The equation of a circle is given below. Identify the radius and center.

x^2 + y^2 + 6x -8y +21 = 0

Answer:

r=<1,0,9>+t<1,2,1>

and

x=1+t

y=2t

z=9+t

Step-by-step explanation:

A vector perpendicular to the plane :

[tex]ax+by+cz+d=0[/tex]

is given by (a,b,c)

So a vector perpendicular to given plane will have :

(1,2,1)

[tex]The\ parametric\ equation\ of\ a\ line\ through\ (1,0,9)\ and\ parallel\ to\ vector\ (a,b,c) is\ given\ by:\\x=x_0+ta\\y=y_0+tb\\z=z_0+tb\\x=1+t(1)\\x=1+t\\y=0+t(2)\\y=2t\\z=9+t(1)\\z=9+t\\The\ vector\ form\ is:\\r=<1,0,9>+t<1,2,1>[/tex] ..

The vector equation r(t) is (1, 0, 9) + t(1, 2, 1) and the parametric equations are x = 1 + t, y = 2t, z = 9 + t.

To find the vector equation and parametric equations for the line that passes through the point (1, 0, 9) and is perpendicular to the plane given by x + 2y + z = 7, follow these steps :

1. Find the Normal Vector to the Plane :

2. Vector Equation of the Line :

3. Parametric Equations :

Extract the parametric equations from the vector equation :

Thus, the parametric equations for the line are :

Answer:

[tex]x=2\cos(t)[/tex] and [tex]y=-2\sin(t)+1[/tex]

Step-by-step explanation:

[tex](x-h)^2+(y-k)^2=r^2[/tex] has parametric equations:

[tex](x-h)=r\cos(t) \text{ and } (y-k)=r\sin(t)[/tex].

Let's solve these for x and y  respectively.

[tex]x-h=r\cos(t)[/tex] can be solved for x by adding h on both sides:

[tex]x=r\cos(t)+h[/tex].

[tex]y-k=r \sin(t)[/tex] can be solve for y by adding k on both sides:

[tex]y=r\sin(t)+k[/tex].

We can verify this works by plugging these back in for x and y respectively.

Let's do that:

[tex](r\cos(t)+h-h)^2+(r\sin(t)+k-k)^2[/tex]

[tex](r\cos(t))^2+(r\sin(t))^2[/tex]

[tex]r^2\cos^2(t)+r^2\sin^2(t)[/tex]

[tex]r^2(\cos^2(t)+\sin^2(t))[/tex]

[tex]r^2(1)[/tex] By a Pythagorean Identity.

[tex]r^2[/tex] which is what we had on the right hand side.

We have confirmed our parametric equations are correct.

Now here your h=0 while your k=1 and r=2.

So we are going to play with these parametric equations:

[tex]x=2\cos(t)[/tex] and [tex]y=2\sin(t)+1[/tex]

We want to travel clockwise so we need to put -t and instead of t.

If we were going counterclockwise it would be just the t.

[tex]x=2\cos(-t)[/tex] and [tex]y=2\sin(-t)+1[/tex]

Now cosine is even function while sine is an odd function so you could simplify this and say:

[tex]x=2\cos(t)[/tex] and [tex]y=-2\sin(t)+1[/tex].

We want to find [tex]\theta[/tex] such that

[tex]2\cos(t-\theta_1)=2 \text{ while } -2\sin(t-\theta_2)+1=1[/tex] when t=0.

Let's start with the first equation:

[tex]2\cos(t-\theta_1)=2[/tex]

Divide both sides by 2:

[tex]\cos(t-\theta_1)=1[/tex]

We wanted to find [tex]\theta_1[/tex] for when [tex]t=0[/tex]

[tex]\cos(-\theta_1)=1[/tex]

Cosine is an even function:

[tex]\cos(\theta_1)=1[/tex]

This happens when [tex]\theta_1=2n\pi[/tex] where n is an integer.

