All categories

3 years ago

which of the following are among the five basic postulates of euclidean geometry? check all that apply

1 answer:

7 0

The five essential hypothesizes of Geometry, additionally alluded to as Euclid's proposes are the accompanying:
1.) A straight line section can be drawn joining any two focuses.
2.) Any straight line portion can be expanded uncertainly in a straight line.
3.) Given any straight line fragment, a circle can be drawn having the portion as a span and one endpoint as the inside.
4.) All correct points are harmonious.
5.) If two lines are drawn which meet a third such that the total of the internal points on one side is under two right edges (or 180 degrees), then the two lines unavoidably should converge each other on that side if reached out sufficiently far.

You might be interested in

Use cylindrical coordinates to evaluate the triple integral ∭ where E is the solid bounded by the circular paraboloid z = 9 - 16

4vir4ik [10]

Answer:

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

Step-by-step explanation:

The Cylindrical coordinates are:

x = rcosθ, y = rsinθ and z = z

From the question, on the xy-plane;

$9 -16 (x^2 + y^2) = 0 \\ \\ 16 (x^2 + y^2) = 9 \\ \\ x^2+y^2 = \dfrac{9}{16}$

$x^2+y^2 = (\dfrac{3}{4})^2$

where:

0 ≤ r ≤ $\dfrac{3}{4}$ and 0 ≤ θ ≤ 2π

∴

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} \int ^{9-16r^2}_{0} \ r \times rdzdrd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 z|^{z= 9-16r^2}_{z=0} \ \ \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 ( 9-16r^2}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} ( 9r^2-16r^4}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( \dfrac{9r^3}{3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3r^3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) d \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) \theta |^{2 \pi}_{0}$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{64}}-\dfrac{243}{320}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{160}})2 \pi$

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

4 0

3 years ago

Ron Prentice bought goods from Shelly Katz. On May 8, Shelly gave Ron a time extension on his bill by accepting a $3,000, 8%, 18

Bad White [126]

Answer:

vsfyvsy stop! me off do is ask if em is everything ok well if the if the if the yr

7 0

3 years ago

What is the probability that if you multiply two randomly-selected two-digit whole numbers, the result is greater than 100 and l

ohaa [14]

Answer:

4/675

Step-by-step explanation:

There can be 90 two-digit numbers ranging from 10 to 99. There will be

90 x 90= 8100 possibilities of randomly selecting and combining 2 entire two-digit numbers, if we find ax b to be distinct from bx a. When 10 is first chosen, there may be 9 two-digit numbers that could be combined within the required range for a product When 11 is chosen first, then the second two-digit number has 9 possibilities. 12 has seven options; 13 has six options; 14 has five options; 15 has four options; 16 has three options; 17 has two options; 18 has 2 options; and 19 has one option. It provides us 48 total choices so the likelihood that the combination of two randomly chosen two-digit whole numbers is one of theses these possibilities is thus 48/8100 = 4/675.

8 0

3 years ago

3. B С A D E How many rays intersect at point o?

swat32

Oneeeeeeeeeeeeeeeeeee

8 0

3 years ago

Read 2 more answers

Please could I have some help :)

nirvana33 [79]

Answer:

a) x = 128 degrees

b) Angle APD is the arc angle, which is equal to the central angle x subtended by the arc. Therefore angle APD = 128 degrees (and not 116 degrees)

Step-by-step explanation:

Given:

attached diagram

ABC is a straight line

Solution:

a) Find x

ABC is a straight line

angle ABD = supplement of CBD = 180-CBD = 180-116 = 64 degrees.

x is the central angle of the arc APD

so angle ABD is the inscribed angle which equals half of the arc angle =>

angle ABD = x/2 = 64 degrees

Solve for x

x/2 = 64

x = 2*64

x = 128 degrees

Angle APD is the arc angle, which is equal to the central angle x subtended by the arc. Therefore angle APD = 128 degrees (and not 116 degrees)

5 0

3 years ago

Read 2 more answers