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3 years ago

Find the length of side x to the nearest tenth.

Mathematics

1 answer:

Inessa [10]3 years ago

7 0

The answer is the square root of 18 because of Pythagorean theorem

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F(x)=x²-x-1 What is the average rate of change of f over the interval -1 ≤ x ≤ 1?

Elena-2011 [213]

Answer: -1/2

Step-by-step explanation:

The average rate of change is $\frac{f(b)-f(a)}{b-a}$ . Since a=-1 and b=1. We can plug them into the equation and solve.

$\frac{[1^2-1-1][(-1)^2-(-1)-1)]}{1-(-1)} =\frac{(-1)(1)}{2} =-\frac{1}{2}$

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 :

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8 0

1 year ago

Subtract polynomials

antoniya [11.8K]

Answer:

$2a^{2} +4$

Step-by-step explanation:

we have:

7a^2+4a+8 -5a^2 -4a -4

2a^2 +4

5 0

4 years ago

The tangent ration! helpp!!

Vitek1552 [10]

Answer:

13.5

Step-by-step explanation:

Given

tan34° = $\frac{x}{20}$ ( multiply both sides by 20 )

20 × tan34° = x , then

x ≈ 13.5 ( to the nearest tenth )

7 0

3 years ago

Points A, B, and C are collinear and AB:AC=2:5. Point A is located at (-3,-4), point B is located at (m,n) and point C is locate

patriot [66]

Answer:

m = -11/7

n = -8/7

Step-by-step explanation:

Here, what we want to calculate is the coordinates of the point B

Let’s consider the straight line as A B C

we can see that actually, the point B divides the line into the ratio 2:5 since AB:AC is 2:5

Thus, by using the internal division formula (section formula) with the ratio of division, we can get the coordinates of point B

Mathematically, the formula for internal division is given as follows;

B = ( (mx2 + nx1) /m + n , (my2 + ny1)/( m + n))

From the question;

m:n = 2:5

Thus m = 2, n = 5

(x1,y1) are coordinates of A

x1 = -3 , y1 = -4

(x2, y2) are coordinates of C

x2 = 2 , y2 = 6

Thus, substituting these values into the equation we have;

B ={ 2(2) + 5(-3)}/(2+5) , {2(6) + 5(-4)}/(2+5)

B = (-11/7, -8/7)

6 0

3 years ago