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3 years ago

X+y+2=-5 -2x-y+2=-1 x-2y-2=0 solve algebraically

Mathematics

1 answer:

Rina8888 [55]3 years ago

7 0

Answer:

x+y+2=-5 -2x-y+2=-1 x-2y-2=0 solve algebraically

x+y+2=-5 equation 1

-2x-y+2=-1 equation 2

x-2y-2=0 equation 3

Add equation 1 + 2+ 3

x+y+2=-5 + -2x-y+2+1 + x-2y-2

x+y+2= -7+2+1-2x+x+y

x+y+2= -4-x+y

x+y+2+4+x-y= 0

2x+6= 0

2x= -6

divide both side by 2

x= -3

from equation 1 insert x

x+y+2=-5

since x=-3

then, -3+2+y= -5

-1+y= -5

y= -5+1

y= -4

Step-by-step explanation:

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Find the vertical and horizontal asymptotes, domain, range, and roots of f (x) = -1 / x-3 +2.

gladu [14]

Answer:

Vertical asymptote: $x=3$

Horizontal asymptote: $f(x) =2$

Domain of f(x) is all real numbers except 3.

Range of f(x) is all real numbers except 2.

Step-by-step explanation:

Given:

Function:

$f (x) = -\dfrac{1 }{ x-3} +2$

One root, $x = 3.5$

To find:

Vertical and horizontal asymptote, domain, range and roots of f(x).

Solution:

First of all, let us find the roots of f(x).

<em>Roots of f(x) means the value of x where f(x) = 0</em>

$0= -\dfrac{1 }{ x-3} +2\\\Rightarrow 2= \dfrac{1 }{ x-3}\\\Rightarrow 2x-2 \times 3=1\\\Rightarrow 2x=7\\\Rightarrow x = 3.5$

One root, $x = 3.5$

Domain of f(x) i.e. the values that we give as input to the function and there is a value of f(x) defined for it.

For x = 3, the value of f(x) $\rightarrow \infty$

For all, other values of $x$ , $f(x)$ is defined.

Hence, Domain of f(x) is all real numbers except 3.

Range of f(x) i.e. the values that are possible output of the function.

f(x) = 2 is not possible in this case because something is subtracted from 2. That something is $\frac{1}{x-3}$ .

Hence, Range of f(x) is all real numbers except 2.

Vertical Asymptote is the value of x, where value of f(x) $\rightarrow \infty$ .

$-\dfrac{1 }{ x-3} +2 \rightarrow \infty$

It is possible only when

$x-3=0\\\Rightarrow x=3$

$\therefore$ vertical asymptote: $x=3$

Horizontal Asymptote is the value of f(x) , where value of x $\rightarrow \infty$ .

$x\rightarrow \infty \Rightarrow \dfrac{1 }{ x-3} \rightarrow 0\\\therefore f(x) =-0+2 \\\Rightarrow f(x) =2$

$\therefore$ Horizontal asymptote: $f(x) =2$

Please refer to the graph of given function as shown in the attached image.

5 0

3 years ago

Find the rectangular coordinates of the point with the polar coordinates ordered pair 7 comma 2 pi divided by 3.

Morgarella [4.7K]

Answer:

$\left(-\dfrac{7}{2},\dfrac{7\sqrt{3}}{2}\right)$ .

Step-by-step explanation:

The given point is

$\left(7,\dfrac{2\pi}{3}\right)$

We need to find the rectangular coordinates of the given point.

If a polar coordinate is $(r,\theta)$ , then

$x=r\cos theta$

$y=r\sin theta$

In the given point $\left(7,\dfrac{2\pi}{3}\right)$ ,

$r=7,\theta=\dfrac{2\pi}{3}$

Now,

$x=7\cos \dfrac{2\pi}{3}$

$x=7\cos \left(\pi-\dfrac{\pi}{3}\right)$

$x=-7\cos \left(\dfrac{\pi}{3}\right)$

$x=-7\left(\dfrac{1}{2}\right)$

$x=-\dfrac{7}{2}$

and,

$y=7\sin \dfrac{2\pi}{3}$

$y=7\sin \left(\pi-\dfrac{\pi}{3}\right)$

$y=7\sin \left(\dfrac{\pi}{3}\right)$

$y=7\left(\dfrac{\sqrt{3}}{2}\right)$

$y=\dfrac{7\sqrt{3}}{2}$

Therefore, the required point is $\left(-\dfrac{7}{2},\dfrac{7\sqrt{3}}{2}\right)$ .

