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what is 6,393,203 in word form

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Definition and examples of Word Form of a Number | Define ...

Complete information about Word Form of a Number, definition of an Word Form of a Number, examples of an Word Form of a Number, ...
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Writing a Number in Expanded Form - YouTube

Aug 29, 2012 · Wondering how to write a number in expanded form? Follow this short tutorial! Wondering how to write a number in expanded form? Follow this short tutorial!

6 007 200 in word form -

Six million seven thousand two hundred would be the way to write 6 007 200 in word form. Minor ...
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100 203 in word form -

100 203 in word form? SAVE CANCEL. already exists. Would you like to merge this question into it? MERGE CANCEL. already exists as an alternate of this ...
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Create a fillable form - Word -

To create a form in Word that others can fill out, start with a template and add content controls. Content controls include things like check boxes, ...
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How to write 100203 in word form? | Wikianswers | Fandom ...

How to write 100203 in word form. How to write 100203 in word form. FANDOM. Skip to Content Skip to Wiki Navigation Skip to Site Navigation. Games ...
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Standard, Expanded and Word Form -

Standard, Expanded and Word Form. This page contains links to free math worksheets for Standard, Expanded and Word Form problems. Click one of the buttons below to ...
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Read and write numbers in word form | LearnZillion

Read and write numbers in word form Created by: Ginny Baldwin Standards: Tags:
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how do you write 100,203 in word form -

how do you write 100,203 in word form - 98617. 1. Log in Join now Katie; a few seconds ago; Hi there! ... Please help me Numbers 6 and 8 Answer Mathematics;
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expanded form of 900

In word form: 9 hundreds, 0 tens, & 0 ones Number form: 900+ 0+ 0
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what is the standard form of x2+y2-4y+3=0

i recognise this as the equation of a circle, and this is also hinted at in the question because of the words "centre" and "radius". The coefficients of x and y are the same, being 1, which confirms a circle rather than an ellipse. What we have to do is to complete the square for y because we have a y term but no x term. So: x^2 + (y^2-4y+4) -4+3=0; x^2+(y-2)^2-1=0; x^2+(y-2)^2=1 is the standard form. On the right we have the radius squared so the radius is 1. The centre is (0,2), because, when x=0 y is 3 or 1, and when y=2, x=1 or -1. The centre lies in between these extremes: halfway between 1 and -1 is x=0, and halfway between 3 and 1 is y=2.
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using 0-9, how many 4 digit sets can be formed, all digits can be repeated, ie 2576, 9471,

If you mean digits can be repeated in the four digit number and all digits from 0 to 9 are available, then numbers can range from 0000 to 9999, that's 10,000 different numbers. If different digits must make up the numbers and permutations are significant, in other words, the order of the digits is important, then after picking the first digit we have 9 digits left to choose from; after picking the second digit we have 8 left to choose from, and so on. Therefore, the calculation is 10*9*8*7=5040 for 4-digit numbers.
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what is the least 10 digit whole number? in word form and standard form

Zero.  0000000000 = 0 Or 10 million.  10000000000 = 1 x 10^10
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Find the value of a. Could anyone smart teach me please?tks

Given ax+2y+8=0, 4x+(a+2)y-6=0 and y=0 The value of a=2 Which is non-zore and the doesn't make triangle Click for image Algebra Word Problems -
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Translating a word problem into a quadratic expression

speed = distance/time, so distance = speed * time s = speed, t = time First train 480 = st or s = 480/t Second train 480 = (s - 8)(t + 3) Plug the first equation into the second for s (I'm also going to switch sides of the second equation) (s - 8)(t + 3) = 480 (480/t -  8)(t + 3) = 480 t*480/t + 3*480/t - 8t - 8*3 = 480 480 +1440/t - 8t - 24 = 480 The 480's cancel on each side, so we have 1440/t - 8t - 24 = 0 Multiply the whole equation by t (I'm going to move stuff around on the left also) -8t² - 24t + 1440 = 0 All 3 terms are divisible by 8, so you can divide that out.  Dividing by -8 gets rid of negative in front of your t² term. t² + 3t - 180 = 0 To factor, what two numbers have a difference of 3 & a product of 180?  12 & 15, so (t + 15)(t - 12) = 0 For that to be true, t = -15 or 12. You can't have negative time, so the answer is 12 hours. speed = distance/time = 480 km/12 hr = 40 km/hr
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Determine whether the line and plane are parallel. x=-5-4t;y=1-t;z=3+2t; x + 2y +3z -9 =0

The parametric equation for a line passing through two points is p = p1 + t(p2 - p1) where t is in [0,1]. Since we have x = -5 - 4t, y = 1 - t, z = 3 + 2t, we get p1 = (-5, 1, 3), p2 = (-9, 0, 5) and p2 - p1 = (-4, -1, 2) = A where A is the direction vector of the line. Given the plane equation x + 2y + 3z -9 = 0 or x + 2y + 3z = 9, this is in the form X dot N = P dot N so that the normal to the plane is N = (1,2,3). The line and plane defined above are parallel if the dot product of N and A is equal to zero. In other words the normal to the plane and the line direction vector are perpendicular to each other. Taking the dot product, we get N dot A = 1*(-4) + 2*(-1) + 3*2 = -4 - 2 + 6 = 0 Therefore the line and plane defined above are parallel!
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How are the locations of vertical asymptotes and holes different, and what role do limits play?

