Answer:
Given that,
The number of grams A of a certain radioactive substance present at time, in years
from the present, t is given by the formula

a) To find the initial amount of this substance
At t=0, we get


We know that e^0=1 ( anything to the power zero is 1)
we get,

The initial amount of the substance is 45 grams
b)To find thehalf-life of this substance
To find t when the substance becames half the amount.
A=45/2
Substitute this we get,


Taking natural logarithm on both sides we get,







Half-life of this substance is 154.02
c) To find the amount of substance will be present around in 2500 years
Put t=2500
we get,




The amount of substance will be present around in 2500 years is 0.000585 grams
The total of can counts - 397
The total of cannot counts - 110
Total number of boys - 263
Total number of girls - 244
The FULL total - 507
6 units
12 is also a diameter because it goes through the whole circle.
Divide it by 2 to find 6.
Answer:
(3,-2)
Step-by-step explanation:
Given equations of line
3x-2y=13
2y+x+1=0
=> x = -1 -2y
Point of intersection will coordinates where both equation have same value of (x,y)
top get that we have to solve the both equations by using method of substitution of simultaneous equation.
using this value of x in 3x-2y=13, we have
3(-1-2y) -2y = 13
=> -3 -6y-2y = 13
=> -8y = 13+3 = 16
=> y = 16/-8 = -2
x = -1 - 2y = -1 -2(-2) = -1+4= 3
Thus, point of intersection of line is (3,-2)
Answer:
graph{3x+5 [-10, 10, -5, 5]}
x
intercept:
x
=
−
5
3
y
intercept:
y
=
5
Explanation:
For a linear graph, the quickest way to sketch the function is to determine the
x
and
y
intercepts and draw a line between the two: this line is our graph.
Let's calculate the
y
intercept first:
With any function,
y
intercepts where
x
=
0
.
Therefore, substituting
x
=
0
into the equation, we get:
y
=
3
⋅
0
+
5
y
=
5
Therefore, the
y
intercept cuts through the point (0,5)
Let's calculate the
x
intercept next:
Recall that with any function:
y
intercepts where
x
=
0
.
The opposite is also true: with any function
x
intercepts where
y
=
0
.
If we substitute
y
=
0
, we get:
0
=
3
x
+
5
Let's now rearrange and solve for
x
to calculate the
x
intercept.
−
5
=
3
x
−
5
3
=
x
Therefore, the
x
intercept cuts through the point
(
−
5
3
,
0
)
.
Now we have both the
x
and
y
intercepts, all we have to do is essentially plot both intercepts on a set of axis and draw a line between them
The graph of the function
y
=
3
x
+
5
:
graph{3x+5 [-10, 10, -5, 5]}