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2 years ago

Make t as the subject!! HELP MEE

Mathematics

1 answer:

Umnica [9.8K]2 years ago

6 0

Answer:

t = (p^2/mn) - 1/n

Step-by-step explanation:

Here, we want to make t the subject of the formula

we start by equating both sides so as to remove the root

we have this as;

m(t + n)/t = p^2

m(t + n) = tp^2

mt + mn = tp^2

mn = tp^2 - mt

mn = t(p^2-m)

t = (p^2 - m)/mn

t = p^2/mn - 1/n

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In this problem we must apply some algebraic handling and some known limits to determine whether the limit exists or not. The limit exists if and only if the result exists.

$\lim_{x \to \infty} \frac{4\cdot x - 1}{7\cdot x + 3}$

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To learn more on limits: brainly.com/question/12207558

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1 year ago

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Step-by-step explanation:

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2 years ago

Read 2 more answers

A 1/17th scale model of a new hybrid car is tested in a wind tunnel at the same Reynolds number as that of the full-scale protot

Olegator [25]

Answer:

The ratio of the drag coefficients $\dfrac{F_m}{F_p}$ is approximately 0.0002

Step-by-step explanation:

The given Reynolds number of the model = The Reynolds number of the prototype

The drag coefficient of the model, $c_{m}$ = The drag coefficient of the prototype, $c_{p}$

The medium of the test for the model, $\rho_m$ = The medium of the test for the prototype, $\rho_p$

The drag force is given as follows;

$F_D = C_D \times A \times \dfrac{\rho \cdot V^2}{2}$

We have;

$L_p = \dfrac{\rho _p}{\rho _m} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_m} \right)^2 \times L_m$

Therefore;

$\dfrac{L_p}{L_m} = \dfrac{\rho _p}{\rho _m} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_m} \right)^2$

$\dfrac{L_p}{L_m} =\dfrac{17}{1}$

$\therefore \dfrac{L_p}{L_m} = \dfrac{17}{1} =\dfrac{\rho _p}{\rho _p} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_p} \right)^2 = \left(\dfrac{V_p}{V_m} \right)^2$

$\dfrac{17}{1} = \left(\dfrac{V_p}{V_m} \right)^2$

$\dfrac{F_p}{F_m} = \dfrac{c_p \times A_p \times \dfrac{\rho_p \cdot V_p^2}{2}}{c_m \times A_m \times \dfrac{\rho_m \cdot V_m^2}{2}} = \dfrac{A_p}{A_m} \times \dfrac{V_p^2}{V_m^2}$

$\dfrac{A_m}{A_p} = \left( \dfrac{1}{17} \right)^2$

$\dfrac{F_p}{F_m} = \dfrac{A_p}{A_m} \times \dfrac{V_p^2}{V_m^2}= \left (\dfrac{17}{1} \right)^2 \times \left( \left\dfrac{17}{1} \right) = 17^3$

$\dfrac{F_m}{F_p} = \left( \left\dfrac{1}{17} \right)^3$ = (1/17)^3 ≈ 0.0002

The ratio of the drag coefficients $\dfrac{F_m}{F_p}$ ≈ 0.0002.

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2 years ago

A gun charges a $50 activation fee and $17 per month for a membership. If you spend $356, for how many months do you have a gym

Alex787 [66]

= $356 - $50
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3 years ago

You want to put a fence around your field. The cost of fencing is $60 per meter. The

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Cost per meter(C) = $60/m
Length of rectangular field(L) = 50m
L = 2W
-> W = L/2
-> W = 50/2
-> Width of rectangular field = 25m
Cost of one field length(l) = L x C
-> l = 50 x 60
-> l = $3000
Two of the lengths of the field = 2 x l
-> 2 x $3000
-> $6000

Cost of one field width(w) = W x C
-> w = 25 x 60
-> w = $1500
Two of the widths of the field = 2 x w
-> 2 x $1500
-> $3000

Cost of fencing entire field = $6000+$3000
Hence, total field cost = $9000

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2 years ago