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3 years ago

WILL BE MARKED BRAINLIEST AND GET POINTS!!!!!

Mathematics

2 answers:

tatiyna3 years ago

8 0

Answer:

a) 980

b) 770

Step-by-step explanation:

Given: A machine produces 280 appliances per hour, and the total number of appliances produced in x hours can be represented as a function:

f(x)=280x

The number of appliances that can be produced in 3.5 hours is given by:

f(x)=280*3.5=980

Therefore, the number of appliances that can be produced in 3.5 hours is 980.

The problem asks for the difference between the appliances produced between 7 hours and 4 1/4 hours.

First, we substitute the values to the function f(x)=280x.

f(x₁)=280x=280(7)=1960

f(x2)=280x=280(4.25)=1190

f(x₁)- f(x2)=1960-1190=770

Therefore, there are 770 more appliances produced.

PolarNik [594]3 years ago

6 0

a) For this first part, all we need to do is "plug in" 3.5 for $x$ , which gives us a value of $280(3.5) = 980$ .

b) For this second part, we will find the number of appliances produced in 7 hours and then subtract the number of appliances produced in 4.25 hours:

$280(7) - 280(4.5) = 1960 - 1190 = 770$

Our answers are a) 840 and b) 770.

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

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.

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