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3 years ago

Who already learned algebra ?

2 answers:

hjlf3 years ago

4 0

The equation is if f(1)=7 and f(n)=f(n-1)-4, then find f(6)
basically we see that to get a term (f(n)), we subtraction 4 from the prevous term (f(n-1))

so see how many times we need to subtract

6th term, so 5 times (don't subtract of first tiem)
5*4=20
7-20=-13

f(6)=-13

beks73 [17]3 years ago

3 0

I did im in algebra 1 now

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Let C be the collection of all lines in the plane. Define a relation on C by saying that two lines are equivalent if and only if

pav-90 [236]

Answer:

Recall that a equivalence relation must satisfy three conditions: it must be symmetric, reflexive and transitive. Let us study each one of them.

Reflexive: Here we have to prove that an element is in relation with itself. In this particular case, we need to show that every line is parallel or equal to itself, which is obvious.

Symmetric: Here we have to show that if one element is in relation with another, the reciprocal is also true. So, in this case, we want to show that if the line r is in relation with s, then the line s is in relation with r. So, as r is in relation with s, they are equal or parallel.

If they are equal, r=s implies s=r directly. If they are parallel, from elementary geometry we know that r║s means that s║r too. So, the relation is reflexive.

Transitive: Here we want to show that if r is in relation with s, and s is in relation with t, then r is in relation with t. Let us assume that the three lines are parallel, because if we have equality between two of them the statement is true directly from the above example. If the three are parallel and distinct, we know from elementary geometry that if r║s and s║t then r║t. So, the relation is transitive.

Therefore, as the relation is transitive, symmetric and reflexive is an equivalence relation.

6 0

4 years ago

4. Ms. Renner is having a cake party for her

fgiga [73]

27 divided by 5 = 5.4 so if you round up you get 6. She’ll need to buy 6 cakes for everyone to have a piece of cake

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3 years ago

Two equations are given below:

Ostrovityanka [42]

A - 3b = 4 . . . . (1)
a = b - 2 . . . . . (2)
Putting (2) into (1), we have
b - 2 - 3b = 4
-2b = 4 + 2 = 6
b = 6/-2 = -3
From (1), a = b - 2 = -3 - 2 = -5
Hence, solution = (-5, -3)

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3 years ago

What is the slope of the line of (1,0) and (-1,-3)

Alinara [238K]

Slope formula: m = $\frac{(y2-y1)}{(x2-x1)}$ (Knowing that m represents the slope)

Substitute (1,0) for (x1,y1), and (-1,-3) for (x2,y2)

Slope of the line of (1,0) and (-1,-3) is:

m = $\frac{(-3-0)}{(-1-1)}$ = $\frac{-3}{-2}$ = $\frac{3}{2}$

(Simplify)

Slope of the line of (1,0) and (-1,-3) is $\frac{3}{2}$

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3 years ago

Let G denote the centroid of triangle ABC. If triangle ABG is equilateral with side length 2, then determine the perimeter of tr

zhannawk [14.2K]

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$AC^2=AM^2+GM^2=BM^2+GM^2=BC^2$
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4 years ago