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Alex_Xolod [135]
3 years ago
10

A boat travels for 4 hours at a constant speed of 35.25 km/h.

Mathematics
2 answers:
Shkiper50 [21]3 years ago
7 0

Answer: C. 141

Step-by-step explanation:

I took the test literally 5 minutes ago. Mark brainliest plz

Black_prince [1.1K]3 years ago
3 0
Again set a proportion
x/4=35.25/1
=x=4(35.25)
x=141
Therefore the boat traveled 141 kilometres.
The answer is C.
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Kisachek [45]

In the first few steps for deriving the quadratic formula left side of the equation due to the distributive property for balancing the equation.

<h3>What is quadratic equation?</h3>

A quadratic equation is the equation in which the highest power of the variable is two.

Here, The first few steps in deriving the quadratic formula are shown in the table.

Use the substitution property of equality to solve it further,

-c=-ax² +bx

Now factor out the term a, to solve further as,

    - c = a (x² + b/a x)

for half of the b value and square it to determine the constant of the perfect square trinomial as,

     (b/2a)² = b²/4a²

Now  the distributive property needs to be applied to determine the value to add to the left side of the equation to balance the sides of the equation.

  -c + b²/4a² = a(x² + b/a x + b²/4a² )

Thus, the distributive property needs to be applied to determine the value to add to the left side of the equation to balance the sides of the equation.

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2 years ago
Slope of (-2,-1) and (5,-1) in fraction form
dangina [55]

Slope is 0

Step-by-step explanation:

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⇒ m = (-1 - -1)/(5 - -2)

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Determine,in each of the following cases, whether the described system is or not a group. Explain your answers. Determine what i
zheka24 [161]

Answer:

(a) Not a group

(b) Not a group

(c) Abelian group

Step-by-step explanation:

<em>In order for a system <G,*> to be a group, the following must be satisfied </em>

<em> (1) The binary operation is associative, i.e., (a*b)*c = a*(b*c) for all a,b,c in G </em>

<em>(2) There is an identity element, i.e., there is an element e such that a*e = e*a = a for all a in G </em>

<em> (3) For each a in G, there is an inverse, i.e, another element a' in G such that a*a' = a'*a = e (the identity) </em>

<em> </em>

If in addition the operation * is commutative (a*b = b*a for every a,b in G), then the group is said to be Abelian

(a)  

The system <G,*> is not a group since there are no identity.  

To see this, suppose there is an element e such that  

a*e = a

then  

a-e = a which implies e=0

It is easy to see that 0 cannot be an identity.

For example  

2*0 = 2-0 = 2

Whereas

0*2 = 0-2 = -2

So 2*0 is not equal to 0*2

(b)

The system <G,*> is not a group either.

If A is a matrix 2x2 and the determinant of A det(A)=0, then the inverse of A does not exist.

(c)

The table of the operation G is showed in the attachment.

It is evident that this system is isomorphic under the identity map, to the cyclic group

\mathbb{Z}_{5}

the system formed by the subset of Z, {0,1,2,3,4} with the operation of addition module 5, which is an Abelian cyclic group

We conclude that the system <G,*> is Abelian.

Attachment: Table for the operation * in (c)

4 0
3 years ago
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