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11 months ago

11

help meeeeeeeeeeeeeeeeeee pleaseee!!!!help meeeeeeeeeeeeeeeeeee pleaseee!!!!help meeeeeeeeeeeeeeeeeee pleaseee!!!!help meeeeeeee

eeeeeeeeeee pleaseee!!!!help meeeeeeeeeeeeeeeeeee pleaseee!!!!help meeeeeeeeeeeeeeeeeee pleaseee!!!!

1 answer:

sineoko [7]11 months ago

6 0

Answer:

15095.39

Step-by-step explanation:

$13000(1+(0.03)/4)^{(4)(5)} \approx 15095.39$

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Solve the following 3 × 3 system. Enter the coordinates of the solution below.

defon

9514 1404 393

Answer:

(x, y, z) = (-1, 0, -3)

Step-by-step explanation:

We notice that the coefficients of z are such that elimination of the z term from the equations is made easy.

Adding equations 1 and 2:

(2x -3y -2z) +(x +3y +2z) = (4) +(-7)

3x = -3

x = -1

Adding equations 2 and 3:

(x +3y +2z) +(-4x -4y -2z) = (-7) +(10)

-3x -y = 3

Substituting for x, we get ...

(-3)(-1) -y = 3

0 = y . . . . . . . . . . . add y-3 to both sides

Then z can be found from any equation. Substituting for x and y in the second equation gives ...

-1 +2z = -7

2z = -6 . . . . . add 1

z = -3 . . . . . .divide by 2

The solution is (x, y, z) = (-1, 0, -3).

3 0

3 years ago

Plz help only have ten minutes

Anarel [89]

Answer:

- $\frac{1}{2}$

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

How many solutions does this equation have 10+5x/15=4

-Dominant- [34]

I think it only has One

8 0

2 years ago

if the measure of the third angle of the triangle is 20 more than three times the measure of either of the other two angles, fin

valentina_108 [34]

You should multiply 20 time 8 to get your answer

5 0

2 years ago

Use the Midpoint Rule with n = 5 to estimate the volume V obtained by rotating about the y-axis the region under the curve y = 1

velikii [3]

Using the shell method, the volume is given exactly by the definite integral,

$2\pi\displaystyle\int_0^1x(1+9x^3)\,\mathrm dx$

Splitting up the interval [0, 1] into 5 subintervals gives the partition,

[0, 1/5], [1/5, 2/5], [2/5, 3/5], [3/5, 4/5], [4/5, 5]

with left and right endpoints, respectively, for the $i$ -th subinterval

$\ell_i=\dfrac{i-1}5$

$r_i=\dfrac i5$

where $1\le i\le5$ . The midpoint of each subinterval is

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}{10}$

Then the Riemann sum approximating the integral above is

$2\pi\displaystyle\sum_{i=1}^5m_i(1+9{m_i}^3)\frac{1-0}5$

$\dfrac{2\pi}5\displaystyle\sum_{i=1}^5\left(\frac{2i-1}{10}+9\left(\frac{2i-1}{10}\right)^4\right)$

$\dfrac{2\pi}{5\cdot10^4}\displaystyle\sum_{i=1}^5\left(16i^4-32i^3+24i^2+1992i-999\right)=\frac{112,021\pi}{25,000}\approx\boxed{14.08}$

(compare to the actual value of the integral of about 14.45)

3 0

3 years ago