Answer:
1
Step-by-step explanation:
say you don't have a dog.. your parents get you one dog, and then get you another, you now have 2 :)
Answer:
y = 3
Step-by-step explanation:
We need to find the slope. We do so by choosing any two points and dividing the change in the y-coordinates (their difference) by the change in the x-coordinates (their difference). Let's just choose (1, 3) and (0, 3). The slope is: . So, the slope is 0.
We want to write a line in slope-intercept form, which is: y = mx + b, where m is the slope and b is the y-intercept (where the line crosses the y-axis). Here, the slope m = 0. Looking at the graph, we see that the y-intercept is (0, 3), so b = 3. Then, our line is: y = 0x + 3 ⇒ y = 3.
Another note is that this is a horizontal line. One thing to remember is that all horizontal lines have slopes of 0, so their function is simply y = k, where k is a constant through which the line cuts through.
Answer:
Roots are not real
Step-by-step explanation:
To prove : The roots of x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0 are real for all real values of k ?
Solution :
The roots are real when discriminant is greater than equal to zero.
i.e. b^2-4ac\geq 0b
2
−4ac≥0
The quadratic equation x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0
Here, a=1, b=1-k and c=k-3
Substitute the values,
We find the discriminant,
D=(1-k)^2-4(1)(k-3)D=(1−k)
2
−4(1)(k−3)
D=1+k^2-2k-4k+12D=1+k
2
−2k−4k+12
D=k^2-6k+13D=k
2
−6k+13
D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))
For roots to be real, D ≥ 0
But the roots are imaginary therefore the roots of the given equation are not real for any value of k.
Answer:
I don't know
<em>bcddghhffyyhfdcdssghttfxzzvbkmvfu</em>
Step-by-step explanation:
if we are taking in terms of domain.. is d set of number that we can put to the function with get division by zero .. squaroot of negative number,log of zero e.t.c
A) p(x)/m(x)= 1/squaroot(x) ÷ x²-4 which we know that x²-4 must not be equal to zero
because if is equal to zero we will have division by zero, which have contradicted the hypothesis of law of domain
x²-4 =! 0
(x-2)(x+2)=!0
x=!2 or x=!-2 because if is equal to that it's has contradicted d hypothesis
the answer is all real number except -2 and -
2
B)p(m(x))=p(x²-4) =1/squaroot of x²-4
note they are two things involve we must not get division by zero and root of negative number
we have all real number expect from -2 to 2
C)m(p(x))=m(1/root of (x))=(1/root of (x))²_4)
we have 1/x-4
so the answer is all real number expect zero