Let's do the second equation:

[tex]-2\sin(t-\theta_2)+1=1[/tex]

Subtract 2 on both sides:

[tex]-2\sin(t-\theta_2)=0[/tex]

Divide both sides by -2:

[tex]\sin(t-\theta_2)=0[/tex]

Recall we are trying to find what [tex]\theta_2[/tex] is when t=0:

[tex]\sin(0-\theta_2)=0[/tex]

[tex]\sin(-\theta_2)=0[/tex]

Recall sine is an odd function:

[tex]-\sin(\theta_2)=0[/tex]

Divide both sides by -1:

[tex]\sin(\theta_2)=0[/tex]

[tex]\theta_2=n\pi[/tex]

So this means we don't have to shift the cosine parametric equation at all because we can choose n=0 which means [tex]\theta_1=2n\pi=2(0)\pi=0[/tex].

We also don't have to shift the sine parametric equation either since at n=0, we have [tex]\theta_2=n\pi=0(\pi)=0[/tex].

So let's see what our equations look like now:

[tex]x=2\cos(t)[/tex] and [tex]y=-2\sin(t)+1[/tex]

Let's verify these still work in our original equation:

[tex]x^2+(y-1)^2[/tex]

[tex](2\cos(t))^2+(-2\sin(t))^2[/tex]

[tex]2^2\cos^2(t)+(-2)^2\sin^2(t)[/tex]

[tex]4\cos^2(t)+4\sin^2(t)[/tex]

[tex]4(\cos^2(t)+\sin^2(t))[/tex]

[tex]4(1)[/tex]

[tex]4[/tex]

It still works.

Now let's see if we are being moving around the circle once around for values of t between [tex]0[/tex] and [tex]2\pi[/tex].

This first table will be the first half of the rotation.

t                  0                      pi/4                pi/2               3pi/4               pi  

x                  2                     sqrt(2)             0                  -sqrt(2)            -2

y                  1                    -sqrt(2)+1          -1                  -sqrt(2)+1            1

Ok this is the fist half of the rotation.  Are we moving clockwise from (2,1)?

If we are moving clockwise around a circle with radius 2 and center (0,1) starting at (2,1) our x's should be decreasing and our y's should be decreasing at the beginning we should see a 4th of a circle from the point (x,y)=(2,1) and the point (x,y)=(0,-1).

Now after that 4th, the x's will still decrease until we make half a rotation but the y's will increase as you can see from point (x,y)=(0,-1) to (x,y)=(-2,1).  We have now made half a rotation around the circle whose center is (0,1) and radius is 2.

Let's look at the other half of the circle:

t                pi               5pi/4                  3pi/2            7pi/4                     2pi

x               -2              -sqrt(2)                0                 sqrt(2)                      2

y                1                sqrt(2)+1             3                  sqrt(2)+1                   1

So now for the talk half going clockwise we should see the x's increase since we are moving right for them.  The y's increase after the half rotation but decrease after the 3/4th rotation.

We also stopped where we ended at the point (2,1).

The parametric equations for the path of a particle moving along the circle x^2 + (y - 1)^2 = 4 in a clockwise direction, starting at (2, 1), are x = 2 + 2sin(-t) and y = 1 + 2cos(-t).

The parametric equations for the path of a particle moving along the circle x2 + (y - 1)2 = 4 in a clockwise direction, starting at (2, 1), can be found using trigonometric functions. From the given equation of the circle, we can determine that the center of the circle is (0, 1) and the radius is 2. Therefore, the parametric equations are:

.

https://brainly.com/question/29187193

#SPJ2

Only two steps needed:

11,655 = 1*11,340 + 315

11,340 = 36*315 + 0

This shows that [tex]\mathrm{gcd}(11,655,\,11,340)=315[/tex].

The Greatest Common Divisor (GCD) of the numbers 11655 and 11340 is found using the Euclidean Algorithm by first dividing 11655 by 11340 to get a remainder of 315. This remainder becomes the divisor in the next step and the process repeats until the remainder is zero. Hence, the GCD of 11655 and 11340 is 315.

To calculate the Greatest Common Divisor (GCD) of the numbers 11655 and 11340 using the Euclidean Algorithm, follow this process:

Therefore, the Greatest Common Divisor of the numbers 11655 and 11340 using the Euclidean Algorithm is 315.

https://brainly.com/question/23270841

#SPJ12

Answer:

Probability: [tex]\frac{1}{2}[/tex] = 0.5 = 50%

Step-by-step explanation:

Based on the question one coin is chosen at random and tossed. That coin then lands and is heads. Since the coin landed on heads we can eliminate the possibility of the coin that was chosen being the coin with double tails.