To talk about asymptotes and holes, you need pictures. These pictures are graphs of functions. The simplest function containing vertical and horizontal asymptotes is y=1/x, where x is the horizontal axis and y the vertical axis. The vertical asymptote is in fact the y axis, because the graph has no values that would quite plot onto the y axis, although the curve for 1/x gets very, very close. The reason is that the y axis represents x=0, and you can't evaluate 1/x when x=0. You' d have to extend the y axis to infinity both positively and negatively. You can see this if you put a small positive or negative value for x into the function. If x=1/100 or 0.01, y becomes 100. If x=-1/100 or -0.01, y becomes -100. If the magnitude of x decreases further y increases further. That's the vertical asymptote. It represents the inachievable. What about the horizontal asymptote? The same graph has a horizontal asymptote. As x gets larger and larger in magnitude, positively or negatively, the fraction 1/x gets smaller and smaller. This means that the curve gets closer and closer to the x axis, but can never quite touch it. So, like the y axis, the axis extends to infinity at both ends. What does the graph look like? Take two pieces of thick wire that can be bent. The graph comes in two pieces. Bend each piece of wire into a right angle like an L. Because the wire is thick it won't bend into a sharp right angle but will form a curved angle. Bend the arms of the L out a bit more so that they diverge a little. Your two pieces of wire represent the curve(s) of the function. The two axes divide your paper into four squares. Put one wire into the top right square and the other into the bottom left and you get a picture of the graph, but make sure neither piece of wire actually touches either axis, because both axes are asymptotes. The horizontal axis represents the value of x needed to make y zero, the inverse function x=1/y. Hence the symmetry of the graph. Any function in which an expression involving a variable is in the denominator of a fraction potentially generates a vertical asymptote if that expression can ever be zero. If the same expression can become very large for large magnitude values of the variable, potentially we would have horizontal asymptotes. I use the word "potentially" because there's also the possibility of holes under special circumstances. Asymptotes and holes are both no-go zones, but holes represent singularities and they're different from asymptotes. Take the function y=x/x. It's a very trivial example but it should illustrate what a hole is. Like 1/x, we can't evaluate when x=0. However, you might think you can just say y=1 for all values of x, since x divides into x, cancelling out the fraction. That's a horizontal line passing through the y axis at y=1. Yes, it is such a line except where x=0. We mustn't forget the original function x/x. So where the line crosses the y axis there's a hole, a very tiny hole with no dimensions, a singularity. So a hole can occur when the numerator and denominator contain a common factor. If this common factor can be zero for a particular value of x, then a hole is inevitable. Effectively it's an example of the graphical result of dividing zero by zero. With functions we can't simply cancel common factors as we normally do in arithmetic. Asymptotes and holes are examples of limits. Asymptotes can show where functions converge to a particular value without ever reaching it. Asymptotes can be slanted, they don't have to be horizontal or vertical, and they can be displaced from both axes. Graphs can aid in the solution of mathematical and physics problems and can reveal where limitations and limits exist for complicated and complex functions. Knowing where the limits are by inspection of functions also aids in drawing the graph. This helps in problems where the student may be asked to draw a graph to show the key features without plotting it formally.
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The Rectangle Problem

There is some difficulty interpreting this question. The best interpretation I can offer is that, if the perimeter of two triangles is 144cm, what is the maximum difference in area between the two triangles, if the dimensions of the triangles are whole numbers? The maximum area of a triangle is when the triangle is equilateral so its area is a/2 * a√3/2=a^2√3/4, where a is the side length, which is 144/3=48cm. The area is 576√3=997.66 sq cm approx. The cosine formula can be used to relate the lengths of the sides: a^2=b^2+c^2-2bccosA. If all the sides are integers, angle A must have a rational cosine. The only angles to do so are 0, 60, 90, 180 degrees. 0 and 180 would not make a triangle, so we are left with 60 and 90. If A=60 we have a^2=b^2+c^2-bc and a+b+c=144cm. So (144-(b+c))^2=b^2+c^2-bc, from which we get: 144^2-288(b+c)+b^2+c^2+2bc=b^2+c^2-bc; and c=(6912-96b)/(96-b)=96(72-b)/(96-b). When 96-b is a factor of 96 we will get an integer result for c: so b=64, 48 are obvious solutions, giving us c=24, 48. That gives us a=56, 48. When b=60, c=96*12/36=32 and a=52. If A=90, 144^2-288(b+c)+2bc=0, c=144(72-b)/(144-b), and we get (a,b,c)=(65,63,16) and (60,48,36). So now we have all the possible triangles: (56,64,24), (48,48,48), (52,60,32), (65,63,16) and (60,48,36). We identified the equilateral triangle area, so we are left with 4. The one with the shortest side is right triangle (65,63,16) and this one has the smallest area: 8*63 (half base of 16 times height of 63)=504 sq cm. the difference between the greatest and smallest area is 576√3-504=72(8√3-7)=493.66 sq cm. FURTHER INTERPRETATION After presenting this possible solution, I pondered over the use of the word "rectangle". Perhaps it was a loose translation of "right angle" implying that the only triangles to be considered were right triangles. In my solution, two such triangles were identified: (60,48,36) and (65,63,16), with areas respectively 864 sq cm and 504 sq cm. the greater area is (60,48,36). The areas of the associated rectangles (the triangles are formed by splitting the rectangles into two using the diagonals) are 1728 sq cm and 1008 sq cm, so the difference in the areas of the two triangles is 360 sq cm.
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how do I write exponents expanded forms?

what is shot word form for 61420185
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