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3 years ago

Which equation has no solution?

Mathematics

1 answer:

lord [1]3 years ago

5 0

Answer:

Equation 5 + 2(3 + 2x) = x + 3(x + 1) has no solution.

Step-by-step explanation:

We are looking at two lines.

4(x + 3) + 2x = 6(x +2)

4x + 12 + 2x = 6x + 12

6x + 12 = 6x + 12

These are two identical lines, with an infinite number of solutions. (All points on the lines are the exactly the same).

5 + 2(3 + 2x) = x + 3(x + 1)

5 + 6 + 4x = x + 3x + 3

4x + 11 = 4x + 3

Both lines have the same gradient but have a different incline with the y axis. By definition, they are parallel to each other and there fore have zero solutions. Equation 5 + 2(3 + 2x) = x + 3(x + 1) has no solution, which is the answer we are looking for.

5(x + 3) + x = 4(x +3) + 3

5x + 15 + x = 4x + 12 + 3

6x + 15 = 4x + 15

These are two different lines with exactly one solution.

4 + 6(2 + x) = 2(3x + 8)

4 + 12 + 6x = 6x + 16

6x + 16 = 6x + 16

These are two identical lines, with an infinite number of solutions. (All points on the lines are the exactly the same).

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3. One of the graphs represents

faltersainse [42]

Answer:

3x-2y=6

Step-by-step explanation:

Subtract 3x from both sides of the equation

Divide each term by −2 and simplify.

y=-3+ $\frac{3x}{2}$

Rewrite in slope-intercept form

y= $\frac{3}{2} *-3$

Use the slope-intercept form to find the slope and y-intercept.

Find the values of m and b using the form y=mx+b

$m=\frac{3}{2}$

$b=-3$

The slope of the line is the value of m, and the y-intercept is the value of b.

Slope: $\frac{3}{2}$

y-intercept: -3

Any line can be graphed using two points. Select two x values, and plug them into the equation to find the corresponding y values.

Choose 0 to substitute in for x to find the ordered pair.

(0,−3)

Choose 1 to substitute in for x to find the ordered pair.

$(1,-\frac{3}{2} )$

Graph the line using the slope and the y-intercept, or the points.

slope: $\frac{3}{2}$

y-intercept:-3

5 0

3 years ago

Name/ Uid:1. In this problem, try to write the equations of the given surface in the specified coordinates.(a) Write an equation

Gemiola [76]

To find:

(a) Equation for the sphere of radius 5 centered at the origin in cylindrical coordinates

(b) Equation for a cylinder of radius 1 centered at the origin and running parallel to the z-axis in spherical coordinates

Solution:

(a) The equation of a sphere with center at (a, b, c) & having a radius 'p' is given in cartesian coordinates as:

$(x-a)^{2}+(y-b)^{2}+(z-c)^{2}=p^{2}$

Here, it is given that the center of the sphere is at origin, i.e., at (0,0,0) & radius of the sphere is 5. That is, here we have,

$a=b=c=0,p=5$

That is, the equation of the sphere in cartesian coordinates is,

$(x-0)^{2}+(y-0)^{2}+(z-0)^{2}=5^{2}$

$\Rightarrow x^{2}+y^{2}+z^{2}=25$

Now, the cylindrical coordinate system is represented by $(r, \theta,z)$

The relation between cartesian and cylindrical coordinates is given by,

$x=rcos\theta,y=rsin\theta,z=z$

$r^{2}=x^{2}+y^{2},tan\theta=\frac{y}{x},z=z$

Thus, the obtained equation of the sphere in cartesian coordinates can be rewritten in cylindrical coordinates as,

$r^{2}+z^{2}=25$

This is the required equation of the given sphere in cylindrical coordinates.

(b) A cylinder is defined by the circle that gives the top and bottom faces or alternatively, the cross section, & it's axis. A cylinder running parallel to the z-axis has an axis that is parallel to the z-axis. The equation of such a cylinder is given by the equation of the circle of cross-section with the assumption that a point in 3 dimension lying on the cylinder has 'x' & 'y' values satisfying the equation of the circle & that 'z' can be any value.

That is, in cartesian coordinates, the equation of a cylinder running parallel to the z-axis having radius 'p' with center at (a, b) is given by,

$(x-a)^{2}+(y-b)^{2}=p^{2}$

Here, it is given that the center is at origin & radius is 1. That is, here, we have, $a=b=0,p=1$ . Then the equation of the cylinder in cartesian coordinates is,

$x^{2}+y^{2}=1$

Now, the spherical coordinate system is represented by $(\rho,\theta,\phi)$

The relation between cartesian and spherical coordinates is given by,

$x=\rho sin\phi cos\theta,y=\rho sin\phi sin\theta, z= \rho cos\phi$

Thus, the equation of the cylinder can be rewritten in spherical coordinates as,

$(\rho sin\phi cos\theta)^{2}+(\rho sin\phi sin\theta)^{2}=1$

$\Rightarrow \rho^{2} sin^{2}\phi cos^{2}\theta+\rho^{2} sin^{2}\phi sin^{2}\theta=1$

$\Rightarrow \rho^{2} sin^{2}\phi (cos^{2}\theta+sin^{2}\theta)=1$

$\Rightarrow \rho^{2} sin^{2}\phi=1$ (As $sin^{2}\theta+cos^{2}\theta=1$ )

Note that $\rho$ represents the distance of a point from the origin, which is always positive. $\phi$ represents the angle made by the line segment joining the point with z-axis. The range of $\phi$ is given as $0\leq \phi\leq \pi$ . We know that in this range the sine function is positive. Thus, we can say that $sin\phi$ is always positive.

Thus, we can square root both sides and only consider the positive root as,

$\Rightarrow \rho sin\phi=1$

This is the required equation of the cylinder in spherical coordinates.

Final answer:

(a) The equation of the given sphere in cylindrical coordinates is $r^{2}+z^{2}=25$

(b) The equation of the given cylinder in spherical coordinates is $\rho sin\phi=1$

7 0

3 years ago

After *triangle* ACE is dilated by a factor of 5, It has an area of 100 square inches. What was its area before dilation?

Anarel [89]

<h3>Answer: 4 square inches</h3>

Explanation:

Square the linear scale factor to get 5^2 = 25

This means that,

new area = 25*(old area)

We take this idea in reverse to find the old area

old area = (new area)/25

old area = (100 sq inches)/25

old area = 4 square inches

6 0

2 years ago

What is the domain and range of each function graohed below?

andriy [413]

Domain is the x values you can use
range is the y values

a.
the domain
hmm, seems to be all real numbers except for at x=0, it gets really close tho
so D=(-∞,0)U(0,∞)

range is all real numbers except for y=0, it gets really close tho
R=(-∞,0)U(0,∞)

b.
domain
that empty circle means something like < or >
goes from 5 to 6, not including 5
so domain is D=(5,6]

range is from -4 to 2, not including -4
range is R=(-4,2]

5 0

3 years ago

Read 2 more answers

PLEASE HELP WILL GIVE BRAINLIEST

Marta_Voda [28]

Answer:

The bird will have traveled 403 meters in 13 seconds if it maintained a speed of 31 m/sec

Explanation:

The equation that relates velocity, distance and time is:

$velocity = \frac{distance}{time}$

This means that:

distance = velocity * time

In the given, we have:

velocity = 31 m/sec

time = 13 sec

Substitute with the givens in the above equation and solve for the distance as follows:

distance = velocity * time

distance = 31 * 13

distance = 403 meters

Hope this helps :)

8 0

3 years ago

Which equation has no solution?​

Which equation has no solution?