The following possibilities are that the coin has double heads or is a regular coin with both tails and heads. Seeing as the coin landed on heads, there are only two possible out comes for the other side of the coin

The other side is either Heads or Tails. That gives us a 50% chance of the other side being tails.

[tex]\frac{1}{2}[/tex] = 0.5 = 50%

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

The symmetric equations of the line are: x/(-2) = y/(-6) = z/1 and the x(t) = 2t, y(t) = 6t and z(t) = -t are the parametric equations.

a parametric equation defines a group of quantities as functions of one or more independent variables called parameters.

To find the parametric equations for the line through the origin and the point (2, 6, -1), we can use the vector equation of a line:

r(t) = r₀ + tv

where r(t) = (x(t), y(t), z(t)) is the position vector of a point on the line,

r₀ = (0, 0, 0) is the position vector of the origin,

t is a parameter, and v is the direction vector

v = (2, 6, -1) - (0, 0, 0) = (2, 6, -1)

Now r(t) = (0, 0, 0) + t(2, 6, -1) = (2t, 6t, -t)

Therefore, the parametric equations of the line are:

x(t) = 2t

y(t) = 6t

z(t) = -t

To find the symmetric equations of the line, we can eliminate the parameter t from the parametric equations.

t = -z

Substituting this into the x(t) and y(t) equations, we get:

x = 2t = -2z

y = 6t = -6z

Therefore, the symmetric equations of the line are:

x/(-2) = y/(-6) = z/1

Hence, the symmetric equations of the line are: x/(-2) = y/(-6) = z/1

To learn more on Parametric equation click:

https://brainly.com/question/27959049

#SPJ2

Final answer:

Parametric equations for a line passing through the origin and a point are found by using the components of the point as coefficients for the parameter t. Symmetric equations are obtained by equating the ratios of each component to their corresponding coefficients in the direction vector.

Explanation:

To find the parametric equations for the line passing through the origin (0,0,0) and the point (2, 6, −1), we can use the position vector of the point (2,6,−1) and multiply it by the parameter t. This gives us the parametric equations:

The symmetric equations of the line can be obtained by eliminating the parameter t. Since t = x/2 = y/6 = z/(−1), the symmetric equations are:

x/2 = y/6 = z/(−1).

Answer:

The answer is true.

Step-by-step explanation:

Sample methods that embody random sampling are often termed probability sampling methods.

Yes this is a true statement.

Random sampling means picking up the samples randomly from a whole population with each sample having an equal chance of getting selected.

For example- selecting randomly 10 students from each class of a school, to survey for the food quality in school's cafeteria.

And this is a type of probability sampling methods. Other types are stratified sampling, cluster sampling etc.

Answer:   t= 2.032

Step-by-step explanation:

Given : Sample size : [tex]n=20[/tex]

Sample mean : [tex]\overline{x}=121[/tex]

Standard deviation : [tex]\sigma= 11[/tex]

Claim : The IQ scores of statistics professors are normally​ distributed, with a mean greater than 116.

Let [tex]\mu [/tex] be the mean scores of statistics professors.

Then the set of hypothesis for the given situation will be :-

[tex]H_0:\mu\leq116\\\\H_1:\mu>116[/tex]

As the alternative hypothesis is right tailed , thus the test would be right tail test.

Since the sample size is less than 30, therefore the test would be t-test .

The test statistics for the given situation will be :-

[tex]t=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

[tex]\Rightarrow\ t=\dfrac{121-116}{\dfrac{11}{\sqrt{20}}}=2.03278907045\approx2.032[/tex]

Hence, the value of the test statistic : t= 2.032

Answer:

Step-by-step explanation:

To prove a hypothesis, we have to use test statisticians like the z-value which is used in normally distributed data, and this is the case.

To calculate the z-value we use: [tex]z=\frac{x-u}{\frac{o}{\sqrt{n} } }[/tex]; where x is the sample mean, u is the population mean, o is the standard deviation and n the sample size.

Replacing all values:

[tex]z=\frac{121-116}{\frac{11}{\sqrt{20} } } =\frac{5}{2.44}=2.05[/tex]

Therefore the value of the test statistic is 2.05.

(It's important to clarify that the problem isn't asking about the hypothesis, or the probability value, it's just asking for the test parameter, which in this case is just a z-value).

Answer:

D. $850.64; $91.14

Step-by-step explanation:

Joyce Meadow pays her three workers $160, $470, and $800, respectively, per week.

