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2 years ago

Find the x and y intercepts Y-2xy+4x=1

Mathematics

1 answer:

liubo4ka [24]2 years ago

7 0

Answer:

$x = \frac{1}{4}$

$y= 1$

Step-by-step explanation:

Given

$y - 2xy + 4x = 1$

Solving (a) The x intercept

Let y = 0

$y - 2xy + 4x = 1$

$0 - 2x*0 + 4x = 1$

$0 - 0 + 4 x = 1$

$4x = 1$

Divide both sides by 4

$x = \frac{1}{4}$

Let x = 0

$y - 2xy + 4x = 1$

$y - 2y * 0 + 4*0 = 1$

$y - 0 + 0 = 1\\$

$y= 1$

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Chang has 2 shirts: a white one and a black one. He also has 2 pairs of pants, one blue and one tan. What is the probability, if

weqwewe [10]

Answer:

The probability that Chang gets dressed with a white shirt and tan pants is 25%.

Step-by-step explanation:

Given that Chang has 2 shirts, a white one and a black one, and he also has 2 pairs of pants, one blue and one tan, to determine what is the probability, if Chang gets dressed in the dark, that he winds up wearing the white shirt and tan pants the following calculation must be performed:

Each shirt = 50% chance

Each pants = 50% chance

0.50 x 0.50 = X

0.25 = X

Therefore, the probability that Chang gets dressed with a white shirt and tan pants is 25%.

5 0

3 years ago

Whats the answer? Please help

PolarNik [594]

Answer:

Answer: y=3/2x+1

To find a line that is perpendicular, invert the slope and multiply by -1

so the original slope is -2/3,

invert it to get -3/2 then multiply by -1 to get 3/2

Put the given point (-2,-2) and the new slope into point intercept form: y+2=3/2(x+2)

Rewrite in standard form to get y=3/2x+1

7 0

2 years ago

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.

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