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3 years ago

T is the midpoint of MV. If MT = 6x and TV = 3x + 3, find the value of x, MT, TV, and MV.

Mathematics

2 answers:

Troyanec [42]3 years ago

7 0

Answer:

x=1

MT = 6

TV= 6

MV = 12

Step-by-step explanation:

M------T------V

6x 3x+3

6 6

Dvinal [7]3 years ago

5 0

Answer:

TV = 3x+9 - (14+2x)

TV = 3x + 9 -14 - 2x

TV = x - 5

U = ((3x + 9) + (14 + 2x)) / 2

U = (5x + 23) / 2

TU = (2(3x + 9) / 2) - (5x+23)/2

TU = (6x + 18 - 5x -23) /2

TU = (x - 5) / 2

Since U is the midpt, UV = TU, so UV = (x-5) / 2

TU + UV = TV

(x-5) / 2 + (x-5) /2 = 2 (x-5) /2 = x-5 = TV

Use a number for x. Let's use 10.

T is 39, V = 34

U = (39 + 34)/2 = 73/2 = 36.5

TV = 39-34 = 5

TV from our equation is x-5. 10 - 5 = 5.

TU and UV = (x-5) / 2. (10-5) /2 = 5/2.

(39-34)/2 = 5/2

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3 years ago

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3 years ago

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Viktor [21]

Answer:

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2 years ago

Solving separable differential equation DY over DX equals xy+3x-y-3/xy-2x+4y-8

Ivanshal [37]

It looks like the differential equation is

$\dfrac{dy}{dx} = \dfrac{xy + 3x - y - 3}{xy - 2x + 4y - 8}$

Factorize the right side by grouping.

$xy + 3x - y - 3 = x (y + 3) - (y + 3) = (x - 1) (y + 3)$

$xy - 2x + 4y - 8 = x (y - 2) + 4 (y - 2) = (x + 4) (y - 2)$

Now we can separate variables as

$\dfrac{dy}{dx} = \dfrac{(x-1)(y+3)}{(x+4)(y-2)} \implies \dfrac{y-2}{y+3} \, dy = \dfrac{x-1}{x+4} \, dx$

Integrate both sides.

$\displaystyle \int \frac{y-2}{y+3} \, dy = \int \frac{x-1}{x+4} \, dx$

$\displaystyle \int \left(1 - \frac5{y+3}\right) \, dy = \int \left(1 - \frac5{x + 4}\right) \, dx$

$\implies \boxed{y - 5 \ln|y + 3| = x - 5 \ln|x + 4| + C}$

You could go on to solve for $y$ explicitly as a function of $x$ , but that involves a special function called the "product logarithm" or "Lambert W" function, which is probably beyond your scope.

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1 year ago