This is the concept of algebra, we are required to solve the given expression;
32^(2c)=8^(c+7)
this can be written as:
2^[5(2c)]=2^[3(c+7)
applying the natural logs we shall get
[5(2c)]ln2=[3(c+7)]ln2
The ln 2 will cancel on from both sides and we shall have:
5(2c)=3(c+7)
10c=3c+21
To make c the subject we begin by subtracting 3c from both sides
10c-3c=3c+21-3c
7c=21
dividing both sides by 7 we get:
(7c)/7=21/7
c=3
The answer is c=3

The Correct option is ~ B


Let's find the slope (m) using points (2 , 1) and (0 , -3)
hence, slope = 2
now, by Observing the given graph we can infer that the given line cuts the y - axis at point (0 , -3), so value of y - intercept (c) = - 3
[ y - coordinate of a point when x - coordinate is equal to 0 is the value of y - intercept of a line ]
And, we know the general equation of line in slope - intercept form is ~
now, let's plug the value of slope (m) and y - intercept (c) in the general equation to find the equation of line in slope intercept form ~
Not completely sure of what this question is asking, but the orientation is the same. The actual triangle moved diagonally.
1.Drawn a straight line AB=7 cm with the help of ruler.
2.With the help of compass drawn an arc from A and at the point where it cuts AB from that point made another arc drawn an arc cutting the previous arc.
3.From A drawn a straight line joining the arc and extend it to M.
4.With the help of ruler measured 5 cm and mark it as AC.
5.Joined BC and we get the required triangle.
6.From C drawn an arc and make it cut on AC and BC and from the point it cuts AC and BC drawn arc cutting each other and extend a line from point C extend a line to the point point of intersection of two arc.
7.Similarly we do for A and the point where the two line intersect denoted as O.
8.Made a perpendicular from O on AB this perpendicular will be radius and taking O as centre we draw a circle this is our incircle.
9.And AN is our locus of points equidistant from two lines AB and AC.
We need to construct a circle inscribed in triangle that is incircle it can be done by making angle bisector of two sides the point where it intersect will be incentre. The centre of required circle.
The angle bisector is the locus where points are equidistant from two sides.
Answer:
-27/7
Step-by-step explanation:
Replace x with 19,
f(19)=3/(19+2) - root(19-3)
=3/21 - root(16)
=1/7 - 4
=1/7 - 28/7
= -27/7