So for 1 quarter, he will pay :

[tex]13\times160[/tex] = $2080

[tex]13\times470[/tex] = $6110

[tex]13\times800[/tex] = $10400

Given is that the base is $7,000.

It implies that the unemployment rate is to be paid on the first $7000 only.

Given rates are :

State rate of 5.6% and a federal rate of 0.6%.

State unemployment = [tex](0.056\times2080)+(0.056\times6110)+(0.056\times7000)[/tex] = $850.64

Federal unemployment = [tex](0.006\times2080)+ (0.006\times6110)+ (0.006\times7000)[/tex]  = $91.14

So, the correct option is D.

Answer:

[tex]a_{n}[/tex]=209.09 mg

Step-by-step explanation:

given: material= radium

half life= 1590 years

initial mass [tex]a_{0}[/tex] =500mg

we know that to calculate the amount left we use

[tex]a_{n}[/tex] = [tex]a_{0}[/tex][tex]\left ( 0.5\right )^{n}[/tex]

[tex]n=\frac{2000}{1590} = 1.2578[/tex]

therefore

[tex]a_{n}[/tex] = [tex]500\times0.5^{1.2578}[/tex]

[tex]a_{n}[/tex]=209.09058407921 mg

[tex]a_{n}[/tex]=209.09 mg amount left after 2000 years

Answer:

A) 2/10 or 20%

B) 3/10 or 30%

Step-by-step explanation:

There are 10 ice pops.

10 is going to be your denominator

Your numerator is going to be whatever amount of ice pops is.

After finding your fraction, divide the numerator by the denominator.

Lastly turn your decimal into a fraction.

2/10=0.2=20%

Hope I helped you.

Answer:

Samantha will need 13 boxes.

Step-by-step explanation:

Samantha wants to sort her greeting cards in boxes.

Each box can hold 24 cards.

Samantha has 312 greeting cards.

To calculate the number of boxes she need we have to divide total number of greeting cards to 24.

Therefore, = [tex]\frac{312}{24}[/tex]

                 = 13 boxes

Samantha will need 13 boxes.

Answer:

The scientific notation of 8,000,000 is 8 × 10^6

Step-by-step explanation:

* Lets explain the meaning of the scientific notation

- Scientific notation is a way of writing very large or very small numbers

- A number is written in scientific notation when a number between 1

 and 10 is multiplied by a power of 10

- Ex:  650,000,000 can be written in scientific notation as

        6.5 × 10^8

- We put a decimal points to make the number between 1 and 10 and

 then count how many places from right to left until the decimal point

 The decimal point between 6 and 5 to make the number 6.5 and

 there are 8 places from the last zero to the decimal point

* Lets solve the problem

∵ The exact value of 400,000 × 200 = 8,000,000

- Put the decimal point before 8 and count how many places from

 the last zero to the decimal point

∵ There are six places from last zero to the decimal point

∴ The scientific notation of 8,000,000 is 8 × 10^6

Answer:

Problem 1:

[tex]r=\frac{3V}{2 \pi h^2}[/tex]

Problem 2:

[tex]h=\frac{3V}{b^2}[/tex]

Problem 3:

The radius is [tex]\frac{25}{\pi}[/tex] cm.

Problem 4:

The width is 15 cm.

Step-by-step explanation:

Problem 1:

We want to solve [tex]V=\frac{2\pi rh^2}{3}[/tex] for [tex]r[/tex].

[tex]V=\frac{2\pi rh^2}{3}[/tex]

Multiply both sides by 3:

[tex]3V=2\pi r h^2[/tex]

Rearrange the multiplication using commutative property:

[tex]3V=2\pi h^2 \cdot r[/tex]

We want to get [tex]r[/tex] by itself so divide both sides by what [tex]r[/tex] is being multiplied by which is [tex]2\pi h^2[/tex].

[tex]\frac{3V}{2 \pi h^2}=r[/tex]

[tex]r=\frac{3V}{2 \pi h^2}[/tex]

Problem 2:

We want to solve for [tex]h[/tex] in [tex]V=\frac{b^2h}{3}[/tex].

Multiply both sides by 3:

[tex]3V=b^2h[/tex]

We want [tex]h[/tex] by itself so divide both sides by what [tex]h[/tex] is being multiply by; that is divide both sides by [tex]b^2[/tex].

