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erma4kov [3.2K]
3 years ago
9

Which of the following statements are always true when a transversal crosses parallel lines? 1.Several congruent angles are form

ed. 2.Vertical angles are formed. 3.Complementary angles are formed. 4.Supplementary angles are formed. 5.Obtuse angles are formed.
Mathematics
2 answers:
slavikrds [6]3 years ago
6 0
Statements that are always true are:1. Several congruent angles are formed.4. Supplementary angles are formed.Other statements are not always true when a transversal crosses parallel lines.

Vikentia [17]3 years ago
5 0

Answer:

Several congruent angles are formed.

Vertical angles are formed.

Supplementary angles are formed.

Obtuse angles are formed.

Step-by-step explanation:

1,2,4,5

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X2 + 5x - 14 Select one of the factors of the quadratic expression.
vfiekz [6]
The equation factored would be (x+7)(x-2). As 7-2=5 and (7)(2)=14.
3 0
3 years ago
PLS HELP ASAP IM TAKING A TEST THANK YOU!!! :) <3
KatRina [158]

Answer:

Runner 1

Step-by-step explanation:

He ran 1 mile in 10 mins

3 0
2 years ago
A wireless company offers two cell phone plans. For the month of September, Plan A charges $35 plus $0.25 per minute for calls.
valentina_108 [34]

Answer:

60 minutes

Step-by-step explanation:

Let the number of minutes be represented as x

For Plan A

Plan A charges $35 plus $0.25 per minute for calls.

$35 + $0.25 × x

35 + 0.25x

For Plan B

Plan B charges $20 plus $0.50 per minute for calls.

$20 + $0.50 × x

20 + 0.50x

For what number of minutes do both plans cost the same amount?

This is calculated by equating Plan A to Plan B

Plan A = Plan B

35 + 0.25x = 20 + 0.50x

Collect like terms

35 - 20 = 0.50x - 0.25x

15 = 0.25x

x = 15/0.25

x = 60 minutes.

Hence, the number of minutes that both plans cost the same amount is 60 minutes

7 0
3 years ago
a set v is given, together with definitions of addition and scalar multiplication. determine which properties of a vector space
agasfer [191]

The properties of a vector space are satisfied Properties 1,2, 5(a) and 5(c) are satisfied, the relaxation of the homes aren't legitimate are ifv = x ^ 2 1× v=1^ ×x ^ 2 = 1 #V

Property three does now no longer follow: Suppose that Property three is legitimate, shall we namev = a * x ^ 2 +bx +cthe neuter of V. Since v is the neuter, then O have to be constant with the aid of using the neuted, consequently 0 = O + v = (O  x ^ 2 + Ox + O) + (a × x ^ 2 + bx + c) = c × x ^ 2 + b ^ 2 + a

= 0 If O is the neuter, then it ought to restore x², but 0+ x² = (0x²+0x+zero) + (x²+0x+zero) = 1.This is a contradiction due to the fact x² isn't 1. We finish that V doesnt have a neuter vector. This additionally method that belongings four would not observe either. A set with out 0 cant

have additive inverse

Let r= v ×2x ^ 2 + v × 1x +v0 , w= w ×2x ^ 2 + w × 1x +w0 . We have that\\v+w= (vO + wO) ^  x^ 2 +(vl^ × wl)^  x+ ( v 2^ × w2)• w+v= (wO + vO) ^x^ 2 +(wl^ × vl)x+ ( w 2^ ×v2)

Since the sum of actual numbers is commutative, we finish that v + w = w + v Therefore, belongings 5(a) is valid.

Property 5(b) isn't valid: we are able to introduce

a counter example. we could use z = 1 thenv = x ^ 2 w = x ^ 2 + 1\\(v + w) + z = (x ^ 2 + 2) + 1 = 3x ^ 2 + 1

v + (w + z) = x ^ 2 + (2x ^ 2 + 1) = x ^ 2 + 3

Since 3x ^ 2 +1 ne x^ 2 +3. then the associativity rule doesnt hold.

