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2 years ago

Help Please :/ math problem ! I don’t know the 5 steps !!

Mathematics

1 answer:

Lilit [14]2 years ago

3 0

Answer:

The answer for the step 5 is 52

Step-by-step explanation:

This is because, when a negative number multiply a negative number, it gives a positive answer

 thus ; - × - =+

so from the expression

16-36(-1)

 the negative 36 multiply the negative 1 to give positive 36

.

so the relation then become

16+36= 52

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For her tutoring services, Thuy charged Demi $10 per hour and $8 for books and supplies. Demi paid a total of $48.00. Which equa

Fudgin [204]

To find the final amount, you would have to multiply her rate per hour (10) by how many hours she tutored (h). Then, you need to add her flat rate of 8.

10h + 8 = 48, where h is how many hours she tutored.

7 0

2 years ago

b. Sixty-five pounds of candy was divided into four different boxes. The second box contained twice the amount of the first box.

Ainat [17]

First box = 14 pounds

Second box = 2x = 28 pounds

Third Box = x+2= 16 pounds

Fourth box = x/2 = 7 pounds

Step-by-step explanation:

Here we have , Sixty-five pounds of candy was divided into four different boxes. The second box contained twice the amount of the first box. The third box contained two more pounds than the first box. The last box contained one-fourth the amount in the second box. We need to find How much candy was in each box. Let's find out:

We have a total of 65 pounds of candy ! Let in first box we have x pounds so , second box contained twice the amount of the first box i.e.

⇒ $2x$

The third box contained two more pounds than the first box i.e.

⇒ $x+2$

The last box contained one-fourth the amount in the second box i.e.

⇒ $(\frac{1}{4})2x = \frac{x}{4}$

Therefore , Sum of pounds of candy are :

⇒ $\frac{x}{2} +x+2+2x+x=65$

⇒ $\frac{x}{2} +4x=63$

⇒ $\frac{9x}{2}=63$

⇒ $x=63(\frac{2}{9} )$

⇒ $x=14$

Therefore , Candy in each box is :

First box = 14 pounds

Second box = 2x = 28 pounds

Third Box = x+2= 16 pounds

Fourth box = x/2 = 7 pounds

8 0

3 years ago

2. Given AMAR, solve for the m&M. Round to the

Naddik [55]

Answer:

what is the question

Step-by-step explanation:

8 0

3 years ago

Which number rounds to 15,700,000 when rounded to the nearest hundred thousand?. A) 15,000,000. B) 15,579,999. C) 15,649,999. D)

Artyom0805 [142]

The correct answer is option D. i.e. 15,659,999. Now, this number is closest to the given number i.e. 15,700,000. Becuase when 659 is rounded to the nearest number of hiher value then its value will be 700. Thus,15,659, 999 rounds to 15,700,000 when rounded to the nearest hundred thousand.

6 0

3 years ago

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a

Alja [10]

Answer:

t = 460.52 min

Step-by-step explanation:

Here is the complete question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution

Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.

inflow = 0 (since the incoming water contains no dye)

outflow = concentration × rate of water inflow

Concentration = Quantity/volume = Q/200

outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.

So, Q' = inflow - outflow = 0 - Q/100

Q' = -Q/100 This is our differential equation. We solve it as follows

Q'/Q = -1/100

∫Q'/Q = ∫-1/100

㏑Q = -t/100 + c

$Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c} = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}$

when t = 0, Q = 200 L × 1 g/L = 200 g

$Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}$

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

$2 = 200e^{(-t/100)}\\\frac{2}{200} = e^{(-t/100)}$

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

6 0

3 years ago