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2 years ago

Factor 21a–27.

Mathematics

2 answers:

Ganezh [65]2 years ago

8 0

Answer:

3(7a - 9)

Step-by-step explanation:

21a–27

3*7*a - 3*9

Factor out 3

3(7a - 9)

vlada-n [284]2 years ago

4 0

Answer:

3(7a-9)

Step-by-step explanation:

The greatest common factor between 21a and 27 is 3. Hope this helps!

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Neeeeeed the help please

lions [1.4K]

The correct answer is -5/7. Hope this helps.

8 0

3 years ago

Can someone please help me please !

TEA [102]

Answer:

Step-by-step explanation:

GH : √(8-4)^2 + (2-5)^2 = √16+9 = √25 = 5

HI : √(-6-2)^2 + (2-8)^2 = √64+36 = √100 = 10

IJ : √(-2-2)^2 + (-3+6)^2 = √16 + 9 = √25 = 5

JH : √(-2-4)^2 + (-3-5)^2 = √36 + 64 = √100 = 10

Slope of the line that contains GH

(2-5)/(8-4) = -3/4

Slope of the line that contains HI

(-6-2) / (2-8) = 8/6 = 4/3

I calculated the distance between points. Thanks to that I noticed that the opposite sides are congruent, so the quadrilateral can be a rectangle or a parallelogram. So I found the slope of the lines that contain two consecutive sides and I discovered that are perpendicular. So the quadrilateral is a rectangle because its angles are all of 90 degrees

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3 years ago

Please help me with all three and explain it for brainiest and extra points! Please help me I don’t get it

Inessa05 [86]

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3 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.

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