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Step2247 [10]
2 years ago
7

20 POINTS!!!! If eight is added to twice a number and the sum is multiplied by five result is the same as if the number is multi

plied by -8 and four is added to the product. What is the number?
Mathematics
1 answer:
Sauron [17]2 years ago
6 0
<h3><u>The value of x is equal to -2.</u></h3>

5(2x + 8) = -8x + 4

<em><u>Distributive property.</u></em>

10x + 40 = -8x + 4

<em><u>Add 8x to both sides.</u></em>

18x + 40 = 4

<em><u>Subtract 40 from both sides.</u></em>

18x = -36

<em><u>Divide both sides by 18</u></em>

x = -2

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Faith reads each night before bed. She read 623 pages last week. If she read the same number of pages every night, which equatio
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Answer:

 x= 623/7      x =89              89 pages every night

Step-by-step explanation:

A week has 7 days

She reads 623 pages in one week(623 in 7 days)

5 0
3 years ago
Can someone help explain please
Vitek1552 [10]

Answer:

step by step^^ hope this helps

3 0
2 years ago
The airline that Vince is using has a
emmainna [20.7K]

Answer:

Each bag must NOT exceed 62 inches (157.48 centimeters) in overall dimensions (length + width + height) and CANNOT exceed 70 pounds (31.75 kg). Overweight bag fees will apply to bags weighing over 50 pounds (22.73 kg).

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Step-by-step explanation:

4 0
2 years ago
Suppose that a spherical droplet of liquid evaporates at a rate that is proportional to its surface area: where V = volume (mm3
Alex

Answer:

V = 20.2969 mm^3 @ t = 10

r = 1.692 mm @ t = 10

Step-by-step explanation:

The solution to the first order ordinary differential equation:

\frac{dV}{dt} = -kA

Using Euler's method

\frac{dVi}{dt} = -k *4pi*r^2_{i} = -k *4pi*(\frac {3 V_{i} }{4pi})^(2/3)\\ V_{i+1} = V'_{i} *h + V_{i}    \\

Where initial droplet volume is:

V(0) = \frac{4pi}{3} * r(0)^3 =  \frac{4pi}{3} * 2.5^3 = 65.45 mm^3

Hence, the iterative solution will be as next:

  • i = 1, ti = 0, Vi = 65.45

V'_{i}  = -k *4pi*(\frac{3*65.45}{4pi})^(2/3)  = -6.283\\V_{i+1} = 65.45-6.283*0.25 = 63.88

  • i = 2, ti = 0.5, Vi = 63.88

V'_{i}  = -k *4pi*(\frac{3*63.88}{4pi})^(2/3)  = -6.182\\V_{i+1} = 63.88-6.182*0.25 = 62.33

  • i = 3, ti = 1, Vi = 62.33

V'_{i}  = -k *4pi*(\frac{3*62.33}{4pi})^(2/3)  = -6.082\\V_{i+1} = 62.33-6.082*0.25 = 60.813

We compute the next iterations in MATLAB (see attachment)

Volume @ t = 10 is = 20.2969

The droplet radius at t=10 mins

r(10) = (\frac{3*20.2969}{4pi})^(2/3) = 1.692 mm\\

The average change of droplet radius with time is:

Δr/Δt = \frac{r(10) - r(0)}{10-0} = \frac{1.692 - 2.5}{10} = -0.0808 mm/min

The value of the evaporation rate is close the value of k = 0.08 mm/min

Hence, the results are accurate and consistent!

5 0
3 years ago
Simplify the expression r^-3 s^5 t^2/r^2 st^-2
Hoochie [10]
R/s^5t^r2st-2/r^3
hope this helps










5 0
3 years ago
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