[tex]\frac{3V}{b^2}=h[/tex]

[tex]h=\frac{3V}{b^2}[/tex]

Problem 3:

The circumference formula for a circle is [tex]2\pi r[/tex]. We are asked to solve for the radius when the circumference is [tex]50[/tex] cm.

[tex]2\pi r=50[/tex]

Divide both sides by what r is being multiply by; that is divide both sides by [tex]2\pi[/tex]:

[tex]r=\frac{50}{2\pi}[/tex]

Reduce fraction:

[tex]r=\frac{25}{\pi}[/tex]

The radius is [tex]\frac{25}{\pi}[/tex] cm.

Problem 4:  

The perimeter of a rectangle is [tex]2w+2L[/tex] where [tex]w[/tex] is the width and [tex]L[/tex] is the length.

We are asked to find w, the width, for when L, the length, is 5, and the perimeter is 40.

So we have this equation to solve for w:

[tex]40=2w+2(5)[/tex]

Simplify the 2(5) part:

[tex]40=2w+10[/tex]

Subtract both sides by 10:

[tex]30=2w[/tex]

Divide both sides by 2:

[tex]\frac{30}{2}=w[/tex]

Simplify the fraction:

[tex]15=w[/tex]

The width is 15 cm.

Answer:

So you have the vertical line passing through is x=4 and the horizontal line passing through is y=-6.

Step-by-step explanation:

In general the horizontal line passing through (a,b) is y=b and the vertical line passing through (a,b) is x=a.

So you have the vertical line passing through is x=4 and the horizontal line passing through is y=-6.

Answer:Yes

Step-by-step explanation:

Given

n=390 x=312

[tex]\hat{p}=\frac{312}{390}=0.8[/tex]

Confidence level=99 %

[tex] Z_{\frac{\alpha }{2}}=2.575[/tex]

Standard error(S.E.)=[tex]\sqrt{\frac{\hat{p}\left ( 1-\hat{p}\right )}{n}}[/tex]

S.E.=[tex]\sqrt{\frac{0.8\times 0.2}{390}}[/tex]

S.E.=0.0202

Confidence interval

[tex]p\pm \left [ z_{\frac{\alpha }{2}}\cdot S.E.\right ][/tex]

[tex]0.8 \pm 0.0521[/tex]

[tex]\left ( 0.7479,0.8521 \right )[/tex]

Since 0.5 does not lie in interval therefore method appear to be effective

a. The moduli are coprime, so you can apply the Chinese remainder theorem directly. Let

[tex]x=4\cdot5+3\cdot5+3\cdot4[/tex]

[tex]x=4\cdot5\cdot2+3\cdot5+3\cdot4[/tex]

[tex]x=4\cdot5\cdot2+3\cdot5\cdot3\cdot2+3\cdot4[/tex]

[tex]x=4\cdot5\cdot2+3\cdot5\cdot3\cdot2+3\cdot4\cdot3\cdot3[/tex]

[tex]\implies x=238[/tex]

By the CRT, we have

[tex]x\equiv238\pmod{3\cdot4\cdot5}\implies x\equiv-2\pmod{60}\implies\boxed{x\equiv58\pmod{60}}[/tex]

i.e. any number [tex]58+60n[/tex] (where [tex]n[/tex] is an integer) satisifes the system.

b. The moduli are not coprime, so we need to check for possible contradictions. If [tex]x\equiv a\pmod m[/tex] and [tex]x\equiv b\pmod n[/tex], then we need to have [tex]a\equiv b\pmod{\mathrm{gcd}(m,n)}[/tex]. This basically amounts to checking that if [tex]x\equiv a\pmod m[/tex], then we should also have [tex]x\equiv a\pmod{\text{any divisor of }m}[/tex].

[tex]x\equiv4\pmod{10}\implies\begin{cases}x\equiv4\equiv0\pmod2\\x\equiv4\pmod5\end{cases}[/tex]

[tex]x\equiv8\pmod{12}\implies\begin{cases}x\equiv0\pmod2\\x\equiv2\pmod3\end{cases}[/tex]

[tex]x\equiv6\pmod{18}\implies\begin{cases}x\equiv0\pmod2\\x\equiv0\pmod3\end{cases}[/tex]

The last congruence conflicts with the previous one modulo 3, so there is no solution to this system.

