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2 years ago

14t-8=6t solve this equation

Mathematics

2 answers:

lesantik [10]2 years ago

7 0

14t - 8 = 6t /+8

14t = 6t + 8 /-6t

8t = 8 /:8

t = 8

ANSWER: t = 8

labwork [276]2 years ago

4 0

$14t-8=6t \\ \\ -8 = 6t - 14t \\ \\ -8 = -8t \\ \\ 1 = t \\ \\ t = 1 \\ \\$

The final result is: t = 1.

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Which of the following statements about ratio's is true

AlekseyPX

All the above hope i helped

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2 years ago

R+sy=t solving for y

motikmotik

Answer:

y =(t-r)/s

Step-by-step explanation:

r + sy = t

Subtract r from both the sides,

sy = t - r

Divide both the sides by s,

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2 years ago

A concentric-tube heat exchanger cools hot water owing at 1 kg/s from 80C to 30C. It uses air owing at 5 kg/s to perform this co

nordsb [41]

Answer:

The log-mean-temperature-difference is 24.03⁰C

Step-by-step explanation:

First we need to know if the heat exchanger is in parallel flow or counter-flow. However, counter flow arrangement is best used to recover heat.

L.M.T.D for counter flow is given as;

$L.M.T.D =\frac{(T_h_f_1 -T_c_f_2)-(T_h_f_2 -T_c_f_1)}{2.3log[\frac{T_h_f_1 -T_c_f_2}{T_h_f_2 -T_c_f_1}]}$

where;

Thf₁ is the initial temperature of the hot fluid = 80°C

Tcf₂ is the final temperature of the cold fluid = 51.5°C

Thf₁ - Tcf₂ = 80 - 51.5 = 28.5⁰C

Thf₂ is the final temperature of the hot fluid = 30°C

Tcf₁ is the initial temperature of the cold fluid = 10°C

Thf₂ - Tcf₁ = 30 - 10 = 20⁰C

$L.M.T.D = \frac{28.5 -20}{2.3Log[\frac{28.5}{20}]} \\\\L.M.T.D = \frac{8.5}{0.3538} =24.03^oC$

Therefore, the log-mean-temperature-difference is 24.03⁰C

3 0

2 years ago

Read 2 more answers

Rectangle fence two sides will have lengths o 28 feet each and Ricky has 100 feet of fencing what is the greatest possible lengt

liraira [26]

Answer: 22

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if each side has 28 feet then 28* 2 = 56. We take 100 - 56 = 44. Then we take 44/2 to find one remaining side = 22.

8 0

2 years ago

Consider the following graph of two functions,S(1) + 8(2)Step 1 of 2: Finds (1) + 8(2)Enable Zoom/Pan8(x) = x2 - 2x - 1S(x) = -x

kumpel [21]

.Given:

$\begin{gathered} f(x)=-x+1 \\ g(x)=x^2-2x-1 \end{gathered}$

ToDetermine: The value of

$f(1)+g(2)$

Solution

$\begin{gathered} f(x)=-x+1 \\ f(1)=-1+1 \\ f(1)=0 \end{gathered}$ $\begin{gathered} g(x)=x^2-2x-1 \\ g(2)=(2)^2-2(2)-1 \\ g(2)=4-4-1 \\ g(2)=-1 \\ f(1)+g(2)=0+(-1) \\ f(1)+g(2)=-1 \end{gathered}$

Hence, f(1) + g(2) = -1

3 0

8 months ago