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3 years ago

What is the equation of the following line? Be sure to scroll down first to see all answer options. (2,10) (0,0)

Mathematics

1 answer:

kherson [118]3 years ago

7 0

Answer: y=5x

Step-by-step explanation: Slope= 10/2=5 Plugged it in equation y=mx+b and solved for b to get b=0

You might be interested in

Which of the following is NOT a solution to the linear equation y=3x+2?

Ipatiy [6.2K]

Answer:

(3,10)

Step-by-step explanation:

When x is 3

${ \bf{y = 3x + 2}} \\ { \bf{y = (3 \times 3) + 2}} \\ { \bf{y = 11}}$

y is 11, and this is invalid because it is not at accord.

6 0

3 years ago

Read 2 more answers

(a) If G is a finite group of even order, show that there must be an element a = e, such that a−1 = a (b) Give an example to sho

Dahasolnce [82]

Answer:

See proof below

Step-by-step explanation:

First, notice that if a≠e and a^-1=a, then a²=e (this is an equivalent way of formulating the problem).

a) Since G has even order, |G|=2n for some positive number n. Let e be the identity element of G. Then A=G\{e} is a set with 2n-1 elements.

Now reason inductively with A by "pairing elements with its inverses":

List A as A={a1,a2,a3,...,a_(2n-1)}. If a1²=e, then we have proved the theorem.

If not, then a1^(-1)≠a1, hence a1^(-1)=aj for some j>1 (it is impossible that a^(-1)=e, since e is the only element in G such that e^(-1)=e). Reorder the elements of A in such a way that a2=a^(-1), therefore a2^(-1)=a1.

Now consider the set A\{a1,a2}={a3,a4,...,a_(2n-1)}. If a3²=e, then we have proved the theorem.

If not, then a3^(-1)≠a1, hence we can reorder this set to get a3^(-1)=a4 (it is impossible that a^(-1)∈{e,a1,a2} because inverses are unique and e^(-1)=e, a1^(-1)=a2, a2^(-1)=a1 and a3∉{e,a1,a2}.

Again, consider A\{a1,a2,a3,a4}={a5,a6,...,a_(2n-1)} and repeat this reasoning. In the k-th step, either we proved the theorem, or obtained that a_(2k-1)^(-1)=a_(2k)

After n-1 steps, if the theorem has not been proven, we end up with the set A\{a1,a2,a3,a4,...,a_(2n-3), a_(2n-2)}={a_(2n-1)}. By process of elimination, we must have that a_(2n-1)^(-1)=a_(2n-1), since this last element was not chosen from any of the previous inverses. Additionally, a_(2n1)≠e by construction. Hence, in any case, the statement holds true.

b) Consider the group (Z3,+), the integers modulo 3 with addition modulo 3. (Z3={0,1,2}). Z3 has odd order, namely |Z3|=3.

Here, e=0. Note that 1²=1+1=2≠e, and 2²=2+2=4mod3=1≠e. Therefore the conclusion of part a) does not hold

7 0

3 years ago

A line passes through the point (-10,8) and has a slope of 1/2.

timofeeve [1]

Answer:

the answer is y=1/2x+13

6 0

3 years ago

How many trucks carried only late variety?

murzikaleks [220]

9514 1404 393

Answer:

late only: 15
extra-late only: 24
one type: 43
total trucks: 105

Step-by-step explanation:

It works well when making a Venn diagram to start in the middle (6 carried all three), then work out.

For example, if 10 carried early and extra-late, then only 10-6 = 4 of those trucks carried just early and extra-late.

Similarly, if 30 carried early and late, and 4 more carried only early and extra-late, then 38-30-4 = 4 carried only early. In the attached, the "only" numbers for a single type are circled, to differentiate them from the "total" numbers for that type.

a) 15 trucks carried only late

b) 24 trucks carried only extra late

c) 4+15+24 = 43 trucks carried only one type

d) 38+67+56 -30-28-10 +6 +6 = 105 trucks in all went out