Is (4,-2) (5,-3), (6,-2), (7,-3) a function??? need answers quick pls(ᗒᗣᗕ)՞

Solve by both sides variable josh has two leaking pipes in his basement. while waiting for the plumber to come, josh puts a buck

aleksandrvk [35]

First off, you misspelled some words, second WHICH BUCKET does josh's brother take? That's important!

6 0

3 years ago

The side of the base of a square prism is decreasing at a rate of 7 kilometers per minute and the height of the prism is increas

Ber [7]

Answer:

dV = - 5.73*10⁹ m³/s

Step-by-step explanation:

Question: What is the rate of change of the volume of the prism at that instant (in cubic meters per second) ?

A function can be dependent on one or more variables. The change in the function due to a change in one o its variables is given by the functions derivative with respect to that variable. For functions that are composed of products of its variables, we may use the product rule to determine its derivative.

The volume of a square prism with base a and height h is given by

V = a²h

When the base and height are changing, we have

dV = 2ah(da/dt) + a²(dh/dt)

Given

a = 4 Km

h = 9 Km

da/dt = - 7 Km/min

dh/dt = 10 Km/min

we have

dV = 2(4 Km)(9 Km)(- 7 Km/min) + (4 Km)²(10 Km/min)

⇒ dV = - 504 Km³/min + 160 Km³/min = - 344 Km³/min

⇒ dV = - 5.73*10⁹ m³/s

8 0

3 years ago

Which any quality represents the statement the sum of -3 and a number is at least 5

Artyom0805 [142]

5 is less than or equal to -3+x

5 0

2 years ago

Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a

8_murik_8 [283]

Answer:

a) 3/5·((-2)^n + 4·3^n)
b) 3·2^n - 5^n
c) 3·2^n + 4^n
d) 4 - 3 n
e) 2 + 3·(-1)^n
f) (-3)^n·(3 - 2n)
g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

The quadratic formula tells us values of r that satisfy this are ...

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

_____

Using these formulas on the first recurrence relation, we get ...

a)

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

__

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

$a[n]=(p+qn)r^n$