Answer:

6

Step-by-step explanation:

Sample space of the experiment

first number in the bracket is white die and second number in the bracket is black

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

As it can be seen that the first numbers in the bracket are (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

∴1 on the white die can be obtained in 6 ways

In the case of rolling two dice and trying to obtain a 1 on the white die, there are 6 ways to accomplish this because the black die outcome is irrelevant and it can show any number from 1 to 6 while pairing with a 1 on the white die.

The question asks about the probability of getting a specific result when rolling two fair dice, which is a problem in the realm of simple probability within mathematics.

Specifically, the question is focused on finding the number of ways to obtain a 1 on the white die.

When rolling two dice, there are a total of 6 different possible outcomes for the black die (since a standard die has 6 faces), and 1 specific outcome we're looking for on the white die, which is a 1.

Each outcome on the black die can be paired with a 1 on the white die, resulting in the combinations (1,1), (2,1), (3,1), (4,1), (5,1), and (6,1).

This gives us a total of 6 ways to achieve a 1 on the white die, regardless of what the black die shows.

Answer:

C. That order is taken into account

Answer:

D. That order is taken into account​ (consider rearrangements of the same items to be different​ sequences).

Step-by-step explanation:

[tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

For example:

Combinations of 2 from a set a,b,c: ab=ba, ac = ca, bc = cb, meaning that the order is not important(ab and ba is the same sequence).

aa, bb, cc are not valid, since there is no replacement.

So the answer for this qustion is:

D. That order is taken into account​ (consider rearrangements of the same items to be different​ sequences).

Answer:

(a) True , (b) False , (c) True , (d) False , (e) False , (f) True , (g) False

(h) True , (i) True , (j) False , (k) True

Step-by-step explanation:

* Lets explain how to solve the problem

(a) Two lines parallel to a third line are parallel (True)

- Their direction vectors are scalar multiplies of the direction of the 3rd

 line, then they are scalar multiples of each other so they are parallel

(b) Two lines perpendicular to a third line are parallel (False)

- The x-axis and the y-axis are ⊥ to the z-axis but not parallel to

  each other

(c) Two planes parallel to a third plane are parallel (True)

- Their normal vectors parallel to the  normal vector of the 3rd plane,

  so these two normal vectors are parallel to each other and the

  planes are parallel

(d) Two planes perpendicular to a third plane are parallel (False)

- The xy plane and yz plane are not parallel to each other but both

 ⊥ to xz plane

(e) Two lines parallel to a plane are parallel (False)

- The x-axis and y-axis are not parallel to each other but both parallel

  to the plane z = 1

(f) Two lines perpendicular to a plane are parallel (True)

- The direction vectors of the lines parallel to the normal vector of

  the plane, then they parallel to each other , so the lines are parallel

(g) Two planes parallel to a line are parallel (False)

- The planes y = 1 and z = 1 are not parallel but both are parallel to

  the x-axis

(h) Two planes perpendicular to a line are parallel (True)

- The normal vectors of the 2 planes are parallel to the direction of

  line, then they are parallel to each other so the planes are parallel

(i) Two planes either intersect or are parallel (True)

(j) Two lines either intersect or are parallel (False)

- They can be skew

(k) A plane and a line either intersect or are parallel (True)

- They are parallel if the normal vector of the plane and the direction

  of the line are ⊥ to each other , otherwise the line intersect the plane

  at the angle 90° - Ф

This question is based on the properties of lines and planes. Therefore, (a) True , (b) False , (c) True , (d) False , (e) False , (f) True , (g) False

, (h) True , (i) True , (j) False , (k) True.

We have to choose correct statement and marked true or false.

Lets solve the problem.

(a) Two lines parallel to a third line are parallel. (True)

Reason - The direction vectors are scalar multiple of the direction of the third  line, then they are scalar multiple of each other. So, they are parallel.

(b) Two lines perpendicular to a third line are parallel. (False)

Reason- As we know that, x-axis and the y-axis are perpendicular to the z-axis but not parallel to  each other.

(c) Two planes parallel to a third plane are parallel (True)

Reason- The normal vectors of planes are parallel to the normal vector of the third  plane. So, these two normal vectors are parallel to each other and the  planes are parallel.

