MATHEMATICS HIGH SCHOOL

Bety compró 16 pares de calcetas, entre blancas y negras, y pagó $339. Los pares de calcetas blancas cuestan $24 y las negras $19. Plantea el sistema de ecuaciones y resuelve por el método que creas conveniente. Después toma foto y súbela en donde se indica.

Answers

Answer 1

Answer:

Answer:

Sistema de ecuaciones:

x+y= 16

24x+19y=339

Respuesta:

x=7, y=9

Step-by-step explanation:

Con la información del enunciado puedes plantear las siguientes ecuaciones:

x= cantidad de calcetas blancas

y= cantidad de calcetas negras

x+y= 16 (1)

24x+19y=339 (2)

Para resolver las ecuaciones, debes iniciar despejando x en 1:

y=16-x (3)

Después, debes reemplazar (3) en (2) y despejar x para encontrar su valor:

24x+19(16-x)=339

24x+304-19x=339

5x=339-304

x=35/5

x=7

Luego, puedes reemplazar el valor de x en (3) para encontrar y:

y=16-x

y=16-7

y=9

De acuerdo a esto, x= 7, y=9.

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Bob Ross has a charge card. Ross's previous monthly balance if $167.42. He made a payment of $159.15. There is no finance charge. This month Ross has no new purchases. What is the new balance?

Answers

Answer:

According to your question, the new balance is $8.27

Step-by-step explanation:

$167.42 - $159.15 = $8.27

What is the sum of the finite arithmetic series?4+8+12+16+...+76

a. 760
b. 780
c. 800
d. 860

Answers

The answer is A. I think. Hope this helped :)

The answer is:

a. 760

What is the solution of the inequality 9 - x^2 < 0?(a) {x | -3 < x < 3}
(b) {x | x > 3 or x < -3}
(c) {x | x | x > 3}
(d) {x | x < -3}

please show how you got the answer. thank you!

Answers

9 - x² < 0
-9    -9 to remove 9 deduct it from both sides
- x² < -9
   ÷ -1 ÷-1 to remove the negative sign, divide both sides by -1
       x² < 9
   √x² < √9    to remove the exponent of 2, find the square root of both sides.
   x    < 3    Choice A.

To check:

9 - x² < 0
9 - 3²   < 0
9 - 9 < 0
0 < 0

$\9-x^2\ \textless \ 0\n3^2-x^2\ \textless \ 0\n(3-x)(3+x)\ \textless \ 0\n\{3-x\ \textless \ 0\ \ \wedge\ \ 3+x\ \textgreater \ 0\}\ \ \ \vee\ \ \ \{3-x\ \textgreater \ 0\ \ \wedge\ \ 3+x\ \textless \ 0\}\n\{-x\ \textless \ -3\ \ \wedge\ \ x\ \textgreater \ -3\}\ \ \ \ \ \ \vee\ \ \ \ \ \{-x\ \textgreater \ -3\ \ \wedge\ \ x\ \textless \ -3\}\n~\ \ \ \{x\ \textgreater \ 3\ \ \ \wedge\ \ \ x\ \textgreater \ -3\}\ \ \ \ \ \ \vee\ \ \ \ \ \ \{x\ \textless \ 3\ \ \ \ \wedge\ \ \ x\ \textless \ -3\}\n~\qquad \qquad \ \ x\ \textgreater \ 3\ \qquad \ \ \ \ \ \vee\ \ \ \ \ \qquad \ x\ \textless \ -3\$

The solution is (b) {x | x > 3 or x < -3}

How do I write 3.83 in decimal form