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andrey2020 [161]
3 years ago
7

An ice cream store charges 75 cents for each scoop of ice cream and 25 cents for each cone. How many scoops would be in one ice

cream cone if it cost a total of $4?
Mathematics
2 answers:
GREYUIT [131]3 years ago
5 0

0.25 +0.75x = 4
0.75x = 4 - 0.25
0.75x = 3.75
x =3.75/0.75
x = 5

answer: In one ice cream cost $4, there will be 5 scoops .

double check:

5(0.75) = 3.75 (amount pay for scoops)
            + 0.25 (one cone)
  ------------------
                $4.00 total cost

Aliun [14]3 years ago
4 0
There will be 4 scoops of ice cream
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Jen picked 3/4 of a gallon of strawberries in half an hour. If she keeps picking strawberries at the same rate, how many gallons
Ratling [72]
Hello!

3/4 = 0.75 as a decimal

We know that Jen picks 0.75 of a gallon of strawberries in half an hour, and now we want to find out how many gallons she'll pick in 2 hours.

Half an hour = 0.5 2 ÷ 0.5 = 4 There are four half hours in 2 hours.

0.75 × 4 = 3

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6 0
3 years ago
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
   x⁴ + 3x² - 4 = 0
   (x² + 4)(x² - 1) = 0
   (x² + 4)(x-1)(x+1) = 0

Therefore, x = ±1.
Leave alone (x² + 4) as it gives no real solutions.

Solve for y:

   y = -3 - x² ... (II)
   y = -3 - (±1)²
   y = -3 - 1
   y = -4

So x = ±1 and y = -4. We can confirm this results in conjugates by substituting into the expressions:

   -3 + ix²y 
   = -3 + i(±1)²(-4)
   = -3 - 4i

   x² + y + 4i
   = (±1)² - 4 + 4i
   = 1 - 4 + 4i
   = -3 + 4i

They result in conjugates
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cupoosta [38]
Correct answer:
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