, with an SG of 2.15, will only weigh 0.61 carats. (If you know your gemstone density values well, you can estimate the weight of a standard size gem by sight).

There are subtle variations to these values. For example, some charts list a 0.25-carat round diamond as measuring 4.1 mm. Just keep this in mind: don't take these figures too literally. Use this information as a starting point for estimating weights.

Standard Gem Sizes

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For more information on these gem cuts, consult our Guide to Gem Cutting Terms.