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Elanso [62]
3 years ago
9

What is thissssssssssssssssss pls help

Mathematics
2 answers:
stiv31 [10]3 years ago
6 0

Answer:

add 5 to both sides

Step-by-step explanation:

thats it the reason why

lesantik [10]3 years ago
5 0

Answer:

1st option will be the answer.

You might be interested in
Sin(60-theta)sin(60+theta)​
ehidna [41]

Step-by-step explanation:

\sin(60 -  \theta)  \sin(60 +  \theta)  \\  =  \{ \sin(60)  \cos( \theta)  -  \sin( \theta)   \cos(60)  \} \{ \sin(60)  \cos( \theta)  +  \cos(60)  \sin( \theta)  \} \\  =  \{ \frac{ \sqrt{3} }{2}  \cos( \theta)  -  \frac{1}{2}  \sin( \theta)  \} \{ \frac{ \sqrt{3} }{2}  \cos( \theta)  +  \frac{1}{2}  \sin( \theta)  \} \\  \\

from difference of two squares:

{ \boxed{(a - b)(a + b) = ( {a}^{2} -  {b}^{2} ) }}

therefore:

=  \{ {( \frac{ \sqrt{3} }{2}) }^{2}  { \cos }^{2}  \theta \} -  \{ {( \frac{ \sqrt{3} }{2} )}^{2}  { \sin }^{2}  \theta \} \\  \\  =  \frac{3}{4}  { \cos }^{2}  \theta -  \frac{3}{4}  { \sin}^{2}  \theta

factorise out ¾ :

=  \frac{3}{4} ( { \cos }^{2}  \theta  -  { \sin}^{2}   \theta) \\  \\  = { \boxed{ \frac{3}{4}  \cos(2 \theta) }}

3 0
3 years ago
Read 2 more answers
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
Convert 75 percent to a decimal
Crank

To convert to a decimal, divide by 100.

75 / 100 = .75

4 0
4 years ago
Read 2 more answers
HeLP!! will mark brainlest!
shtirl [24]
Perimeter=54 a.2ft b.9 missing length=4 a.light rectangle=15 dark rectangle=10 b. sum=25 5*5=25
4 0
3 years ago
Please help me!
Taya2010 [7]

Answer:

  • x = 12
  • Side = 23 units

Step-by-step explanation:

<u>STU is an equilateral triangle. And:</u>

  • ST = 2x - 1
  • SU = 5x - 37
  • TU = x + 11

<u>All sides are equal. Then take one pair and solve for x:</u>

  • 2x - 1 = 5x - 37
  • 3x = 36
  • x = 12

<u>Substitute x to find the side length:</u>

  • 2x - 1 = 5x - 37 = x + 11 = 23
8 0
3 years ago
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