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Kamila [148]
3 years ago
14

X-y+z=-4 3x+2y-z=5 -2x+3y-z=15 How do I solve this?

Mathematics
1 answer:
Lera25 [3.4K]3 years ago
5 0
1)\ \ \ x-y+z=-4\ \ \ \Rightarrow\ \ \ z=-4-x+y\\\\2)\ \ \ 3x+2y-z=5 \\.\ \ \ \  \Rightarrow\ \ 3x+2y-(-4-x+y)=5 \\.\ \ \ \  \Rightarrow\ \ 3x+2y+4+x-y=5 \\.\ \ \ \  \Rightarrow\ \ 4x+y=1\\\\3)\ \ \ -2x+3y-z=15\\.\ \ \ \  \Rightarrow\ \ -2x+3y-(-4-x+y)=15\\.\ \ \ \  \Rightarrow\ \ -2x+3y+4+x-y=15\\.\ \ \ \  \Rightarrow\ \ -x+2y=11\\--------------------\\

z=-4-x+y\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ and\\ \left \{ {{4x+y=1\ \ \ \ } \atop {-x+2y=11\ /\cdot4}} \right. \\\\\left \{ {{4x+y=1\ \ \ \ } \atop {-4x+8y=44}} \right. \\-------\\y+8y=1+44\\9y=45\ /:9\\y=5\\\\-x+2y=11\ \ \ \Rightarrow\ \ \ x=2y-11\ \ \ \Rightarrow\ \ \ x=2\cdot5-11=-1\\\\z=-4-x+y=-4-(-1)+5=-4+1+5=2\\\\Ans.\ x=-1\ \ \ and\ \ \ y=5\ \ \ and\ \ \ z=2
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a line whose perpendicular distance from the origin is 4 units and the slope of perpendicular is 2÷3. Find the equation of the l
GrogVix [38]

Answer:

\huge\boxed{y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{2}{3}x+\dfrac{4\sqrt{13}}{3}}

Step-by-step explanation:

The equation of a line:

y=mx+b

We have

m=\dfrac{2}{3}

substitute:

y=\dfrac{2}{3}x+b

The formula of a distance between a point and a line:

General form of a line:

Ax+By+C=0

Point:

(x_0,\ y_0)

Distance:

d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+b^2}}

Convert the equation:

y=\dfrac{2}{3}x+b     |<em>subtract y from both sides</em>

\dfrac{2}{3}x-y+b=0    |<em>multiply both sides by 3</em>

2x-3y+3b=0\to A=2,\ B=-3,\ C=3b

Coordinates of the point:

(0,\ 0)\to x_0=0,\ y_0=0

substitute:

d=4

4=\dfrac{|2\cdot0+(-3)\cdot0+3b|}{\sqrt{2^2+(-3)^2}}\\\\4=\dfrac{|3b|}{\sqrt{4+9}}

4=\dfrac{|3b|}{\sqrt{13}}\qquad|    |<em>multiply both sides by \sqrt{13}</em>

4\sqrt{13}=|3b|\iff3b=-4\sqrt{13}\ \vee\ 3b=4\sqrt{13}   |<em>divide both  sides by 3</em>

b=-\dfrac{4\sqrt{13}}{3}\ \vee\ b=\dfrac{4\sqrt{13}}{3}

Finally:

y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{4\sqrt{13}}{3}

4 0
3 years ago
An automobile manufacturer has given its car a 46.7 miles/gallon (MPG) rating. An independent testing firm has been contracted t
erastova [34]

Answer:

z=\frac{46.5-46.7}{\frac{1.1}{\sqrt{150}}}=-2.23    

The p value would be given by:

p_v =2*P(z  

For this case since th p value is lower than the significance level of0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean for this case is significantly different from 46.7 MPG

Step-by-step explanation:

Information given

\bar X=46.5 represent the mean

\sigma=1.1 represent the population standard deviation

n=150 sample size  

\mu_o =46.7 represent the value to verify

\alpha=0.05 represent the significance level for the hypothesis test.  

z would represent the statistic

p_v represent the p value

Hypothesis to test

We want to test if the true mean for this case is 46.7, the system of hypothesis would be:  

Null hypothesis:\mu = 46.7  

Alternative hypothesis:\mu \neq 46.7  

Since we know the population deviation the statistic is given by:

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

Replacing we got:

z=\frac{46.5-46.7}{\frac{1.1}{\sqrt{150}}}=-2.23    

The p value would be given by:

p_v =2*P(z  

For this case since th p value is lower than the significance level of0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean for this case is significantly different from 46.7 MPG

7 0
3 years ago
The measure of the supplement of an angle is 84° less than the measure of the angle. Find the measures of the angles
babunello [35]

Let the measure of angle be 'x'

Since, the measure of the supplement of an angle is 84° less than the measure of the angle.

Therefore, measure of supplement of x = x - 84°

The measure of two supplementary angles is 180 degrees.

x + x -84^\circ = 180^\circ

2x -84^\circ = 180^\circ

2x =84^\circ + 180^\circ

2x =264^\circ

x =132^\circ

So, the measure of angle =x =132^\circ

Its supplement = x - 84°  = 132°  - 84°

=  48°

So, the measure of supplement of angle is 48 degrees.

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3 years ago
1. (5x2-2x+3) + (x2+x+2)
antiseptic1488 [7]
The answer is: x^2-x+15
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Write a polynomial, f(x), with degree 5 that has zeros 1, √2, -√2, √3, -√3.
just olya [345]

The get (x-1)(x-√2)(x+√2)(x-√3)(x+√3).


Expanding, you get x^5-x^4-5x^3+5x^2+6x-6.


3 0
3 years ago
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