Answers:
- Standard form =
![\frac{3}{5}m^5 + 7m^3 - 11m-9](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B5%7Dm%5E5%20%2B%207m%5E3%20-%2011m-9)
- Degree = 5
- Leading coefficient = 3/5
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Explanation:
The largest exponent term is what will go first. Then the term with the second largest goes next, and so on. We count the exponents down. The exponents will count down in this case as 5, 3, 1, 0
Note: -9 = -9*1 = -9m^0 and -11m = -11m^1
So that's why standard form is ![\frac{3}{5}m^5 + 7m^3 - 11m-9](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B5%7Dm%5E5%20%2B%207m%5E3%20-%2011m-9)
We can write this as (3/5)m^5 + 7m^3 - 11m - 9
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The leading term is the first term after getting everything in standard form. The coefficient of the leading term is the leading coefficient. So in this case, it would be 3/5.
The degree is the largest exponent. So the degree is 5. This is a quintic polynomial.
Answer:
3√2
Step-by-step explanation:
The congruent length of the triangle are the lengths with equal lengths. For an isosceles triangle, two of its sides are equal. In order to know the length of one of the congruent legs of JLK, we will find the distance between the adjacent point of the triangle.
Given triangle J L K has points J(2, 4), K(5, 1) andL (2, -2).
Using the formula to calculating distance between two points
D = √(x2-x1)²+(y2-y1)²
For side JK,
J(2, 4), K(5, 1)
JK = √(5-2)²+(1-4)²
JK = √3²+-3²
JK = √18
JK = 3√2
For side KL,
K(5,1), L(2, -2)
KL = √(2-5)²+(-2-1)²
KL = √-3²+-3²
KL = √18
KL = 3√2
For side JL
J(2, 4), L(2, -2)
JL= √(2-2)²+(-2-4)²
JL = √0²+-6²
JL = √36
JL = 6
SINCE JK = KL = 3√2, the length of one of the congruent legs is 3√2
Answer:
The equation of line is given by
![y=\frac{3}{2}x+5](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B3%7D%7B2%7Dx%2B5)
Step-by-step explanation:
We have been given the points (–2, 2) and (0, 5) and we have to find the equation of the line.
Let us first find the slope of the line using the formula
![m=\frac{y_2-y_1}{x_2-x_2}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7By_2-y_1%7D%7Bx_2-x_2%7D)
On substituting the values, we get
![m=\frac{5-2}{0+2}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B5-2%7D%7B0%2B2%7D)
![m=\frac{3}{2}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B3%7D%7B2%7D)
Now, using the point slope form, the equation of line is given by
![y-y_1=m(x-x_1)\\y-2=\frac{3}{2} (x+2)\\\\y-2=\frac{3}{2}x+3\\\\y=\frac{3}{2}x+5](https://tex.z-dn.net/?f=y-y_1%3Dm%28x-x_1%29%5C%5Cy-2%3D%5Cfrac%7B3%7D%7B2%7D%20%28x%2B2%29%5C%5C%5C%5Cy-2%3D%5Cfrac%7B3%7D%7B2%7Dx%2B3%5C%5C%5C%5Cy%3D%5Cfrac%7B3%7D%7B2%7Dx%2B5)
Therefore, the equation of line is given by
![y=\frac{3}{2}x+5](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B3%7D%7B2%7Dx%2B5)
2x²-3x-15=5
2x²-3x-15-5=0
2x²-3x-20=0
½(2x+5)(2x-8)
-2x=-5
x=-5/2
2x-8=0
2x=8
x=4
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