3 0

3 years ago

What is the volume of the sphere?

nataly862011 [7]

The volume of the sphere in the given image, to the nearest tenth, is: 2,143.6 cm³.

<h3>How to calculate the Volume of a Sphere?</h3>

A sphere is a solid shape that is round. The volume of a sphere is the amount of space it contains, which is calculated by the formula expressed as:

Volume = 4/3 × π × r³, where r is the radius of the sphere.

The parameter of the sphere is given below:

Radius of the sphere (r) = 8 cm.

π = 3.14

Substitute r = 8 into 4/3 × π × r³ to find the volume of the sphere

Volume of the sphere = 4/3 × π × r³ = 4/3 × 3.14 × 8³

Volume of the sphere = 4/3 × 3.14 × 512

Volume of the sphere = (4 × 3.14 × 512)/3

Volume of the sphere = (6,430.72)/3

Volume of the sphere = 2,143.6 cm³

In summary, the volume of the sphere in the given image, to the nearest tenth, is: 2,143.6 cm³.

Learn more about the volume of sphere on:

brainly.com/question/22807400

#SPJ1

5 0

1 year ago

Using the distance formula, d = √(x2 - x1)2 + (y2 - y1)2, what is the distance between point (0, 5) and point (3, -1) rounded to

miss Akunina [59]

$\sqrt{(3-0)^2+(-1-5)^2}=\sqrt{3^2+(-6)^2}=\sqrt{45}\approx 6.7$

answer (a)

5 0

3 years ago

Need the answers plz

Shkiper50 [21]

1)
         x^2 + 4x - 5
y =   --------------------
              3x^2 - 12

Vertcial asym => find the x-values for which the equation is not defined and check whether the limit goes to + or - infinite

=> 3x^2 - 12 = 0 => 3x^2 = 12

=> x^2 = 12 / 3 = 4

=> x = +/-2

Limit of y when x -> 2(+) = ( 2^2 + 4(2) - 5) / 0 = - 1 / 0(+) = ∞

Limit of y when x -> 2(-) = - 1 / (0(-) = - ∞

Limit of y when x -> - 2(+) = +∞

Limit of y when x -> - 2(-) = -

=> x = 2 and x = - 2 are a vertical asymptotes

Horizontal asymptote => find whether y tends to a constant value when x -> infinite of negative infinite

Limi of y when x -> - ∞ = 1/3

Lim of y when x -> +∞ = 1/3

=> Horizontal asymptote y = 1/3

x - intercept => y = 0

=>
          x^2 + 4x - 5
0 =   ------------------ => x ^2 + 4x - 5 = 0
              3x^2 - 12

Factor x^2 + 4x - 5 => (x + 5) (x - 1) = 0 => x = - 5 and x = 1

=> x-intercepts x = - 5 and x = 1

Domain: all the real values except x = 2 and x = - 2

2) y = - 2 / (x - 4) - 1

using the same criteria you get:

Vertical asymptote: x = 4

Horizontal asymptote: y = - 1

Domain:all the real values except x = 4

Range: all the real values except y = - 1

3)

x/ (x + 2) + 7 / (x - 5) = 14 / (x^2 - 3x - 10)

factor x^2 - 3x - 10 => (x - 5)(x + 2)

Multiply both sides by (x - 5) (x + 2)

=> x(x - 5) + 7( x + 2) = 14

=> x^2 - 5x + 7x + 14 = 14

=> x^2 + 2x = 0

=> x(x + 2) = 0 => x = 0 and x = - 2 but the function is not defined for x = - 2 so it is not a solution => x = 0

Answer: x = 0

4 0

3 years ago