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Sonja [21]
3 years ago
11

What is the value of c in this linear system? A) 1 B) -3 C) -1 D) 2

Mathematics
1 answer:
Alika [10]3 years ago
4 0

9514 1404 393

Answer:

   B)  -3

Step-by-step explanation:

There are methods for finding only c. Cramer's rule is one of them. It involves finding two determinants and taking their ratio. Here, we choose a more <em>ad hoc</em> approach. It appears that the value of b can be found by combining the last two equations.

  (1/2)(2a +4b -2c) -(a -3b -c) = (1/2)(12) -(-4)

  5b = 10

  b = 2

Now, we can substitute this value into the first two equations. This gives ...

  5a +c = -8

  a - c = 2

Subtracting 5 times the second from the first gives ...

  (5a +c) -5(a -c) = (-8) -5(2)

  6c = -18 . . . . simplify

  c = -3 . . . . . . divide by 6

The value of c is -3.

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The maximum number of shrubs (x) = a(area) divided by 15.9

a(area)=37.1*15
37.1*15= 556.5
556.5/15.9=x
556.5/15.9=35
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6 0
3 years ago
At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel
Alexeev081 [22]
Use Law of Cooling:
T(t) = (T_0 - T_A)e^{-kt} +T_A
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
42 = (80-20)e^{-3k} +20 \\  \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\  \\ k = -\frac{1}{3} ln (\frac{11}{30})

Substituting k back into cooling equation gives:
T(t) = 60(\frac{11}{30})^{t/3} + 20

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\  \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80
Solve for x:
x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80
Sub back into original cooling equation, x = T(t)
-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20
Solve for t:
60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3}  = 60 \\  \\  (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\  \\  \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\  \\ \\  t = 4.959

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM
8 0
3 years ago
Evaluate the determination for the following matrix [144] [522] [155]
Anettt [7]
Matrix :

1    4    4

5    2    2

1    5    5

=> Determinant =

1 * (2*5 - 2*5) - 4 * (5*5 - 1*2) + 4 * ( 5*5 - 2*1)

= 1*(10 - 10) - 4*(25 - 2) + 4*(25 -2) = 1*0 + 4* 23 + 4*(25 - 2) = 0 - 4*23 + 4*23 = 0

Answer: 0
8 0
3 years ago
(MULTIPLE CHOICE QUESTION)
weeeeeb [17]

Answer:

Transitive property is used in step 4.

Step-by-step explanation:

Transitive property states that if

a = b \\b = c\\ then\\a = c

Here we have

\angle 5 \cong \angle 1\\\angle 1 \cong \angle 3\\then\\\angle 3 \cong \angle 5

3 0
3 years ago
Read 2 more answers
A wall has an area of 40 square feet. Its height is 8 feet. Which equation could you use to find its width?
lord [1]

Answer:

40/8=5

Other side length = 5

Step-by-step explanation:

7 0
3 years ago
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