(d) Two planes perpendicular to a third plane are parallel. (False)

Reason- x-y plane and y-z plane are not parallel to each other. But they are perpendicular to x-z plane.

(e) Two lines parallel to a plane are parallel. (False)

Reason - Both  x-axis and y-axis are not parallel to each other. But, parallel  to the plane z = 1.

(f) Two lines perpendicular to a plane are parallel. (True)

Reason - The direction vectors of the lines parallel to the normal vector of  the plane, then they parallel to each other , so the lines are parallel.

(g) Two planes parallel to a line are parallel. (False)

Reason- The planes y = 1 and z = 1 are not parallel, but they are parallel to the x-axis.

(h) Two planes perpendicular to a line are parallel. (True)

Reason- The normal vectors of the two planes are parallel to the direction of  line. So, they are parallel to each other. Hence, they are parallel.

(i) Two planes either intersect or are parallel (True)

(j) Two lines either intersect or are parallel (False)

Reason- They can also be  skew.

(k) A plane and a line either intersect or are parallel (True)

Reason- They are parallel, if the normal vector of the plane and the direction  of the line are perpendicular to each other, otherwise the line intersect the plane  at the angle 90° [tex]\theta[/tex].

For more details, please prefer this link:

https://brainly.com/question/24569174

Answer:

the solution is x = 0

Step-by-step explanation:

We know the fact that if the column of the square matrix is linearly independent then the determinant of the matrix is non zero.

Now, since the determinant of the matrix is not zero then the inverse of the matrix must exists.

Therefore, we have

[tex]A^{-1}A=I....(i)[/tex]

Now, for the equation Ax =0

multiplying both sides by [tex]A^{-1}[/tex]

[tex]A^{-1}Ax=A^{-1}\cdot0[/tex]

From equation (i)

[tex]Ix=0\\\\x=0[/tex]

Therefore, the solution is x = 0

The dad had 90 cents.

Multiply the 90 cents by 5%:

90 x 0.05 = 4.5 cents.

Subtract that from 90:

90 - 4.5 = 85.5 cents.

The lowest guess the son could say was 86 cents to be within 5%

Since the son guessed lower than that he did not get the money.

1. A dad holds five coins in his hand. He tells his son that if he can guess the amount of money he is holding within 5% error, he can have the money. The son guesses that dad is holding 81 cents. The dad opens his hand and displays 90 cents. Did the son guess close enough to get the money?

yes

Answer:

The interval is : (7.206 , 7.794)

Step-by-step explanation:

The mean is = 7.5

Standard deviation = 3

n = 400

At 95% confidence interval, the z score is 1.96

[tex]7.5+1.96(\frac{3}{\sqrt{400} } )[/tex]

And [tex]7.5-1.96(\frac{3}{\sqrt{400} } )[/tex]

[tex]7.5+0.294[/tex] and [tex]7.5-0.294[/tex]

So, the interval is : (7.206 , 7.794)

Answer:

. Let fX,Y(x,y) = 10xy^2 for 0 < x < y < 1 be the joint density function of the random pair (X, Y). (a) Obtain the marginal density f(y) of Y. (b) Obtain the conditional density fx|y(x|y) of X given Y = y. (c) Evaluate the conditional expectation of X, given Y=y

Y = y.

Step-by-step explanation:

Answer:

Cereal B

Step-by-step explanation:

Given are two different rates for cereals A and B.

as Cereal A: 15oz for $2.95 or Cereal B: 32oz for $5.95

As such we cannot compare unless we make it unit rate for same number of units

Let us find unit oz rates

Cereal A per oz= [tex]\frac{2.95}{15} =0.1967[/tex]dollars

Cereal B per oz = [tex]\frac{5.95}{32} =0.1859[/tex]dollars

Comparing unit rates per ounce,

we find that Cereal B per oz is lower.

Answer is Cereal B.

Answer:

214 ft

Step-by-step explanation:

Height of building = 94.9 ft

The angle of elevation of the top of the flagpole = θ₁ = 35.3°

The angle of elevation of the bottom of the flagpole = θ₂ = 26.2°

Let,

Height of building = x

Distance from observation point to base of building = y

[tex]tan 26.2 =\frac{x}{y}\\\Rightarrow y=\frac{x}{tan26.2}[/tex]

[tex]tan 35.3 =\frac{94.9+x}{y}\\\Rightarrow tan 35.3 =\frac{94.9+x}{\frac{x}{tan26.2}}\\\Rightarrow \frac{x}{tan26.2}tan35.3=94.9+x\\\Rightarrow \frac{tan35.3}{tan26.2}x-x=94.9\\\Rightarrow x=\frac{94.9}{\frac{tan35.3}{tan26.2}-1}\\\Rightarrow x=214.84/ ft[/tex]

I have used the exact values from the calculator.

