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1 year ago

A shape is rotated 180° about the origin and then its image is reflected

Mathematics

1 answer:

Blizzard [7]1 year ago

4 0

See it up on brainly it should give you your answer that you need!!

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Kenji earned the test scores below in English class.

vitfil [10]

Answer:

Step-by-step explanation:

Mean is all the numbers divided by the amount of numbers listed and Median is what ever number is in the middle (when in order of value). when there's not a exact middle number u add the two numbers in the middle and divide it by two

8 0

3 years ago

Number 10 please please someone please help please please help

Tanzania [10]

Answer: A compound inequality contains at least two inequalities that are separated by either "and" or "or". The graph of a compound inequality with an "and" represents the intersection of the graph of the inequalities. A number is a solution to the compound inequality if the number is a solution to both inequalities.

3 0

2 years ago

Solve this system of linear equations. Separate

Elanso [62]

Answer:

(-3, -3)

Step-by-step explanation:

1.) Rewrite the second equation so 3y is on one side of the equation:

3y=6+5x

2.) Substitute the given value of 3y (replacing 3y with 6+5x, since we know they equal each other) into the equation 17x=-60-3y

Should end up with this:

17x=-60-(6+5x)

3.) Solve 17x=-60-(6+5x)

Calculate Difference: 17x=-66-5x

Combine Like Terms: 22x = -66

Divided both sides by 22 to isolate and solve for x: -3

So We know x=-3, now we got to find the y value. We can use either the first or second equation to find y value, so lets use the second.

3y=6+5x

1.) We know that x=-3, so we can simply substitute x in the equation

3y=6+5x with -3

3y=6+5(-3)

2.) Solve 3y=6+5(-3)

Combine Like Term: 3y=6+-15

Combine Like Term Even More: 3y= -9

Divide by 3 on both sides to isolate and solve for y: y=-3

So now we know y=-3 and once again we know x=-3, so if we format that

(-3,-3)

^ ^

x y

4 0

2 years ago

There is a spinner with 15 equal areas, numbered 1 through 15. If the spinner is spun

Brrunno [24]

Answer:

1/3 and 1.5

Step-by-step explanation:

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15

There are 5 Multiples of 3 on the Spinner and 3 multiples of four.

5/15 is the probability for a multiple of 3 which can be simplified to 1/3

3/15 is the probability for a multiple of 4 which can be simplified to 1/5

Hope this Helps!

5 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.

$· (1+2)^ * (x^ 2 +x)=3^ * (x ^ 2 + x) = 3x + 31^ * (x^ 2 +x)+2^ * (x ^ 2 + x) = (x + 1) + (2x + 2) = 3x ^ 2 + x ( ne 3x + 3 )(1^ * 2)^ * (x^ 2 +x)=2^ * (x ^ 2 + x) = 2x + 21^ * (2^ * (x ^ 2 + x) )=1^ ×* (2x+2)=2x^ 2 +2x( ne2x+2)$