The lines have equal gradients;
m

Equation of line 2;
y=mx+c
y=

x+c
-15=

+c
c=-11
y=4/5x-11
Only the function in C has the range of all real numbers.
Later on, when you will be familiar with more type of functions you will know that when x is on the power range is usually only positive values or negative values if only not shifted, even if shifted it will only add to its range this numbers by which unit it was shifted.
B is shifted parabola. It has the range of {y| -3≤∞}
Answer:
-11
Step-by-step explanation:
Answer:
let A (2, 6) be (x1, y1)
B (-5, 2) be (x2, y2)
distance AB =√[(x2 - x1)² + (y2 - y1)²]
√[(-7)²+ (-4)²]
√65
Answer:
hi your question options is not available but attached to the answer is a complete question with the question options that you seek answer to
Answer: v = 5v + 4u + 1.5sin(3t),
Step-by-step explanation:
u" - 5u' - 4u = 1.5sin(3t) where u'(1) = 2.5 u(1) = 1
v represents the "velocity function" i.e v = u'(t)
As v = u'(t)
<em>u' = v</em>
since <em>u' = v </em>
v' = u"
v' = 5u' + 4u + 1.5sin(3t) ( given that u" - 5u' - 4u = 1.5sin(3t) )
= 5v + 4u + 1.5sin(3t) ( noting that v = u' )
so v' = 5v + 4u + 1.5sin(3t)
d/dt
=
+
Given that u(1) = 1 and u'(1) = 2.5
since v = u'
v(1) = 2.5
note: the initial value for the vector valued function is given as
= ![\left[\begin{array}{ccc}1\\2.5\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%5C%5C2.5%5C%5C%5Cend%7Barray%7D%5Cright%5D)