Power point solving literal equations
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This document discusses solving literal equations, which involves taking an equation containing variables and solving it for a particular variable. Solving literal equations uses the same steps as solving other equations, such as isolating the targeted variable by performing inverse operations like division or multiplication on both sides of the equation. An example problem walks through solving the equation P=4s for the variable s by dividing both sides of the equation by 4. Practice problems are provided to solve equations for different variables.
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The document contains class notes on solving literal and other equations, including tips and examples. It also provides a schedule for Khan Academy work and math classwork for the day. Students are asked to submit outstanding forms, work on literal and other equations, and review topics on Khan Academy. The document gives the dimensions of two-dimensional and three-dimensional objects, the formula for volume of a rectangular prism, and examples of solving various equation types. Warm-up activities are provided for the math periods involving solving literal equations, rewriting equations using correct formulas, and proportion word problems.
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Measuring Area of Irregular ShapesGeneral formula
The general recipe for the surface territory of the chart of a persistently differentiable capacity where and is an area in the xy-plane with the smooth limit:
Considerably more general equation for the range of the diagram of a parametric surface in the vector structure where is a persistently differentiable vector capacity of: [1]
Illustration 1 Determine the surface region of the solid got by pivoting / rotating
,
about the x-axis.
Solution
The formula
since we are turning about the x-pivot and we'll utilize the first ds as a part of this case in light of the fact that our capacity is in the right structure for that ds and we won't increase anything by explaining it for x.
We should first get the subsidiary and the root dealt with.
Below is the integral; S.A,
There is an issue be that as it may. The dx implies that we shouldn't have any y's in the basic. So, before assessing the essential we'll have to substitute in for y also.
The S.A is then,
Area of a Leave
Leaves go about as sun oriented vitality gatherers for plants. Thus, their surface zone is a paramount property. In this segment we utilize our procedures to focus the zone of a rhododendron leaf, demonstrated in Figure 2.4. For straightforwardness of medication, we will first think about a capacity intended to copy the state of the leaf in a straightforward arrangement of units: we will scale removes by the length of the leaf, so its profile is held in the interim 0 ! x ! 1. We later ask how to change this medicine to portray correspondingly bended leaves of subjective length and width, and leaves that are less symmetric. As demonstrated in Figure 2.4, a basic parabola, of the structure y = f(x) = x(1 − x), gives a helpful close estimation to the top edge of the leaf. To watch that this is the situation, we watch that at x = 0 and x = 1, the bend converges the x hub. At 0 < x < 1, the bend is over the hub. Hence, the range between this bend and the x pivot, is one a large portion of the leaf zone.
We set up the calculation of approximating rectangular strips as some time recently, by subdividing the interim of enthusiasm into N rectangular strips. We can set up the count efficiently, as takes after:
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Length of interim = 1− 0 = 1
y
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In this figure we indicate how the territory of a leaf could be approximated by rectangular
strips.
The agent k'th rectangle is demonstrated shaded in Above Figure: Its territory is
Comments
The capacity in this sample might be composed as y = x − x2. For some piece of this outflow, we have seen a comparative figuring in Section 2.2. This case outlines a vital property of entireties, in particular the way that we can rework the terms into more straightforward statements that could be summed separately.
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Power point solving literal equations
- 2. Sometimes you have a formula and you need to solve for some variable other than the "standard" one. Example: Perimeter of a square P=4s It may be that you need to solve this equation for s, so you can plug in a perimeter and figure out the side length.
- 3. This process of solving a formula for a given variable is called "solving literal equations".
- 4. One of the dictionary definitions of "literal" is "related to or being comprised of letters“. Variables are sometimes referred to as literals.
- 5. So "solving literal equations" may just be another way of saying "taking an equation with lots of variables, and solving for one variable in particular.”
- 6. To solve literal equations, you do what you've done all along to solve equations, except that, due to all the variables, you won't necessarily be able to simplify your answers as much as you're used to doing.
- 7. Here's how "solving literal equations" works: Suppose you wanted to take the formula for the perimeter of a square and solve it for ‘s’ (or the side length) instead of using it to solve for perimeter. P=4s How can you get the ‘s’ on a side by itself?
- 8. P=4s Just as when you were solving linear equations, you want to isolate the variable. So, what do you have to do to get rid of the ‘4’?
- 9. P=4s That’s right, you have to divide by ‘4’. You also have to remember to divide both sides by 4.
- 10. This new formula allows us to use the perimeter formula to find the length of the sides of a square if we know the perimeter.
- 11. Let’s look at another example: 2Q-c = d Multiply both sides by 2. Subtract ‘c’ from each side.
- 12. As you can see, we sometimes must do more that one step in order to isolate the targeted variable. You just need to follow the same steps that you would use to solve any other ‘Multi-Step Equation’.
- 13. Work these on your paper. 1. d = rt for ‘r’ 2. P = 2l +2w for ‘w’ 3. for ‘t’
- 14. Check your answers. 1. d = rt for ‘r’ 2. P = 2l +2w for ‘w’ 3. for ‘t’