If k(x) = x2 and p(x) = k(x) + n, what is the value of n?
1 answer:
You might be interested in
Answer:
2a: (c)
5o: (1, 3) and (1,1)
3a: (b)
1a: (d)
4o: (b)
Step-by-step explanation:
2a: the equation of a circle circumference needs to be transformable to the form where
is the center and <em>r</em> is the radius. (a) and (d) can’t be it because they contain non-zero factors on <em>xy</em>. (b) isn’t an equation.
5o: just put the given (<em>x</em>, <em>y</em>) into the equations and see if it holds. (2, 3) isn’t on the circumference of (1) because , (3, 1) isn’t on it either because
.
3a: calculate the value of the left-hand side term of the equation using (<em>x</em>, <em>y</em>) from the given point <em>M</em>. That’s the difference of square distance to the center to the square radius . Thus it’s 0 if the point is on the circumference, negative if inside and positive if outside. You get
, positive, so it’s outside the circle.
1a: see definition from 2a. Here, .
4o: insert the y from the straight line equation (r) (which can be equivalently transformed to ) into the circumference equation. If it yields no solution, that’s outside, it there’s exactly one solution, that’s a tangent and if there are two solutions, it’s a secant.
There are two solutions, so it’s a secant.
For the distance from nikos home to the library, the distance on the map should be
4/6th of an inch
4/6th of an inch
Since in Jimmy's 90 times die roll six appeared 11 times, so the probability of face of sixes appearing is: .
Thus , when the die is rolled 1500 times then it is obvious that the number of times the face of six will appear will also increase proportionately.
This proportionate increase in the number of times the face of six will appear will be given thus:
If six appears 11 times in 90 rolls then to find how many times it will appear in 1500 rolls is calculated as where x is the number of times the face of six will appear.
Thus, expression gives:
Therefore, Jimmy can six's approximately 183 times if he rolled the die 1500 times.