All categories

3 years ago

Can a linear function be continuous but not have a domain and range of all real numbers

1 answer:

5 0

Yes. As long as it is linear, it will be continuous no matter the numbers. Now, I have no idea what a "real" number is, but I hope this helped.

You might be interested in

XY is a diameter of a circle and Z is a point on the circle such that ZY=6. If the area of the triangle XYZ is 18 square root 3

nataly862011 [7]

<h2>Answer:</h2>

4π

<h2>Step-by-step explanation:</h2>

As shown in the diagram, triangle XYZ is a right triangle. Therefore, its area (A) is given by:

A = $\frac{1}{2}$ x b x h -------------(i)

Where;

A = 18 $\sqrt{3}$

b = XZ = base of the triangle

h = YZ = height of the triangle = 6

<em>Substitute these values into equation(i) and solve as follows:</em>

18 $\sqrt{3}$ = $\frac{1}{2}$ x b x 6

18 $\sqrt{3}$ = 3b

<em>Divide through by 3</em>

6 $\sqrt{3}$ = b

Therefore, b = XZ = 6 $\sqrt{3}$

<em>Now, assume that the circle is centered at O;</em>

Triangle XOZ is isosceles, therefore the following are true;

(i) |OZ| = |OX|

(ii) XZO = ZXO = 30°

(iii) XOZ + XZO + ZXO = 180° [sum of angles in a triangle]

=> XOZ + 30° + 30° = 180°

=> XOZ + 60° = 180°

=> XOZ = 180° - 60°

=> XOZ = 120°

Therefore we can calculate the radius |OZ| of the circle using sine rule as follows;

$\frac{sin|XOZ|}{XZ} = \frac{sin|ZXO|}{OZ}$

$\frac{sin120}{6\sqrt{3} } = \frac{sin 30}{OZ}$

$\frac{\sqrt{3} /2}{6\sqrt{3} } = \frac{1/2}{|OZ|}$

$\frac{1}{12} = \frac{1}{2|OZ|}$

$\frac{1}{6} = \frac{1}{|OZ|}$

|OZ| = 6

The radius of the circle is therefore 6.

<em>Now, let's calculate the length of the arc XZ</em>

The length(L) of an arc is given by;

L = θ / 360 x 2 π r ------------------(ii)

Where;

θ = angle subtended by the arc at the center.

r = radius of the circle.

In our case,

θ = ZOX = 120°

r = |OZ| = 6

Substitute these values into equation (ii) as follows;

L = 120/360 x 2π x 6

L = 4π

Therefore the length of the arc XZ is 4π

5 0

2 years ago

1. Consider the rational numbers 4/13 and 3/10.

Murljashka [212]

A)4/13 would be the larger number since if you were to try to add them, 4/13 = 40/130 and 3/10 = 39/130, 40/130 > 39/130

B) A rational number between them would be 39.5/130.

5 0

3 years ago

The coffee pub has cans of coffee that weigh 4 4/5 pounds each. The pub has 8 1/2 cans of coffee left. What is the total weight?

tankabanditka [31]

4.8 x 8.5 = 40.8 lbs total
quick mafs

6 0

2 years ago

Which equation involves a prime quadratic and cannot be solved by factoring?

brilliants [131]

B. x^2 + 5x - 4 = 0

Anyway that you try to factor it will not work

~~hope this helps~~

6 0

3 years ago

Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.2.a. If the distri

zalisa [80]

Answer:

$P(\= X \ge 51 ) =0.0062$

$P(\= X \ge 51 ) = 0$

Step-by-step explanation:

From the question we are told that

The mean value is $\mu = 50$

The standard deviation is $\sigma = 1.2$

Considering question a

The sample size is n = 9

Generally the standard error of the mean is mathematically represented as

$\sigma_x = \frac{\sigma }{\sqrt{n} }$

=> $\sigma_x = \frac{ 1.2 }{\sqrt{9} }$

=> $\sigma_x = 0.4$

Generally the probability that the sample mean hardness for a random sample of 9 pins is at least 51 is mathematically represented as

$P(\= X \ge 51 ) = P( \frac{\= X - \mu }{\sigma_{x}} \ge \frac{51 - 50 }{0.4 } )$

$\frac{\= X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ \= X )$

$P(\= X \ge 51 ) = P( Z \ge 2.5 )$

=> $P(\= X \ge 51 ) =1- P( Z < 2.5 )$

From the z table the area under the normal curve to the left corresponding to 2.5 is

$P( Z < 2.5 ) = 0.99379$

=> $P(\= X \ge 51 ) =1-0.99379$

=> $P(\= X \ge 51 ) =0.0062$

Considering question b

The sample size is n = 40

Generally the standard error of the mean is mathematically represented as

$\sigma_x = \frac{\sigma }{\sqrt{n} }$

=> $\sigma_x = \frac{ 1.2 }{\sqrt{40} }$

=> $\sigma_x = 0.1897$

Generally the (approximate) probability that the sample mean hardness for a random sample of 40 pins is at least 51 is mathematically represented as

$P(\= X \ge 51 ) = P( \frac{\= X - \mu }{\sigma_x} \ge \frac{51 - 50 }{0.1897 } )$

=> $P(\= X \ge 51 ) = P(Z \ge 5.2715 )$

=> $P(\= X \ge 51 ) = 1- P(Z < 5.2715 )$

From the z table the area under the normal curve to the left corresponding to 5.2715 and

=> $P(Z < 5.2715 ) = 1$

$P(\= X \ge 51 ) = 1- 1$

=> $P(\= X \ge 51 ) = 0$

5 0

2 years ago