The length of a median is equal to half the square root of the difference of twice the sum of the squares of the two sides of the triangle that include the vertex the mediam is drawn from and the square of the side of the triangle the median is drawn to.
triangle sides by a, b, c.
ma=122c2+2b2−a2
mb=122c2+2a2−b2
mc=122a2+2b2−c2
140
25+105+90=220
360-220=140
Answer:
(c) y < x^2 -5x
Step-by-step explanation:
A quadratic inequality is one that involves a quadratic polynomial.
<h3>Identification</h3>
The degree of a polynomial is the value of the largest exponent of the variable. When the degree of a polynomial is 2, we call it a <em>quadratic</em>.
For the following inequalities, the degree of the polynomial in x is shown:
- y < 2x +7 . . . degree 1
- y < x^3 +x^2 . . . degree 3
- y < x^2 -5x . . . degree 2 (quadratic)
<h3>Application</h3>
We see that the degree of the polynomial in x is 2 in ...
y < x^2 -5x
so that is the quadratic inequality you're looking for.
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<em>Additional comment</em>
When a term involves only one variable, its degree is the exponent of that variable: 5x^3 has degree 3. When a term involves more than one variable, the degree of the term is the sum of the exponents of the variables: 8x^4y3 has degree 4+3=7.
The factored form of the equation is f(x) = (x - 2)(x + 10), which makes the zeros of the function x = -10 and x = 2.
In order to factor a quadratic like this, you must find factors of the constant (in this case -20). The pairs of factors are listed below.
1 and -20
-1 and 20
2 and -10
-2 and 10
4 and -5
-4 and 5
Now we must pick out the pair that add to the coefficient of x.
1 and -20
-1 and 20
2 and -10
-2 and 10
4 and -5
-4 and 5
Once you've picked out those numbers, you can place each in a parenthesis with x.
f(x) = (x - 2)(x + 10)
Then to find the zeros to the equation, set each parenthesis equal to 0 and solve.
x - 2 = 0
x = 2
x + 10 = 0
x = -10
-3i+3/4+2i-9/3-3i=(3/4-9/3) +(-3i+2i-3i)=(3/4-3)+(-4i)=(3/4-12/4)-4i=
=-9/4-4i
