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IRISSAK [1]
3 years ago
12

Check each graph below that represents a function

Mathematics
2 answers:
jeyben [28]3 years ago
5 0

Answer:

Graphs A, B, and C

Step-by-step explanation:

got it right on e2020

Sholpan [36]3 years ago
5 0

Answer:

the answer is

<em>- </em><u><em>A</em></u><em> </em>

<u><em>- B</em></u>

<u><em>- C</em></u>

Step-by-step explanation:

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What is the equation of the line that goes through the point (6,-1) and is parallel to the line represented by the equation belo
stepladder [879]

\huge{\boxed{y=-\frac{5}{6} x+4}}

Parallel lines share the same slope, so the slope of the parallel line in this case must be -\frac{5}{6}.

Point-slope form is y-y_1=m(x-x_1), where m is the slope and (x_1, y_1) is any known point on the line.

Plug in the values. y-(-1)=-\frac{5}{6} (x-6)

Simplify and distribute. y+1=-\frac{5}{6} x+5

Subtract 1 from both sides. \boxed{y=-\frac{5}{6} x+4}

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3 years ago
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Consider −75/3. Which TWO statements are correct?
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B.) the quotient is -25
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2 years ago
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a set v is given, together with definitions of addition and scalar multiplication. determine which properties of a vector space
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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

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8 0
10 months ago
Kelli and her family went to the beach for a vacation. They drove 293 miles in 7 hours to get there. ABOUT how many miles did th
12345 [234]

Answer:

41.86 miles each hour

Step-by-step explanation:

Kelli and her family went to the beach for a vacation. They drove 293 miles in 7 hours to get there. ABOUT how many miles did they drive each hour?

This Question is calculated as:

7 hours = 293 miles

1 hour = x miles

Cross Multiply

7 hours × x = 1 × 293 miles

x = 293 miles/7 hours

x = 41.857142857 miles per hour

x ≈ 41.86 miles per hour

Hence, they drove 41.86 miles each hour

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2 years ago
Which expression is equivalent to the following complex fraction?<br>1-1/x/2<br>​
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<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>att</em><em>ached</em><em> </em><em>picture</em>

<em>hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em>

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