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NISA [10]
3 years ago
5

D= kA [T2 - T1 / L] Solve for T1

Mathematics
1 answer:
mel-nik [20]3 years ago
5 0
d=\dfrac{k_A(T_2-T_1)}{T}\ \ \ \ |multiply\ both\ sides\ by\ T\\\\k_A(T_2-T_1)=dT\ \ \ \ |divide\ both\ sides\ by\ k_A\\\\T_2-T_1=\dfrac{dT}{k_A}\ \ \ \ |subtract\ T_2\ from\ both\ sides\\\\-T_1=\dfrac{dT}{k_A}-T_2\ \ \ \ \ |change\ signs\\\\\boxed{T_1=T_2-\frac{dT}{k_A}}
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104 tests is what % of 160 tests
Afina-wow [57]
160 test=100%

Therefore:

160 test------------------100%
104 test------------------   x
x=(104 test*100%) / 160 test=65%

Answer: 104 test is the 65% of 160 test
8 0
3 years ago
Miguel wants to wallpaper the back wall of his bedroom. the wall is in the shape of a rectangle. its length is 13 feet and its w
Bingel [31]
First of all lets find the area of the rectangle.
13 ft × 12 ft = 156 square feet.
Now lets find the price which is $8 / square foot. Using proportions we get:
$8/sq ft × 156 sq ft. Simplifying we get
$8 × 156 = $1248.
The wallpaper will cost $1248
3 0
3 years ago
Please answer this question now
lidiya [134]

Answer:

C = (2,2)

Step-by-step explanation:

B = (10 ; 2)

M = (6 ; 2)

C = (x ; y )

|___________|___________|

B (10;2)            M (6;2)             C ( x; y)

So:

dBM = dMC

√[(2-2)^2 + (6-10)^2] = √[(y-2)^2 + (x - 6)^2]

(2-2)^2 - (6-10)^2 = (y-2)^2 + (x - 6)^2

0 + (-4)^2 = (y-2)^2 + (x - 6)^2

16 = (y-2)^2 + (x - 6)^2

16 - (x - 6)^2 = (y-2)^2

Also:

2*dBM = dBC

2*√[(2-2)^2 + (6-10)^2] = √[(y-2)^2 + (x - 10)^2]

4*[(0)^2 + (-4)^2] = (y-2)^2 + (x - 10)^2

4*(16) = (y-2)^2 + (x - 10)^2

64 = (y-2)^2 + (x - 10)^2

64 = 16 - (x - 6)^2 + (x - 10)^2

48 = (x - 10)^2 - (x - 6)^2

48 = x^2 - 20*x + 100 - x^2 + 12*x - 36

48 = - 20*x + 100 + 12*x - 36

8*x = 16

x = 2

Thus:

16 - (x - 6)^2 = (y-2)^2

16 - (2 - 6)^2 = (y-2)^2

16 - (-4)^2 = (y-2)^2

16 - 16 = (y-2)^2

0 =  (y-2)^2

0 = y - 2

2 = y

⇒ C = (2,2)

4 0
3 years ago
Eli invested $ 330 $330 in an account in the year 1999, and the value has been growing exponentially at a constant rate. The val
Cloud [144]

Answer:

The value of the account in the year 2009 will be $682.

Step-by-step explanation:

The acount's balance, in t years after 1999, can be modeled by the following equation.

A(t) = Pe^{rt}

In which A(t) is the amount after t years, P is the initial money deposited, and r is the rate of interest.

$330 in an account in the year 1999

This means that P = 330

$590 in the year 2007

2007 is 8 years after 1999, so P(8) = 590.

We use this to find r.

A(t) = Pe^{rt}

590 = 330e^{8r}

e^{8r} = \frac{590}{330}

e^{8r} = 1.79

Applying ln to both sides:

\ln{e^{8r}} = \ln{1.79}

8r = \ln{1.79}

r = \frac{\ln{1.79}}{8}

r = 0.0726

Determine the value of the account, to the nearest dollar, in the year 2009.

2009 is 10 years after 1999, so this is A(10).

A(t) = 330e^{0.0726t}

A(10) = 330e^{0.0726*10} = 682

The value of the account in the year 2009 will be $682.

4 0
3 years ago
14. Find the value of X round the nearest degree 12 15
satela [25.4K]

Answer:

37

Step-by-step explanation:

6 0
3 years ago
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