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3 years ago

Solve the system by substitution -x - y - z= -8 -4x + 4y + 5z = 7 2x + 2z = 4

Mathematics

1 answer:

Stella [2.4K]3 years ago

5 0

First:

2x + 2z = 4

2(x+z)=4

x+z=4/2

x=2-z

-x - y - z= -8

-(2-z)-y-z= -8 replace x

-2+z-y-z= -8 (cancel z)

-2-y= -8

6 = y

Then:

-4x + 4y + 5z = 7 replace x and y

-4(2-z)+4*6 + 5z = 7

-8 + 4z + 24 + 5z = 7

16 + 9z = 7

9z = 7 - 16

9z = -9

z = -9/9

z= -1

Finally

x= 2-z replace z

x = 2- (-1)

x = 3

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lesya [120]

Answer:

correct option is d. $242.81

Step-by-step explanation:

given data

APR = 25.5% = $\frac{25.5}{12}$ = 2.125

paid = $3,729

solution

we get here finance charge on the 1st month by multiplying 3,729 and now adding it to existing balance

so we get finance charge for the second and third months similarly as

APR ÷ 100 = $\frac{2.125 }{100}$ = 0.02125

so 1st

= $3,729 × 0.02125

= 79.25

and

$3,729 + $79.25 = $3808.24

so for next

= $3808.24 × 0.02125

= 80.93

and

$3808.24 + $80.93 = $3889.17

so for next

= $3889.17 × 0.02125

= 82.64

and

$3889.17 + $ 82.64 = $3971.81

finance charge = 3971.81 - 3729

finance charge = 242.81

so correct option is d. $242.81

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3 years ago

Sam received a 15 percent tip for delivering pizzas. If the cost of the pizzas was $247.62, what was the total amount he collect

saveliy_v [14]

247.62 x 0.15 = 37.143

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3 years ago

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Leya [2.2K]

Answer:

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Step-by-step explanation:

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2 years ago

The weight of a cat is normally distributed with a mean of 9 pounds and a standard deviation of 2 pounds. Using the empirical ru

coldgirl [10]

If the value of the z-score is 1. Then the probability that a cat will weigh less than 11 pounds will be 0.84134.

<h3>What is the z-score?</h3>

The z-score is a statistical evaluation of a value's correlation to the mean of a collection of values, expressed in terms of standard deviation.

The z-score is given as

z = (x - μ) / σ

Where μ is the mean, σ is the standard deviation, and x is the sample.

The weight of a cat is normally distributed with a mean of 9 pounds and a standard deviation of 2 pounds.

Then the probability that a cat will weigh less than 11 pounds will be

The value of z-score will be

z = (11 – 9) / 2

z = 1

Then the probability will be

P(x < 11) = P(z < 1)

P(x < 11) = 0.84134

Thus, the probability that a cat will weigh less than 11 pounds will be 0.84134.

More about the z-score link is given below.

brainly.com/question/15016913

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1 year ago

Can I get help with finding the Fourier cosine series of F(x) = x - x^2

trapecia [35]

Assuming you want the cosine series expansion over an arbitrary symmetric interval $[-L,L]$ , $L\neq0$ , the cosine series is given by

$f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx$

You have

$a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx$
$a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}$
$a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)$
$a_0=-\dfrac{2L^2}3$

$a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx$

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

$\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}$

and so

$a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}$

So the cosine series for $f(x)$ periodic over an interval $[-L,L]$ is

$f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx$

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3 years ago