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rodikova [14]
2 years ago
15

Leticia charges $8 per hour to babysit. She babysat Friday night for 4 hours,

Mathematics
2 answers:
JulsSmile [24]2 years ago
6 0

For this case we have that the variable "x" represents the number of hours that Leticia uses to take care of children on Saturday.

IF on Friday I use 4 hours ($ 8 each) and on Saturday "x" hours ($ 8 each) obtaining a profit of $ 72, we have the following equation:

8 (4 + x) = 72

We apply distributive property:

32 + 8x = 72\\8x = 72-32\\8x = 40\\x = \frac {40} {8}\\x = 5

So, on Saturday she spent 5 hours.

Answer:

8 (4 + x) = 72\\x = 5

FinnZ [79.3K]2 years ago
6 0

Answer:

Option A.

Option  C.

Step-by-step explanation:

Let be "x" the amount of hours  Leticia babysat on Saturday.

We know that she charges $8 per hour to babysit, she babysat Friday night for 4 hours and the total amount of money she earned on those two days was $72. Knowing this, we can set up the followin equation models this situation:

8(4+x)=72

Finally, we must  solve for "x":

8(4+x)=72\\\\32+8x=72\\\\8x=72-32\\\\8x=40\\\\x=\frac{40}{8}\\\\x=5

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Find the real numbers x and y if -3+ix^2y and x^2+y+4i are conjugate of each other. Pls solve with the steps
Firdavs [7]
ANSWER
x = ±1 and y = -4.
Either x = +1 or x = -1 will work

EXPLANATION
If -3 + ix²y and x² + y + 4i are complex conjugates, then one of them can be written in the form a + bi and the other in the form a - bi. In other words, between conjugates, the imaginary parts are same in absolute value but different in sign (b and -b). The real parts are the same

For -3 + ix²y
⇒ real part: -3
⇒ imaginary part: x²y

For x² + y + 4i
⇒ real part: x² + y (since x, y are real numbers)
⇒ imaginary part: 4

Therefore, for the two expressions to be conjugates, we must satisfy the two conditions. 

Condition 1: Imaginary parts are same in absolute value but different in sign. We can set the imaginary part of -3 + ix²y to be the negative imaginary part of x² + y + 4i so that the 

   x²y = -4 ... (I)

Condition 2: Real parts are the same

   x² + y = -3 ... (II)

We have a system of equations since both conditions must be satisfied

   x²y = -4 ... (I)
   x² + y = -3 ... (II)

We can rearrange equation (II) so that we have

   y = -3 - x² ... (II)

Substituting into equation (I)

   x²y = -4 ... (I)
   x²(-3 - x²) = -4
   -3x² - x⁴ = -4
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Therefore, x = ±1.
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Solve for y:

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