I need help with three problems!!!!!

Cuanto es (60 : 10 + 8) x 2

Nataliya [291]

Answer:

28

Step-by-step explanation:

4 0

3 years ago

Which combination of integers can be used to generate the Pythagorean triple (7,24,25)?

allochka39001 [22]

A Pythagorean triple is a set of thre integer numbers, a, b and c that meet the Pythgorean theorem a^2 + b^2 = c^2

Use Euclide's formula for generating Pythagorean triples.

This formula states that given two arbitrary different integers, x and y, both greater than zero, then the following numbers a, b, c form a Pythagorean triple:

a = x^2 - y^2

b = 2xy

c = x^2 + y^2.

From a = x^2 - y^2, you need that x > y, then you can discard options A and D.

Now you have to probe the other options.

Start with option B, x = 4, y = 3

a = x^2 - y^2 = 4^2 - 3^2 = 16 -9 = 7

b = 2xy = 2(4)(3) = 24

c = x^2 9 y^2 = 4^2 + 3^2 = 16 + 9 = 25

Then we could generate the Pythagorean triple (7, 24, 25) with x = 4 and y =3.

If you want, you can check that a^2 + b^2 = c^2; i.e. 7^2 + 24^2 = 25^2

The answer is the option B. x = 4, y = 3

3 0

3 years ago

Read 2 more answers

-1^{2008} + (-1)^{2007} = ? (1 - (-1)^[11])^[2] = ? Questions for fun~

vladimir2022 [97]

1. -2
2. 4
Is the answers to the following questions

7 0

3 years ago

Simplify the algebraic expression -2(x - 3) + 4(-2x + 8)

dangina [55]

The answer to this equation is :-10x+38 because
-2x(x-3)+4(-2x+8)
(-2x+6)+(-8x+32)
(-2x+6)(-8x+32)
-2x+6-8x+32
-10x+38

7 0

3 years ago

Simplify u^2+3u/u^2-9 A.u/u-3, =/ -3, and u=/3 B. u/u-3, u=/-3

VashaNatasha [74]

  The correct answer is: Answer choice: [A]:
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→  " $\frac{u}{u-3}$ " ; " { u $\neq$ ± 3 } " ;

  →  or, write as: " u / (u − 3) " ;  {" u ≠ 3 "}  AND: {" u ≠ -3 "} ;
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Explanation:
__________________________________________________________
We are asked to simplify:

   $\frac{(u^2+3u)}{(u^2-9)}$ ;

Note that the "numerator" —which is: "(u² + 3u)" — can be factored into:
  → " u(u + 3) " ;

And that the "denominator" —which is: "(u² − 9)" — can be factored into:
  → "(u − 3) (u + 3)" ;
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Let us rewrite as:
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→ $\frac{u(u+3)}{(u-3)(u+3)}$ ;

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→ We can simplify by "canceling out" BOTH the "(u + 3)" values; in BOTH the "numerator" AND the "denominator" ;  since:

" $\frac{(u+3)}{(u+3)} = 1$ " ;

→ And we have:
_________________________________________________________

→  " $\frac{u}{u-3}$ " ; that is: " u / (u − 3) " ; { u $\neq 3$ } .
and: { u $\neq-3$ } .

→ which is:  "Answer choice: [A] " .
_________________________________________________________

NOTE:  The "denominator" cannot equal "0" ; since one cannot "divide by "0" ;

and if the denominator is "(u − 3)" ; the denominator equals "0" when "u = -3" ; as such:

"u $\neq$ 3" ;

→ Note: To solve: "u + 3 = 0" ;

Subtract "3" from each side of the equation;

→ " u + 3 − 3 = 0 − 3 " ;

→ u = -3 (when the "denominator" equals "0") ;

→ As such: " u $\neq$ -3 " ;

Furthermore, consider the initial (unsimplified) given expression:

→   $\frac{(u^2+3u)}{(u^2-9)}$ ;

Note:  The denominator is: "(u²  − 9)" .

The "denominator" cannot be "0" ; because one cannot "divide" by "0" ;

As such, solve for values of "u" when the "denominator" equals "0" ; that is:
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→ " u² − 9 = 0 " ;

→ Add "9" to each side of the equation ;

→ u² − 9 + 9 = 0 + 9 ;

→ u² = 9 ;

Take the square root of each side of the equation;
to isolate "u" on one side of the equation; & to solve for ALL VALUES of "u" ;

→ √(u²) = √9 ;

→ | u | = 3 ;

→ " u = 3" ; AND; "u = -3 " ;

We already have: "u = -3" (a value at which the "denominator equals "0") ;

We now have "u = 3" ; as a value at which the "denominator equals "0");

→ As such: " u $\neq 3$ " ; "u $\neq$ -3 " ;

or, write as: " { u $\neq$ ± 3 } " .

_________________________________________________________

6 0

3 years ago