So for this, we will be using synthetic division. To set it up, have the equation so that the divisor is -10 (since that is the solution of k + 10 = 0) and the dividend are the coefficients. Our equation will look as such:
<em>(Note that synthetic division can only be used when the divisor is a 1st degree binomial)</em>
- -10 | 1 + 2 - 82 - 28
- ---------------------------
Now firstly, drop the 1:
- -10 | 1 + 2 - 82 - 28
- ↓
- -------------------------
- 1
Next, you are going to multiply -10 and 1, and then combine the product with 2.
- -10 | 1 + 2 - 82 - 28
- ↓ - 10
- -------------------------
- 1 - 8
Next, multiply -10 and -8, then combine the product with -82:
- -10 | 1 + 2 - 82 - 28
- ↓ -10 + 80
- -------------------------
- 1 - 8 - 2
Next, multiply -10 and -2, then combine the product with -28:
- -10 | 1 + 2 - 82 - 28
- ↓ -10 + 80 + 20
- -------------------------
- 1 - 8 - 2 - 8
Now, since we know that the degree of the dividend is 3, this means that the degree of the quotient is 2. Using this, the first 3 terms are k^2, k, and the constant, or in this case k² - 8k - 2. Now what about the last coefficient -8? Well this is our remainder, and will be written as -8/(k + 10).
<u>Putting it together, the quotient is 
</u>
Answer = 16 square blocks.
There are 16 square blocks in the shape.
I hope this helps.
Answer: $ 4,515
Step-by-step explanation:
The cost of the car she wanted to buy = $35,550
Her savings = $20,700
That means she needs( $35,550 - $20,700) more in order for her to buy the car.
Amount borrowed from the finance company = $35,550 - $20,700
= $ 14,850
Since she now owes the financial company a total of $19,365, then the interest accrued by the loan she borrowed is given by:
$19,365 - $ 14,850
= $4,515
<span>
</span><span>Suppose you have a triangle with sides {6,7,8} — how do you find the height?This is a question some GMAT test takers ask. They know they would need the height to find the area, so they worry: how would I find that height. The short answer is:fuhgeddaboudit! Which height?First of all, the “height” of a triangle is its altitude. Any triangle has three altitudes, and therefore has three heights. You see, any side can be a base. From any one vertex, you can draw a line that is perpendicular to the opposite base — that’s the altitude to this base. Any triangle has three altitudes and three bases. You can use any one altitude-base pair to find the area of the triangle, via the formula A = (1/2)bh.
In each of those diagrams, the triangle ABC is the same. The green line is the altitude, the “height”, and the side with the red perpendicular square on it is the “base.” All three sides of the triangle get a turn. Finding a heightGiven the lengths of three sides of a triangle, the only way one would be able to find a height and the area from the sides alone would involve trigonometry, which is well beyond the scope of the GMAT. You are 100% NOT responsible for knowing how to perform these calculations. This is several levels of advanced stuff beyond the math you need to know. Don’t worry about that stuff.In practice, if the GMAT problem wants you to calculate the area of a triangle, they would have to give you the height. The only exception would be a right triangle — in a right triangle, if one of the legs is the base, the other leg is the altitude, the height, so it’s particularly easy to find the area of right triangles. Some “more than you need to know” caveatsIf you don’t want to know anything about this topic that you don’t absolutely need for the GMAT, skip this section!a. Technically, if you know the three sides of a triangle, you could find the area from something called Heron’s formula, but that’s also more than the GMAT will expect you to know. More than you needed to know!b. If one of the angles of the triangle is obtuse, then the altitudes to either base adjacent to this obtuse angle are outside of the triangle. Super-technically, an altitude is not a segment through a vertex perpendicular to the opposite base, but instead, a segment through a vertex perpendicular to the line containing the opposite base.In the diagram above, in triangle DEF, one of the three altitudes is