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3 years ago

How are numbers related?

1 answer:

3 0

two numbers are related in this way:
the number 1 is subtracted from the first number, the difference is squared, then doubled;the result is equal to the second number. the number 1 is added to the first number, and the sum is squared;the result is equal to 4 minus the second number. determine the numbers.

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The U.S.-based motorcycle manufacturer says that it expects to build $150,000 motorcycles this year, up from $133,000 last year.

Leokris [45]

Answer:

Step-by-step explanation:

(148-134)/134 = x%

14/134 = x%

x = 10.4%

Have a great day

4 0

3 years ago

Read 2 more answers

Which best explains why the equation 3x+8=3x-5 has no solutions?

tekilochka [14]

3x + 8 = 3x - 5
3x - 3x + 8 = -5
8 = -5 (incorrect)
this means there is no solution, because whatever number u sub in for x, it will not come out true

6 0

3 years ago

Read 2 more answers

For integers a, b, and c, consider the linear Diophantine equation ax C by D c: Suppose integers x0 and y0 satisfy the equation;

Dmitrij [34]

Answer:

$x = x_1+r(\frac{b}{gcd(a, b)} )\\y=y_1-r(\frac{a}{gcd(a, b)} )$

b. x = -8 and y = 4

Step-by-step explanation:

This question is incomplete. I will type the complete question below before giving my solution.

For integers a, b, c, consider the linear Diophantine equation

$ax+by=c$

Suppose integers x0 and yo satisfy the equation; that is,

$ax_0+by_0 = c$

what other values

$x = x_0+h$ and $y=y_0+k$

also satisfy ax + by = c? Formulate a conjecture that answers this question.

Devise some numerical examples to ground your exploration. For example, 6(-3) + 15*2 = 12.

Can you find other integers x and y such that 6x + 15y = 12?

How many other pairs of integers x and y can you find ?

Can you find infinitely many other solutions?

From the Extended Euclidean Algorithm, given any integers a and b, integers s and t can be found such that

$as+bt=gcd(a,b)$

the numbers s and t are not unique, but you only need one pair. Once s and t are found, since we are assuming that gcd(a,b) divides c, there exists an integer k such that gcd(a,b)k = c.

Multiplying as + bt = gcd(a,b) through by k you get

$a(sk) + b(tk) = gcd(a,b)k = c$

So this gives one solution, with x = sk and y = tk.

Now assuming that ax1 + by1 = c is a solution, and ax + by = c is some other solution. Taking the difference between the two, we get

$a(x_1-x) + b(y_1-y)=0$

Therefore,

$a(x_1-x) = b(y-y_1)$

This means that a divides b(y−y1), and therefore a/gcd(a,b) divides y−y1. Hence,

$y = y_1+r(\frac{a}{gcd(a, b)})$ for some integer r. Substituting into the equation

$a(x_1-x)=rb(\frac{a}{gcd(a, b)} )\\gcd(a, b)*a(x_1-x)=rba$

$x = x_1-r(\frac{b}{gcd(a, b)} )$

Thus if ax1 + by1 = c is any solution, then all solutions are of the form

$x = x_1+r(\frac{b}{gcd(a, b)} )\\y=y_1-r(\frac{a}{gcd(a, b)} )$

In order to find all integer solutions to 6x + 15y = 12

we first use the Euclidean algorithm to find gcd(15,6); the parenthetical equation is how we will use this equality after we complete the computation.

$15 = 6*2+3\\6=3*2+0$

Therefore gcd(6,15) = 3. Since 3|12, the equation has integral solutions.

We then find a way of representing 3 as a linear combination of 6 and 15, using the Euclidean algorithm computation and the equalities, we have,

$3 = 15-6*2$

Because 4 multiplies 3 to give 12, we multiply by 4

$12 = 15*4-6*8$

So one solution is

$x=-8$ & $y = 4$

All other solutions will have the form

$x=-8+\frac{15r}{3} = -8+5r\\y=4-\frac{6r}{3} =4-2r$

where $r$ ∈ Ζ

Hence by putting r values, we get many (x, y)

3 0

3 years ago

I really need help, please

solong [7]

Answer:

C: $\frac{5}{2} + \frac{1}{2} i$

Step-by-step explanation:

Did this through my TI-89 calculator. ^^

6 0

3 years ago

{3x+2y = -5 {4X-3y = -18

larisa86 [58]

You make an x= ... equation by moving 2y to the other side and /3 (in the first equation) then you fill in x into the second equation and you calculate y. If you have Y you fill it in in the first equation.

6 0

4 years ago