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Elden [556K]
3 years ago
15

Violet is trying to start an Intramural Club at her school. The principal tells her she must get signatures from students to sho

w support. Each filled sheet contains 25 signatures. By Monday, she and her friend already have 6 sheets filled with signatures. The principal tells Violet she must have 7 more sheets filled with signatures. Write an equation for the number of signatures Violet will get. If she fills all of these, how many signatures will she get in all?
Mathematics
1 answer:
sesenic [268]3 years ago
6 0

Answer:

325

Step-by-step explanation:

Let s represent sheets

Let x represent number of signatures needed

s = 25 signatures

6s + 7s = x

Plug in the function

6(25) + 7(25)

= 150 + 175

= 325

x = 325

In total she will need 325 signatures

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marysya [2.9K]

Answer:

12. 5(5x+4)(5x-4)

13. 2(4v+5)(4v-5)

14. 5(3x+2)(3x-2)

Step-by-step explanation:

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
11 months ago
Ok part 2 hey hey it jassie can you answer all my parts and other people to ok here it is
MissTica
0.368 in expanded form would be...

0.000 + 0.300 + 0.060 + 0.008
or simply
.3 + .06 + .008
3 0
2 years ago
Read 2 more answers
Use function notation to write a recursive formula to represent the sequence: 3, 6, 9, …
dlinn [17]

Answer:

f(n) = f(n - 1) + 3

Step-by-step explanation:

Substitute $ n = 1, 2, 3,.. $ to get the recursive formula.

OPTION 1: f(n) = f(n - 1) + 3

Substituting n = 1.

f(1) = f(1 - 1) + 3 = 0 + 3 = 3.

Substituting n = 2.

f(2) = f(2 - 1) + 3 = f(1) + 3 = 3 + 3 = 6.

Substituting n = 3.

f(3) = f(3 - 1) + 3 = f(2) + 3 = 6 + 3 = 9.

The numbers match the given sequence. So, we say the above recursive formula represents the sequence.

OPTION 2: f(n) = f(n - 1) + 2

Substituting n = 1

f(1) = f(0) +  2 $ \ne $ 3.

So, this is eliminated.

Similarly, OPTION 3 and OPTION 4 can be eliminated as well.

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3 years ago
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