DG, which goes from vertex D to the infinite straight line that contains side EF. That’s a technicality the GMAT will not test or expect you to know. Again, more than you needed to know!c. If the three sides of a triangle are all nice pretty positive integers, then in all likelihood, the actual mathematical value of the altitudes will be ugly decimals. Many GMAT prep sources and teachers in general will gloss over that, and for the purposes of easy problem-solving, give you a nice pretty positive integer for the altitude also. For example, the real value of the altitude from C to AB in the 6-7-8 triangle at the top is:Not only are you 100% NOT expected to know how to find that number, but also most GMAT practice question writers will spare you the ugly details and just tell you, for example, altitude = 5. That makes it very easy to calculate the area. Yes, technically, it’s a white lie, but one that spares the poor students a bunch of ugly decimal math with which they needn’t concern themselves. Actually, math teachers of all levels do this all the time — little white mathematical lies, to spare students details they don’t need to know.So far as I can tell, the folks who write the GMAT itself are sticklers for truth of all kinds, and do not even do this “simplify things for the student” kind of white lying. They are more likely to circumvent the entire issue, for example, by making all the relevant lengths variables or something like that. Yet again, more than you needed to know! What you need to knowYou need to know basic geometry. Yes, there is tons of math beyond this, and tons more you could know about triangles and their properties, but you are not responsible for any of that. You just need to know the basic geometry of triangles, including the formula A = (1/2)b*h. If the triangle is not a right triangle, you have absolute no responsibility for knowing how to find the height — it will always be given if you need it. Here’s a free practice question for you.</span>
Answer:
V = 63π / 200 m^3
Step-by-step explanation:
Given:
- The function y = f(x) is revolved around the x-axis over the interval [1,6] to form a spherical surface:
y = √(42*x - x^2)
- The surface is coated with paint with uniform layer thickness t = 1.5 mm
Find:
The volume of paint needed
Solution:
- Let f be a non-negative function with a continuous first derivative on the interval [1,6]. The Area of surface generated when y = f(x) is revolved around x-axis over the interval [1,6] is:
![S = 2*\pi \int\limits^a_b { [f(x)*\sqrt{1 + f'(x)^2} }] \, dx](https://tex.z-dn.net/?f=S%20%3D%202%2A%5Cpi%20%5Cint%5Climits%5Ea_b%20%7B%20%5Bf%28x%29%2A%5Csqrt%7B1%20%2B%20f%27%28x%29%5E2%7D%20%7D%5D%20%5C%2C%20dx)
- The derivative of the function f'(x) is as follows:
- The square of derivative of f(x) is:
- Now use the surface area formula:
![S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2} *\sqrt{1 + \frac{(21-x)^2}{42x-x^2 } }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+(21-x)^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+441-42x+x^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{441} }] \, dx\\S = 2*\pi \int\limits^6_1 { 21} \, dx\\\\S = 42*\pi \int\limits^6_1 { dx} \,\\\\S = 42*\pi [ 6 - 1 ]\\\\S = 42*5*\pi \\\\S = 210\pi](https://tex.z-dn.net/?f=S%20%3D%202%2A%5Cpi%20%5Cint%5Climits%5E6_1%20%7B%20%5B%5Csqrt%7B42x-x%5E2%7D%20%2A%5Csqrt%7B1%20%2B%20%5Cfrac%7B%2821-x%29%5E2%7D%7B42x-x%5E2%20%7D%20%7D%5D%20%5C%2C%20dx%5C%5C%5C%5CS%20%3D%202%2A%5Cpi%20%5Cint%5Climits%5E6_1%20%7B%20%5B%5Csqrt%7B42x-x%5E2%2B%2821-x%29%5E2%7D%20%7D%5D%20%5C%2C%20dx%5C%5C%5C%5CS%20%3D%202%2A%5Cpi%20%5Cint%5Climits%5E6_1%20%7B%20%5B%5Csqrt%7B42x-x%5E2%2B441-42x%2Bx%5E2%7D%20%7D%5D%20%5C%2C%20dx%5C%5C%5C%5CS%20%3D%202%2A%5Cpi%20%5Cint%5Climits%5E6_1%20%7B%20%5B%5Csqrt%7B441%7D%20%7D%5D%20%5C%2C%20dx%5C%5CS%20%3D%202%2A%5Cpi%20%5Cint%5Climits%5E6_1%20%7B%2021%7D%20%5C%2C%20dx%5C%5C%5C%5CS%20%3D%2042%2A%5Cpi%20%5Cint%5Climits%5E6_1%20%7B%20dx%7D%20%5C%2C%5C%5C%5C%5CS%20%3D%2042%2A%5Cpi%20%5B%206%20-%201%20%5D%5C%5C%5C%5CS%20%3D%2042%2A5%2A%5Cpi%20%5C%5C%5C%5CS%20%3D%20210%5Cpi)
- The Volume of the pain coating is:
V = S*t
V = 210*π*3/2000
V = 63π / 200 m^3
