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1 year ago

What's the angle of x?

Mathematics

2 answers:

vichka [17]1 year ago

3 0

Answer:

60 as corresponding and alternate angles r equal

Step-by-step explanation:

romanna [79]1 year ago

3 0

Answer:

x = 60°

Step-by-step explanation:

here, the given lines are parallel,

so, x = 60°

( pair of alternate interior angles )

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Determine whether each of these sets is the power set of a set, where a and b are distinct elements.

Schach [20]

A. $\varnothing$ cannot be the power set of any set. Consult Cantor's theorem, which says that the cardinality of the power set of any set (even the empty set) is strictly greater than the cardinality of the set.

(No part b?)

c. Also not the power set of any set, because any power set must have $2^n$ elements, where $n$ is the cardinality of the original set. The cardinality of this set is 3, but there is no integer $n$ such that $2^n=3$ . This set would be a power set if $\{\varnothing\}$ (that is, the set containing the empty set) were a member of it.

4 0

2 years ago

Solve for z -c + 6z = tz +83

sladkih [1.3K]

-c +6z = tz+83

add c to each side

6z = tz +83 +c

subtract tz from each side

6z-tz = c+83

factor out z

z(6-t) = c+83

divide by (6-t)

z = (c+83)/(6-t)

Answer:z = (c+83)/(6-t)

6 0

2 years ago

Read 2 more answers

A group of three friends cycle from Llanfyllin to Vyrnwy at 12 km/h.

garik1379 [7]

The average speed for the journey from Llanfyllin to Bala is 14 km/h

Average speed is the ratio of total distance travelled to total time taken. Also, average speed can be given as the average of the different speeds.

Let d₁ represent the distance from Llanfyllin to Vyrnwy and t represent the time taken, since the speed is 12 km/h, hence:

12 = d₁/t

d₁ = 12t

Let d₂ represent the distance from Vyrnwy to bala and t represent the time taken. Since they spent the same time, since the speed is 16 km/h, hence:

16 = d₂/t

d₂ = 16t

Average speed = total distance/total time

Average speed = (d₁ + d₂) / (t + t)

Average speed = (12t + 16t) / 2t

Average speed = 28t / 2t

Average speed = 14 km/h

Hence their average speed for the journey from Llanfyllin to Bala is 14 km/h.

Find out more at: brainly.com/question/23774048

6 0

1 year ago

Steve made a sign for his room that is shaped like a triangle. It is 4.5 feet long and 3 feet high. He wants to make another sig

Fed [463]

Answer: PART A : $\frac{4.4}{x}$ = $\frac{3}{4}$

PART B : 5.9 FEET

Step-by-step explanation:

Length of the first sign = 4.4 feet

height of the first sign = 3 feet

Length of the second sign = x feet

height of the second sign = 4 feet

If two shapes are similar , then the ratio of their sides are equal,

That is ;

$\frac{h_{1}}{h_{2} }$ = $\frac{L_{1}}{L_{2} }$

PART A

$\frac{4.4}{x}$ = $\frac{3}{4}$

PART B

$\frac{4.4}{x}$ = $\frac{3}{4}$

cross multiplying , we have

3x = 4.4 x 4

3x = 17.6

Divide through by 3

x = 17.6/3

x = 5.86666666666667

x≈ 5.9 feet

Therefore , the length of the new sign is 5.9 feet

6 0

2 years ago

Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D = {(x, y) |

Bas_tet [7]

Answer:

$M=168k$

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

Step-by-step explanation:

Let's begin with the mass definition in terms of density.

$M=\int\int \rho dA$

Now, we know the limits of the integrals of x and y, and also know that ρ = ky², so we will have:

$M=\int^{9}_{1}\int^{4}_{1}ky^{2} dydx$

Let's solve this integral:

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}21dx$

$M=21k\int^{9}_{1}dx=21k*x|^{9}_{1}$

So the mass will be:

$M=21k*8=168k$

Now we need to find the x-coordinate of the center of mass.

$\bar{x}=\frac{1}{M}\int\int x*\rho dydx$

$\bar{x}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}x*ky^{2} dydx$

$\bar{x}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}x*y^{2} dydx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*\frac{y^{3}}{3}|^{4}_{1}dx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*21 dx$

$\bar{x}=\frac{21}{168}\frac{x^{2}}{2}|^{9}_{1}$

$\bar{x}=\frac{21}{168}*40=5$

Now we need to find the y-coordinate of the center of mass.

$\bar{y}=\frac{1}{M}\int\int y*\rho dydx$

$\bar{y}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}y*ky^{2} dydx$

$\bar{y}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}y^{3} dydx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{y^{4}}{4}|^{4}_{1}dx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{255}{4}dx$

$\bar{y}=\frac{255}{672}\int^{9}_{1}dx$

$\bar{y}=\frac{255}{672}8=\frac{2040}{672}$

$\bar{y}=\frac{85}{28}$

Therefore the center of mass is:

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

I hope it helps you!

3 0

2 years ago