All categories

2 years ago

Can someone help me with this problem

Mathematics

1 answer:

IRISSAK [1]2 years ago

7 0

If your looking for the value of "x" it is 4.

Hope this helped!

You might be interested in

Carlos deposited $7,924 into a savings account 30 years ago. The account has an interest rate of 4.6% and the balance is current

exis [7]

Answer:

n = 1, this means the interest compounds ANNUALLY.

Step-by-step explanation:

Carlos deposited $7,924 into a savings account 30 years ago. The account has an interest rate of 4.6% and the balance is currently $30,541.83. How often does the interest compound?

Compound Interest Formula

: A = P(1 + r/n)^nt

A = Amount after time t

P = Principal (Initial Amount Invested)

r = Interest rate

n = Number of times the interest is compounded

t = time in years

A = $30,541.83

r = 4.6% = 0.046

t = 30

P = $7,924

Hence,

$30,541.83 = $7924(1 + 0.046/n)^30n

Divide both sides by 7924

$30,541.93/$7924 = (1 + 0.046/n)^30n

$30,541.93/$7924 = (n + 0.046/n)^30n

3.8543576477 = (n + 0.046/n)^30n

We take the logarithm of both sides

log 3.8543576477 = log (n + 0.046/n)^30n

Solving for n,

n = 1

Therefore, from the calculation above, since n = 1, this means the interest compounds ANNUALLY.

6 0

2 years ago

ILL GIVE BRAINLEYyyyyyyy

Fed [463]

Answer:

2+10x-x^2, the last answer

Step-by-step explanation:

3 0

3 years ago

What are the possible numbers of positive, negative, and complex zeros of f(x)=x^6-x^5-x^4+4x^3-12x^2+12?

Vadim26 [7]

$f(x)=x^6-x^5-x^4+4x^3-12x^2+12\\\\12:\{\pm1;\ \pm2;\ \pm3;\ \pm4;\ \pm6;\ \pm12\}\\1:\{\pm1\}\\\\Answer:\boxed{\{\pm1;\ \pm2;\ \pm3;\ \pm4;\ \pm6;\ \pm12\}}$

8 0

3 years ago

Find the absolute extrema for f(x,y)=4-x^2-y^4+1/2y^2 over the closed disk D:x^2+y^2 is less than or equal to 1

algol [13]

Find the critical points of $f(x,y)$ :

$\dfrac{\partial f}{\partial x}=-2x=0\implies x=0$

$\dfrac{\partial f}{\partial y}=y-4y^3=y(1-4y^2)=0\implies y=0\text{ or }y=\pm\dfrac12$

All three points lie within $D$ , and $f$ takes on values of

$\begin{cases}f(0,0)=4\\f\left(0,-\frac12\right)=\frac{65}{16}\\f\left(0,\frac12\right)=\frac{65}{16}\end{cases}$

Now check for extrema on the boundary of $D$ . Convert to polar coordinates:

$f(x,y)=f(\cos t,\sin t)=g(t)=4-\cos^2-\sin^4t+\dfrac12\sin^2t=3+\dfrac32\sin^2t-\sin^4t$

Find the critical points of $g(t)$ :

$\dfrac{\mathrm dg}{\mathrm dt}=3\sin t\cos t-4\sin^3t\cos t=\sin t\cos t(3-4\sin^2t)=0$

$\implies\sin t=0\text{ or }\cos t=0\text{ or }\sin t=\pm\dfrac{\sqrt3}2$

$\implies t=n\pi\text{ or }t=\dfrac{(2n+1)\pi}2\text{ or }\pm\dfrac\pi3+2n\pi$

where $n$ is any integer. There are some redundant critical points, so we'll just consider $0\le t< 2\pi$ , which gives

$t=0\text{ or }t=\dfrac\pi3\text{ or }t=\dfrac\pi2\text{ or }t=\pi\text{ or }t=\dfrac{3\pi}2\text{ or }t=\dfrac{5\pi}3$

which gives values of

$\begin{cases}g(0)=3\\g\left(\frac\pi3\right)=\frac{57}{16}\\g\left(\frac\pi2\right)=\frac72\\g(\pi)=3\\g\left(\frac{3\pi}2\right)=\frac72\\g\left(\frac{5\pi}3\right)=\frac{57}{16}\end{cases}$

So altogether, $f(x,y)$ has an absolute maximum of 65/16 at the points (0, -1/2) and (0, 1/2), and an absolute minimum of 3 at (-1, 0).

5 0

3 years ago

Estimate the line of best fit using two points on the line

DaniilM [7]

There is no picture

8 0

2 years ago