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2 years ago

If f(x) is a third degree polynomial function, how many distinct complex roots are possible?

Mathematics

1 answer:

sasho [114]2 years ago

7 0

Answer:

first onei think

Step-by-step explanation: I think sorry if im wrong

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There were 178 old cars at the junkyard. Then a truck brought in some more cars. Now there are a total of 236 cars in the junkya

crimeas [40]

Answer:

<u>The correct answer is 58 cars.</u>

Step-by-step explanation:

Cars in the junkyard before the truck arrived with some = 178

Cars after the truck brought into the junkyard some more= 236

Cars brought in by the truck = 236 - 178

Cars brought into the junkyard by the truck = 58

<u>The number of cars brought by the truck is 58 cars.</u>

5 0

3 years ago

Samantha and her children went into a movie theater and will buy bags of popcorn and candies. Each bag of popcorn costs $6 and e

Juliette [100K]

Multiply the price of popcorn by the number of bags so 6b.

Multiply the price of candy by the number bought, so 3.25c

Add those together to get total :

6b + 3.25c

Now the most she can spend is 50 so set the equation to less than it equal to what she can spend:

6b + 3.25c <= 50

3 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 if $v = x ^ 2 1× v=1^ ×x ^ 2 = 1 #V$

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, 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 then $v = x ^ 2 w = x ^ 2 + 1\\(v + w) + z = (x ^ 2 + 2) + 1 = 3x ^ 2 + 1$

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, consequently $0 = O + v = (O × x ^ 2 + Ox + O) + (a × x ^ 2 + bx + c) = c × x ^ 2 + b ^ 2 + a$

= 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 use $z = 1 then v = x ^ 2 w = x ^ 2 + 1(v + w) + z = (x ^ 2 + 2) + 1 = 3x ^ 2 + 1v + (w + z) = x ^ 2 + (2x ^ 2 + 1) = x ^ 2 + 3\\Since 3x ^ 2 +1 ne x^ 2 +3.$ 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

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8 0

7 months ago

Tamara invests $8000 in two different accounts. The first account has a simple interest rate of 3% and the second account has a

Paraphin [41]

Answer:

$3200 at 3%
$4800 at 2%

Step-by-step explanation:

Interest earned is proportional to the interest rate, so if the interest earned is the same, the amounts invested must be inversely proportional to the interest rates. That is, for the 3% and 2% accounts, the ratio of money invested is 2:3.

In other words, 2/5 of the money ($3200) was invested at 3%, and 3/5 of the money ($4800) was invested at 2%.

_____

If you need an equation, you can let x represent the amount invested at the highest rate. Then 8000-x is the amount invested at the lower rate. For the interest in the two accounts to be equal, we have ...

3%·x = 2%·(8000-x) . . . . . the amounts of interest earned are the same

3/2·x = 8000 -x . . . . . . . . divide by 2%

5/2·x = 8000 . . . . . . . . . . . add x and simplify

x = 8000·(2/5) = 3200 . . . multiply by the inverse of the x coefficient

8000-x = 8000 -3200 = 4800 . . . . the amount invested at the lower rate

Tamara invested $3200 in the 3% account and $4800 in the 2% account.

_____

She earned $96 in each account for the year.

8 0

2 years ago

Please helpppppppp -2p-4=2

OlgaM077 [116]

p = -3

Step-by-step explanation:

-2p-4=2

add 4 both sides

-2p-4+4=2+4

-2p=6

divide -2 both sides

-2p/-2=6/-2

p = -3

hope it's helpful ❤❤❤

THANK YOU.

5 0

3 years ago