(1+2)^ * (x^ 2 +x)=3^ * (x ^ 2 + x) = 3x + 3\\1^ × (x^ 2 +x)+2^ × (x ^ 2 + x) = (x + 1) + (2x + 2) = 3x ^ 2 + x ( ne 3x + 3 )\\(1^ ×2)^ ×(x^ 2 +x)=2^ × (x ^ 2 + x) = 2x + 2\\1^ × (2^ × (x ^ 2 + x) )=1^ × (2x+2)=2x^ 2 +2x( ne2x+2)

Property f doesnt observe because of the switch of variables. for instance, if v = x ^ 2 1 × v=1^ × x ^ 2 = 1 #V

Properties 1,2, 5(a) and 5(c) are satisfied, the relaxation of the homes arent legitimate.

Step-with the aid of using-step explanation:

Note that each sum and scalar multiplication entails in replacing the order from that most important coefficient with the impartial time period earlier than doing the same old sum/scalar multiplication.

Property three does now no longer follow: Suppose that Property three is legitimate, shall we name v = a × x ^ 2 +bx +c the neuter of V. Since v is the neuter, then O have to be constant with the aid of using the neuted, consequently0 = O + v = (O × x ^ 2 + Ox + O) + (a × x ^ 2 + bx + c) = c × x ^ 2 + b ^ 2 + a

= zero If O is the neuter, then it ought to restore x², but zero + x² = (0x²+0x+zero) + (x²+0x+zero) = 1.This is a contradiction due to the fact x² isn't 1. We finish that V doesnt have a neuter vector. This additionally method that belongings four would not observe either. A set with out 0 cant have additive inverse

Let r= v × 2x ^ 2 + v × 1x +v0 , w= w2x ^ 2 + w × 1x +w0 . \\We have thatv+w= (vO + wO) ^ x^ 2 +(vl^ wl)^x+ ( v 2^ w2)w+v= (wO + vO) ^ x^ 2 +(wl^ vl)x+ ( w 2^v2)

Since the sum of actual numbers is commutative, we finish that v + w = w + v Therefore, belongings 5(a) is valid.

Property 5(b) isn't valid: we are able to introduce

a counter example. we could usez = 1 then v = x ^ 2 w = x ^ 2 + 1(v + w) + z = (x ^ 2 + 2) + 1 = 3x ^ 2 + 1v + (w + z) = x ^ 2 + (2x ^ 2 + 1) = x ^ 2 + 3\\Since 3x ^ 2 +1 ne x^ 2 +3.then the associativity rule doesnt hold.

Note that each expressions are same because of the distributive rule of actual numbers. Also, you could be aware that his assets holds due to the fact in each instances we 'switch variables twice.

· (1+2)^ * (x^ 2 +x)=3^ * (x ^ 2 + x) = 3x + 31^ * (x^ 2 +x)+2^ * (x ^ 2 + x) = (x + 1) + (2x + 2) = 3x ^ 2 + x ( ne 3x + 3 )(1^ * 2)^ * (x^ 2 +x)=2^ * (x ^ 2 + x) = 2x + 21^ * (2^ * (x ^ 2 + x) )=1^ ×* (2x+2)=2x^ 2 +2x( ne2x+2)

Read more about polynomials :

brainly.com/question/2833285

#SPJ4

8 0
1 year ago
Complete the square to solve 4x^2 + 24x = 4
dalvyx [7]

Answer:

4x^2+24x=4

Step 1:

make the coefficient of x² as 1, divide both sides by 4

x² +6x =1

Step 2:

take half of coefficient of x, square it and add it to both sides

half of 6 is 3 and 3²=9, so add 9 to both sides

x² +6x+9=1+9

x²+6x+3²=10

(x+3)²=10

now when we take square root we will get both + and - √10 on right side

x+3= +√10

x+3=-√10

solving for x

x=+√10-3

Step-by-step explanation:

4 0
3 years ago
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