∴ Height of the building is 214.84 ft

Answer:

The height of the building is 214.84 ft.

Step-by-step explanation:

Given information:

The height of the flagpole = 94.9 ft.

The angle of elevation of top = θ[tex]_1[/tex] = [tex]35.3^o[/tex]

The angle of elevation of bottom = θ[tex]_2=26.2^o[/tex]

If the height of building is [tex]x[/tex]

Then,

[tex]tan 26.2=x/y\\y=x/(tan26.2)\\[/tex]

And:

[tex]tan 35.3=(94.9+x)/y\\[/tex]

[tex]94.9+x=y \times tan35.3[/tex]

On putting the value in above equation:

[tex]x=\frac{94.9}{\frac{tan35.3}{tan26.2}-1 }[/tex]

solving the above equation:

[tex]x=214.84 ft.[/tex]

Hence, the height of the building is 214.84 ft.

For more information visit:

https://brainly.com/question/15274867?referrer=searchResults

Answer with explanation:

The equation of line is, y= -x +3

→x+y-3=0---------(1)

⇒Equation of line Parallel to Line , ax +by +c=0 is given by, ax + by +K=0.

Equation of Line Parallel to Line 1 is

  x+y+k=0

The Line passes through , (-5,6).

→ -5+6+k=0

→ k+1=0

→k= -1

So, equation of Line Parallel to line 1 is

x+y-1=0

⇒Equation of line Perpendicular  to Line , ax +by +c=0 is given by, bx - a y +K=0.

Equation of Line Perpendicular to Line 1 is

  x-y+k=0

The Line passes through , (-5,6).

→ -5-6+k=0

→ k-11=0

→k= 11

So, equation of Line Parallel to line 1 is

x-y+11=0

Answer:

5.195

Step-by-step explanation:

To solve the different operations of a mathematical expression there are rules that must always be followed, that is, there is a hierarchy of operations that must be respected.

The hierarchy of operations works as follows:

First, operations in brackets, brackets or braces are resolved independently.

Next, the powers and roots are realized.

It continues to solve multiplications and divisions.

Finally, the addition and subtraction are carried out, regardless of the order.

6 x 0.9 + 0.1 / 4 -0.23 =

First, multiplications and divisions:

5.4 + 0.025 - 0.23.

Then, addition and subraction:

5.195

Answer:

the correct answer is master data

Step-by-step explanation:

Enterprise system is a information system which provides a company with a wide integration and coordination regarding the important business processes  and also helps in providing seamless flow of information through out the company.

Master data  is a type of data in the enterprise system which is changed only occasionally , as this data includes all the information related to the customers like name, contact etc which helps a firm in analyzing their behavior and conduct high level research.

Answer:

0.7743

Step-by-step explanation:

Mean of age = u = 26 years

Standard Deviation = [tex]\sigma[/tex] = 4 years

We need to find the probability that the person getting married is in his or her twenties. This means the age of the person should be between 20 and 30. So, we are to find P( 20 < x < 30), where represents the distribution of age.

Since the data is normally distributed we can use the z distribution to solve this problem. The formula to calculate the z score is:

[tex]z=\frac{x-u}{\sigma}[/tex]

20 converted to z score will be:

[tex]z=\frac{20-26}{4}=-1.5[/tex]

30 converted to z score will be:

[tex]z=\frac{30-26}{4}=1[/tex]

So, now we have to find the probability that the z value lies between -1.5 and 1.

P( 20 < x < 30) = P( -1.5 < z < 1)

P( -1.5 < z < 1 ) = P(z < 1) - P(z<-1.5)

From the z-table:

P(z < 1) = 0.8413

P(z < -1.5) =0.067

So,

P( -1.5 < z < 1 ) = 0.8413 - 0.067 = 0.7743

Thus,

P( 20 < x < 30) = 0.7743

So, we can conclude that the probability that a person getting married for the first time is in his or her twenties is 0.7743