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3 years ago

Factor completely 4x2 + 36X + 81

Mathematics

2 answers:

GarryVolchara [31]3 years ago

8 0

Answer:

The answer is (2x+9)(2x+9)

Step-by-step explanation:

Hope that helps:)

Serhud [2]3 years ago

3 0

<h3>I'll teach you how to factor 4x^2 + 36X + 81</h3>

------------------------------------------------------

4x^2 + 36X + 81

Rewrite 4x^2 + 36X + 81 as...

(2x)^2+2*2x*9+9^2

Apply perfect square formula:

(2x+9)^2

Your Answer Is (2x+9)^2

plz mark me as brainliest :)

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What is the length, in centimeters, of the hypotenuse of a right triangle with legs measuring V2 cm and 3 cm?

WINSTONCH [101]

Answer:

$\sqrt{13}$

Step-by-step explanation:

A is the hyp. And b/c are the sides

A²=b²+c²=3²+2²=9+4=13

Now take the square root of 13 as the answer

5 0

3 years ago

H is between points Q and R. QH=23 and HR=12. What is the length of QR

PSYCHO15rus [73]

Answer:

Step-by-step explanation:

Given:

GH = 23

HR = 12

Required:

Length big QR

SOLUTION:

Since H is a point in between points Q and R, points Q, H, R are collinear.

QH = 23

HR = 12

QH + HR = QR (segment addition postulate)

23 + 12 = QR (substitution)

35 = QR

Therefore, the length of QR is 35

3 0

4 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

1 year ago

Please help me <br>ineed it now

yanalaym [24]

Answer:

look at explanation

Step-by-step explanation:

To solve all of these questions you just need to scale. While scaling you can divide or multiply.

1. $\frac{180 km}{4 hrs}$ (divide by 4 to get unit rate)

unit rate = $\frac{45km}{1hr}$

2. If it takes one hour to travel 40 miles, then we don't have to scale or anything. We know $\frac{40 miles}{1 hr}$ , the answer is...

"It'll only take one hour to travel 40 miles. This is correct because the unit rate is $\frac{40 miles}{1 hr}$ . "

3. Knowing that the constant speed of the plane is $\frac{800 km}{1 hour\\}$ , we can scale then add half of the unit rate to get our answer.

$\frac{800 km}{1 hour\\}$ x 3 = $\frac{2400 km}{3 hrs}$ 800 x 3 = 2400

then add half of unit rate which is $\frac{400 km}{30 mins}$ $\sqrt[2]{800}$ = 400

so, our answer is $\frac{2800 km}{3.5 hours}$

4. $\frac{3km}{30 mins}$ <--- ( divided by 2) $\frac{6 km}{1 hr}$ ( multiplied x 2.5) ---> $\frac{15 km}{2.5 hrs}$

To get an easier way to find the answer, if possible you can scale back farther than the 1 hour mark.

Elmer would take 2 and a half hours to ride 15 km.

5. The boys speed is $\frac{4 km}{1 hr}$ . All we need to do is divide by 2. $\sqrt[2]{8}$ = 4

3 0

3 years ago

X + 5 = 24 whar values can be substituted for x to make the equation true

levacccp [35]

Answer:

x=19

Step-by-step explanation:

24-5=19

8 0

3 years ago

Read 2 more answers