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3 years ago

Explain how multiplying with 6 is like multiplying with 3.

2 answers:

5 0

Multiplying with 6 is like multiplying with 3 because, you just double it. Let's say 3 × 4 = 12
Its the same as 6 × 2 = 12 ... which is just 3 + 3 + 3 + 3 . ((:

aliina [53]3 years ago

5 0

It has a common multiple.so pretty much *ALMOST* anything you multiply with either of them you can do for the other one

3 x 10=30
~ GOOD
6 x 5=30

3 x 7= 21
~ BAD
6 x 3=18

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Find the rational zeros of the polynomial function, f(x)= 4x^3-8x^2-19x-7

Dima020 [189]

Answer:

The rational zero of the polynomial are $\pm \frac{7}{4}, \pm \frac{1}{4},\pm \frac{7}{2},\pm \frac{1}{2},\pm 7,\pm 1$ .

Step-by-step explanation:

Given polynomial as :

f(x) = 4 x³ - 8 x² - 19 x - 7

Now the ration zero can be find as

$\dfrac{\textrm factor of P}{\textrm factor Q}$ ,

where P is the constant term

And Q is the coefficient of the highest polynomial

So, From given polynomial , P = -7 , Q = 4

Now , $\dfrac{\textrm factor of \pm P}{\textrm factor of \pm Q}$

I.e $\dfrac{\textrm factor of \pm P}{\textrm factor of \pm Q}$ = $\frac{\pm 7 , \pm 1}{\pm 4 ,\pm 2,\pm 1 }$

Or, The rational zero are $\pm \frac{7}{4}, \pm \frac{1}{4},\pm \frac{7}{2},\pm \frac{1}{2},\pm 7,\pm 1$

Hence The rational zero of the polynomial are $\pm \frac{7}{4}, \pm \frac{1}{4},\pm \frac{7}{2},\pm \frac{1}{2},\pm 7,\pm 1$ . Answer

7 0

3 years ago

2 different questions

spin [16.1K]

Answer:

950 and 950

Step-by-step explanation:

Lets just get this done :

10 x 50 = 500

9 x 50 = 450

500 + 450 =

950

---

20 x 50 = 1000

1000 - 50 = 950

Hope this helps!

5 0

3 years ago

Read 2 more answers

What is x+9=2(x_1)^2 in the form of ax^2+bx +c=0

Naddik [55]

<span>x + 9 = 2(x - 1)^2
x + 9 = 2(x^2 - 2x + 1)
x + 9 = 2x^2 - 4x + 2
</span>2x^2 - 4x + 2 - x - 9 = 0
2x^2 - 5x -7 =0

7 0

3 years ago

WILL MARK BRAINLYIST IF CORRECT!

Nady [450]

Your answer will be C

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.

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