How many solutions will this system of equations have?(2 points)

Mathematics

2 answers:

kodGreya [7K]3 years ago

7 0

The answer would be no solution. Hope this helps :)

Vaselesa [24]3 years ago

4 0

It would have no solution because both lines have the same slope of -3.5 but different y-intercepts, which means that they are parallel and will never intersect.

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the value of a share of stock decreases in value at a rate of $1.20 per hour during the first 3.5 hours of trading. write and so

kvv77 [185]

Well since I don't know what the original value I'll label it as X. So your equation will start out as X - (1.20 x 1) which would be for every hour for a half hour just divide 1.20 / 2 to get your half an hour. (1.20 x 3) + 0.60. Should be easy if they want it in faction form than it would just be 1.20 x 3 1/2.

Hope this helps

7 0

4 years ago

Is my answer correct?

tatyana61 [14]

9514 1404 393

Answer:

yes

Step-by-step explanation:

By the "rule of 72", the amount will be doubled in 72/I years, where I is the annual interest rate in percent. That is, it can be estimated to take 72/4 = 18 years to double the $240 investment to $480. It would take another 18 years to double it again to $960. So, to achieve a balance of $1500 will be expected to take more than 36 years. The only reasonable answer choice is the one you have selected.

The exact solution is ...

log (1500/240)/log(1 +0.04) ≈ 46.72 years ≈ 47 years.

8 0

3 years ago

What is the equation of -7 increased by a number n is 12

777dan777 [17]

-7+n=12
this leaves us with n=19

6 0

3 years ago

How many positive integers $n$ satisfy $127 \equiv 7 \pmod{n}$? $n=1$ is allowed.

Svetllana [295]

Naturally, any integer $n$ larger than 127 will return $127\equiv127\mod n$ , and of course $127\equiv0\mod127$ , so we restrict the possible solutions to $1\le n$ .

Now,

$127\equiv7\mod n$

is the same as saying there exists some integer $k$ such that

$127=nk+7$

We have

$\implies 120=nk$

which means that any $n$ that satisfies the modular equivalence must be a divisor of 120, of which there are 16: $\{1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120\}$ .

In the cases where the modulus is smaller than the remainder 7, we can see that the equivalence still holds. For instance,

$127=21\cdot6+1\iff127\equiv1\equiv7\mod6$

(If we're allowing $n=1$ , then I see no reason we shouldn't also allow 2, 3, 4, 5, 6.)