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Bas_tet [7]
3 years ago
15

Finding the x and y intercepts x+3y=6

Mathematics
1 answer:
Alenkinab [10]3 years ago
8 0
Just substitute 0 in for x and y at different times and work out, gives you the intercept 0 + 3y=6 Y=2 X + 0y=6 X =6 Hope this helps
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Translate the sentence into an inequality. Eight times the sum of a number and 20 is greater than or equal to 25. Use the variab
PtichkaEL [24]

Answer:

8x + 20 ≥ 25

Step-by-step explanation:

Well 8 times the sum of a number "x" plus 20,

So we can write the following,

8x + 20

and that should be greater than or equal to 25

8x + 20 ≥ 25

<em>Thus,</em>

<em>as an inequality it is 8x + 20 ≥ 25.</em>

<em />

<em>Hope this helps :)</em>

3 0
3 years ago
The length of a subway is 385 miles. If 3 cm represents 35 miles on the map, then what is the length of the subway on the map in
Gre4nikov [31]

Answer:

33cm

Step-by-step explanation:

385 / 35 = 11

11 times 3 = 33

7 0
2 years ago
Your teacher says, “There are 100 centimeters in a meter, and this fact is revealed in the unit’s name (centimeter). There are 3
Travka [436]
I think what the teacher wants to express is that SI units are much more convenient to use than English units because the prefixes of the metric system will give you an idea on the conversion factor. Unlike English units, like ft to yard, there is no prefix to help you. You have to know the fact that 1 yard is equal to 3 feet.
8 0
3 years ago
Read 2 more answers
1. Let a; b; c; d; n belong to Z with n &gt; 0. Suppose a congruent b (mod n) and c congruent d (mod n). Use the definition
lukranit [14]

Answer:

Proofs are in the explantion.

Step-by-step explanation:

We are given the following:

1) a \equi b (mod n) \rightarrow a-b=kn for integer k.

1) c \equi  d (mod n) \rightarrow c-d=mn for integer m.

a)

Proof:

We want to show a+c \equiv b+d (mod n).

So we have the two equations:

a-b=kn and c-d=mn and we want to show for some integer r that we have

(a+c)-(b+d)=rn. If we do that we would have shown that a+c \equiv b+d (mod n).

kn+mn   =  (a-b)+(c-d)

(k+m)n   =   a-b+ c-d

(k+m)n   =   (a+c)+(-b-d)

(k+m)n  =    (a+c)-(b+d)

k+m is is just an integer

So we found integer r such that (a+c)-(b+d)=rn.

Therefore, a+c \equiv b+d (mod n).

//

b) Proof:

We want to show ac \equiv bd (mod n).

So we have the two equations:

a-b=kn and c-d=mn and we want to show for some integer r that we have

(ac)-(bd)=tn. If we do that we would have shown that ac \equiv bd (mod n).

If a-b=kn, then a=b+kn.

If c-d=mn, then c=d+mn.

ac-bd  =  (b+kn)(d+mn)-bd

          =    bd+bmn+dkn+kmn^2-bd

          =           bmn+dkn+kmn^2

          =            n(bm+dk+kmn)

So the integer t such that (ac)-(bd)=tn is bm+dk+kmn.  

Therefore, ac \equiv bd (mod n).

//

3 0
3 years ago
Who watching temptation island tonight?
ikadub [295]

Answer:

yes

Step-by-step explanation:

yes and uh yeah totally yeaaahhhh <3

6 0
2 years ago
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