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anyanavicka [17]

3 years ago

12

3 part question. If answer all(brainilest will be given.

1 answer:

docker41 [41]3 years ago

8 0

Answer:

1. 1080

2. It shows the initial debt, i.e., $1080

3.24 months

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Consider the multivariable function f\left(x,y,z\right)=\frac{x^3-2.5y^2+\frac{z}{2}}{z\left(x-y\right)}f ( x , y , z ) = x 3 −

Lelu [443]

Answer:

$f\left(2,3,1\right)=\bold{14}$

Step-by-step explanation:

Given the function:

$f\left(x,y,z\right)=\frac{x^3-2.5y^2+\frac{z}{2}}{z\left(x-y\right)}$

To find:

The value of $f(2, 3, 1)$ = ?

Solution:

$f(2, 3, 1)$ means the values of $x, y\ and\ z$ as:

$x=2\\y=3\\z=1$

Let us put the given values in the given function and let us solve for it:

$\Rightarrow f\left(2,3,1\right)=\dfrac{2^3-2.5\times 3^2+\frac{1}{2}}{1\left(2-3\right)}\\\\\Rightarrow f\left(2,3,1\right)=\dfrac{8-2.5\times 9+0.5}{1\left(-1\right)}\\\\\Rightarrow f\left(2,3,1\right)=\dfrac{8-22.5+0.5}{-1}\\\\\Rightarrow f\left(2,3,1\right)=\dfrac{-14}{-1}\\\\\Rightarrow f\left(2,3,1\right)=\bold{14}$

Therefore, the answer is:

$f\left(2,3,1\right)=\bold{14}$

8 0

2 years ago

The average weight of a 7-year old girl is 49 pounds. The average weight of an 11 year old girl is 77 pounds.

Aneli [31]

I only know D. but I'll try to figure out the rest too. D. 9 years old

8 0

3 years ago

Find an equation in the form y=ax2+bx+c for the parabola passing through the points.(−3,61), (4,96), (2,16)

puteri [66]

Answer:

9657

Step-by-step explanation:

3 0

3 years ago

Find the function of which is the solution of 36y"-48y'-48y=0 with initial conditions y1(0)=1. y1'(0)=0

lilavasa [31]

Answer:

$y=\frac{1}{4} e^{2x}+\frac{3}{4} e^{-\frac{2}{3}x }$

Step-by-step explanation:

Let, $D=\frac{d}{dx}$

Now, us simplify the given differential equation and write it in terms of D,

$36y''-48y'-48y=0$

or, $3y''-4y'-4y=0$

or, $(3D^2-4D-4)y=0$

We have our auxiliary equation:

$3D^2-4D-4=0$

or, $(3D+2)(D-2)=0$

or, $D=2, - \frac{2}{3}$

Therefore our solution is,

$y=Ae^{2x}+Be^{- \frac{2}{3}x}$

and, $y'=2Ae^{2x}-\frac{2}{3}Be^{\frac{2}{3}x }$

Applying the boundary conditions, we get,

$A+B=1$

$2A-\frac{2}{3}B=0$

Solving them gives us,

$A=\frac{1}{4} ,B=\frac{3}{4}$

Hence,

$y=\frac{1}{4} e^{2x}+\frac{3}{4} e^{-\frac{2}{3}x }$

4 0

3 years ago

ale4655 [162]

592,722 is the correct answer

3 0

3 years ago

Read 2 more answers