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Ilya [14]
3 years ago
15

What is the answer to this word problem or how would I solve it

Mathematics
1 answer:
USPshnik [31]3 years ago
5 0
Its a simultaneous equation:
lets make "x" the used games and "y" the new ones

4x + 2y = 84
6x + y = 78. (multiply this by 2 to cancel out y)
12x + 2y = 156

so now we subtract them:
12x - 4x = 156 - 84
x = 9
used games cost $9
new games cost 78 - 6(9) = $24

Janet has $120 but she already bought 3 NEW games so:
120 - 3(24) = $48 left

48/9 = how many used ones she can buy = 5. something

she can buy 5 used
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X-16 the first one

Step-by-step explanation:

The first one is equal to x-4

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Consider the three isosceles triangles.
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True or false??? Please help me and explain!
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irrational keeps going and rational is like 4.34 and it stops

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Let x = a bi and y = c di and z = f gi which statements are true? check all of the boxes that apply. x y = y x (x × y) × z = x ×
Alex73 [517]

A complex number is a part of a number system that includes a real unit and an imaginary unit. All the statements are true except C and E.

<h3>What is a Complex Number?</h3>

A complex number is a part of a number system that includes the real numbers as well as a particular element labelled i, sometimes known as the imaginary unit, and which obeys the equation i² = 1.

As it is given that x=(a+bi), y=(c+di), and z=(f+gi), therefore, we need to plug in the values to check which all are true.

A.) x+y=y+x

x+y=y+x\\a+bi+c+di=c+di+a+bi

As the two sides of the equation are equal, the equation is true.

B.) (x × y) × z = x × (y × z)

(x \times y) \times  z = x \times  (y \times  z)\\\\\ [ (a+bi)  \times (c+di) ]\times (f+gi) = (a+bi)  \times [(c+di) \times (f+gi)]\\\\(ac+adi+bci-bd) \times (f+gi)=  (a+bi)  \times (cf+cgi+fdi-dg)

(ac+adi+bci-bd) \times (f+gi)=  (a+bi)  \times (cf+cgi+fdi-dg)\\\\\\acf+adfi+bcfi-bdf+acgi-adg-bcg-bdgi = acf+acgi+afdi-adg + bcfi-bcg-bfd-bdgi

As the two sides of the equation are equal, the equation is true.

C.) x−y=y−x

x-y=y-x\\\\(a+bi)-(c+di)=(c+di)-(a+bi)\\\\a+bi-c-di=c+di-a-bi

As the two sides of the equation are not equal, the equation is not true.

D.) (x+y)+z=x+(y+z)

(x+y)+z=x+(y+z)\\\\(a+bi+c+di)+f+gi = a+bi+(c+di+f+gi)\\\\a+bi+c+di+f+gi = a+bi+c+di+f+gi

As the two sides of the equation are equal, the equation is true.

E.) (x−y)−z=x−(y−z)

(x-y)-z=x-(y-z)\\\\(a+bi-c-di)-f-gi = a+bi-(c+di-f-gi)\\\\a+bi-c-di-f-gi =a+bi-c-di+f+gi)

As the two sides of the equation are not equal, the equation is not true.

Hence, all the statements are true except C and E.

Learn more about Complex Number:

brainly.com/question/12464608

6 0
2 years ago
Help me you can get 10p
PilotLPTM [1.2K]

Answer:

.45 m

Step-by-step explanation:

6.3/14= x

x= .45

4 0
3 years ago
Read 2 more answers
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