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3 years ago

“solve the literal equation for the variable t:” xt-3=y

Mathematics

2 answers:

natali 33 [55]3 years ago

8 0

Xt-3=y
Rearrange to find t.
xt=y+3
t= y+3/x

crimeas [40]3 years ago

6 0

Answer:

$\frac{y+3}{x}=t$

Step-by-step explanation:

To solve for variable t, you need to isolate t.

$y=xt-3\\y+3=xt\\\frac{y+3}{x}=t$

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The height of a cone is twice the radius of its base.

fredd [130]

Answer: 2/3πx³

Step-by-step explanation:

Let the radius of the cone be represented by x.

Since the height of the cone is twice the radius of its base, the height will be: = 2x

Volume of a cone = 1/3πr²h

where,

r = x

h = 2x

Volume of a cone = 1/3πr²h

= 1/3 × π × x² × 2x

= 1/3 × π × 2x³

= 2/3πx³

Therefore, the correct answer is 2/3πx³.

3 0

4 years ago

108 identical books have a mass of 30 kg. Find

Lynna [10]

Answer:

(i) 540kg

(ii) 5.56 (rounded off to 3 significant figures)

Step-by-step explanation:

$108 \div 30 = 3.6kg$

$(i)3.6 \times 150 = 540kg$

$(ii)3.6 \times x = 20$

$3.6x = 20$

$x = \frac{20}{3.6}$

$x = 5 \frac{5}{9}$

4 0

3 years ago

Helpppp plsssss!!thanks

DedPeter [7]

A
$2x - 6 - (3x + 9)$

5 0

3 years ago

Solve the initial value problem where y′′+4y′−21 y=0, y(1)=1, y′(1)=0 . Use t as the independent variable.

igor_vitrenko [27]

Answer:

$y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

Step-by-step explanation:

y′′ + 4y′ − 21y = 0

The auxiliary equation is given by

m² + 4m - 21 = 0

We solve this using the quadratic formula. So

$m = \frac{-4 +/- \sqrt{4^{2} - 4 X 1 X (-21))} }{2 X 1}\\ = \frac{-4 +/- \sqrt{16 + 84} }{2}\\= \frac{-4 +/- \sqrt{100} }{2}\\= \frac{-4 +/- 10 }{2}\\= -2 +/- 5\\= -2 + 5 or -2 -5\\= 3 or -7$

So, the solution of the equation is

$y = Ae^{m_{1} t} + Be^{m_{2} t}$

where m₁ = 3 and m₂ = -7.

So,

$y = Ae^{3t} + Be^{-7t}$

Also,

$y' = 3Ae^{3t} - 7e^{-7t}$

Since y(1) = 1 and y'(1) = 0, we substitute them into the equations above. So,

$y(1) = Ae^{3X1} + Be^{-7X1}\\1 = Ae^{3} + Be^{-7}\\Ae^{3} + Be^{-7} = 1 (1)$

$y'(1) = 3Ae^{3X1} - 7Be^{-7X1}\\0 = 3Ae^{3} - 7Be^{-7}\\3Ae^{3} - 7Be^{-7} = 0 \\3Ae^{3} = 7Be^{-7}\\A = \frac{7}{3} Be^{-10}$

Substituting A into (1) above, we have

$\frac{7}{3}B e^{-10}e^{3} + Be^{-7} = 1 \\\frac{7}{3}B e^{-7} + Be^{-7} = 1\\\frac{10}{3}B e^{-7} = 1\\B = \frac{3}{10} e^{7}$

Substituting B into A, we have

$A = \frac{7}{3} \frac{3}{10} e^{7}e^{-10}\\A = \frac{7}{10} e^{-3}$

Substituting A and B into y, we have

$y = Ae^{3t} + Be^{-7t}\\y = \frac{7}{10} e^{-3}e^{3t} + \frac{3}{10} e^{7}e^{-7t}\\y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

So the solution to the differential equation is

$y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

6 0

4 years ago

Find the coordinates of the image of the point (0,2) when it is reflected across the line x=3

Firdavs [7]

Answer:

(6,2)

Step-by-step explanation:

because math

8 0

3 years ago