The area of the square increasing at 30 
The rate of change function is defined as the rate at which one quantity is changing with respect to another quantity. In simple terms, in the rate of change, the amount of change in one item is divided by the corresponding amount of change in another
Let the side of a square be x.
So,
Area of square = 
A =
Differentiating with respect to time we get,
dA/dt = 2x dx/dt
When area = 9 
the side becomes 3 cm
At x = 3cm and dx/dt = 5 cm/s (Given)
dA/dt = 2.3.5 = 30
Thus the area of the square increasing at 30
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the center is at the origin of a coordinate system and the foci are on the y-axis, then the foci are symmetric about the origin.
The hyperbola focus F1 is 46 feet above the vertex of the parabola and the hyperbola focus F2 is 6 ft above the parabola's vertex. Then the distance F1F2 is 46-6=40 ft.
In terms of hyperbola, F1F2=2c, c=20.
The vertex of the hyperba is 2 ft below focus F1, then in terms of hyperbola c-a=2 and a=c-2=18 ft.
Use formula c^2=a^2+b^2c
2
=a
2
+b
2
to find b:
\begin{gathered} (20)^2=(18)^2+b^2,\\ b^2=400-324=76 \end{gathered}
(20)
2
=(18)
2
+b
2
,
b
2
=400−324=76
.
The branches of hyperbola go in y-direction, so the equation of hyperbola is
\dfrac{y^2}{b^2}- \dfrac{x^2}{a^2}=1
b
2
y
2
−
a
2
x
2
=1 .
Substitute a and b:
\dfrac{y^2}{76}- \dfrac{x^2}{324}=1
76
y
2
−
324
x
2
=1 .
First, we convert the interest such that it is compounded annually. The formula would be:
ieff = (1 + i/m)^m - 1
where m = 4, since there are 4 quarters in a year
ieff = (1 + 0.025/4)^4 - 1
ieff = 0.0252
Then we use this for this equation:
F = P(1 + i)^n, where F is the future worth, P is the present worth and n is the number of years
F = $600(1 + 0.0252)^15
F = $871.53
Answer:
B.
Step-by-step explanation:
If you rearrange the equation in b, you get
y=5(x-3)+15
This means that your f(1) is equal to f(3), or 15, and for each additional month, it increases by 5, which is what the table shows.
Answer:
The answer is 31 pennies
Step-by-step explanation:
we have to find a number that gives a remainder of 1 when divided by both 2, 3 and 5.
The easiest way to do this is to list the factors of 5, add 1 to it, and test them until we find the one with a remainder of 1 by all three divisors. This is done as follows:
1) factor: 5 (+1) = 6
6 ÷ 2 = 3 remainder 0
6 ÷ 3 = 2 remainder 0
since there is a remainder of 0, when divided by 3, 6 is wrong
2) factor; 10 (+1) = 11
11 ÷ 3 = 3 remainder 2 ( 11 is not the correct answer
3) factor : 15 (+1) = 16
16 ÷ 2 = 8 remainder 0 ( 16 is not the correct answer
4) factor : 20(+1) = 21
21 ÷ 3 = 7 remainder 0 ( 21 is wrong)
5) factor : 25(+1) = 26
26 ÷ 2 = 13 remainder 0 ( 26 is wrong)
6) factor : 30(+1) = 31
31 ÷ 2 = 15 Remainder 1
31 ÷ 3 = 10 Remainder 1
31 ÷ 5 = 6 Remainder 1
since all three divisors give a remainder of 1, the correct answer is 31
Therefore you have 31 pwnnies
