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almond37 [142]
3 years ago
6

3x+1 = 27 answer choices: a) -1/3 b) -6 C) -2/3 d) -3/2

Mathematics
1 answer:
DochEvi [55]3 years ago
8 0
None of those are correct. x= 8 and 2/3
You might be interested in
21/33 Marks
Marrrta [24]

Answer:

ooooooooooooooooooooooooooooooooo

Step-by-step explanation:

la de da de da la de da de deee sudennly you call my name an I lose my weight and I float up to the sky!

8 0
3 years ago
Given QT = SR, QV = SU, and the diagram, prove that triangles QUT and SVR are congruent. Write a paragraph proof.
QveST [7]

Answer:

Triangles QUT and SVR are congruent because the defining two sides and an included angle of triangles QUT and SVR are equal

Step-by-step explanation:

Here we have QT = SR and

QV = SU

Therefore,

QT = √(UT² + QU²)........(1)

RS = √(VS² + RV²)..........(2)

Since QS = QU + SU = QV + VS ∴ QU = VS

Therefore, since SR = QT and QU = VS, then from (1) and (2), we have UT = RV

Hence since we know all sides of the triangles QUT and SVR are equal and we know that the angle in between two congruent sides of the the triangles QUT and SVR that is the angle in between sides QU and UT for triangle QUT and the angle in between the sides RV and VS in triangle SVR are both equal to 90°, therefore triangles QUT and SVR are congruent.

5 0
3 years ago
Based on the data given in the picture, calculate the area of the car track...
Verizon [17]

Refer to the diagram below. We need to find the areas of the green and blue regions, then subtract to get the area of the orange track only.

The larger green region is composed of a rectangle of dimensions 200 meters by 4+42+4 = 50 meters, along with two semicircles that combine to make a full circle. This circle has radius 25 meters.

The green rectangle has area 200*50 = 10000 square meters. The green semicircles combine to form an area of pi*r^2 = pi*25^2 = 625pi square meters. In total, the full green area is 10000+625pi square meters. I'm leaving things in terms of pi for now. The approximation will come later.

The blue area is the same story, but smaller dimensions. The blue rectangle has dimensions 200 meters by 42 meters, so its area is 200*42 = 8400 square meters. The blue semicircular pieces combine to a circle with area pi*r^2 = pi*21^2 = 441pi square meters. In total, the blue region has area 8400+441pi square meters.

After we figure out the green and blue areas, we subtract to get the orange region's area, which is the area of the track only.

orange area = (green) - (blue)

track area = (10000+625pi) - (8400+441pi)

track area = 10000+625pi - 8400-441pi

track area = (10000-8400) + (625pi - 441pi)

track area = 184pi + 1600 is the exact area in terms of pi

track area = 2178.05304826052 is the approximate area when you use the pi constant built into your calculator. If you use pi = 3.14 instead, then you'll get 2177.76 as the approximate answer. I think its better to use the more accurate version of pi. Of course, be sure to listen/follow your teachers instructions.

4 0
3 years ago
(a) Let R = {(a,b): a² + 3b <= 12, a, b € z+} be a relation defined on z+)
grin007 [14]

Answer:

R is an equivalence relation, since R is reflexive, symmetric, and transitive.

Step-by-step explanation:

The relation R is an equivalence if it is reflexive, symmetric and transitive.

The order to options required to show that R is an equivalence relation are;

((a, b), (a, b)) ∈ R since a·b = b·a

Therefore, R is reflexive

If ((a, b), (c, d)) ∈ R then a·d = b·c, which gives c·b = d·a, then ((c, d), (a, b)) ∈ R

Therefore, R is symmetric

If ((c, d), (e, f)) ∈ R, and ((a, b), (c, d)) ∈ R therefore, c·f = d·e, and a·d = b·c

Multiplying gives, a·f·c·d = b·e·c·d, which gives, a·f = b·e, then ((a, b), (e, f)) ∈R

Therefore R is transitive

From the above proofs, the relation R is reflexive, symmetric, and transitive, therefore, R is an equivalent relation.

Reasons:

Prove that the relation R is reflexive

Reflexive property is a property is the property that a number has a value that it posses (it is equal to itself)

The given relation is ((a, b), (c, d)) ∈ R if and only if a·d = b·c

By multiplication property of equality; a·b = b·a

Therefore;

((a, b), (a, b)) ∈ R

The relation, R, is reflexive.

Prove that the relation, R, is symmetric

Given that if ((a, b), (c, d)) ∈ R then we have, a·d = b·c

Therefore, c·b = d·a implies ((c, d), (a, b)) ∈ R

((a, b), (c, d)) and ((c, d), (a, b)) are symmetric.

Therefore, the relation, R, is symmetric.

Prove that R is transitive

Symbolically, transitive property is as follows; If x = y, and y = z, then x = z

From the given relation, ((a, b), (c, d)) ∈ R, then a·d = b·c

Therefore, ((c, d), (e, f)) ∈ R, then c·f = d·e

By multiplication, a·d × c·f = b·c × d·e

a·d·c·f = b·c·d·e

Therefore;

a·f·c·d = b·e·c·d

a·f = b·e

Which gives;

((a, b), (e, f)) ∈ R, therefore, the relation, R, is transitive.

Therefore;

R is an equivalence relation, since R is reflexive, symmetric, and transitive.

Based on a similar question posted online, it is required to rank the given options in the order to show that R is an equivalence relation.

Learn more about equivalent relations here:

brainly.com/question/1503196

4 0
3 years ago
The sum of nine-tenths and g is two-thirds solve for g
Naya [18.7K]

Answer:

- 7/30

Step-by-step explanation:

9/10 + g = 2/3

So subtract 9/10 from both sides of the equation to get g = -7/30

Thank me later :D

4 0
3 years ago
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