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2 years ago

Consider the equation 3/ x = 4/x-2 and solve for x

Mathematics

1 answer:

Goryan [66]2 years ago

6 0

Answer:

x = -6

Step-by-step explanation:

3/ x = 4/(x-2)

Using cross products

3 (x-2) = 4x

Distribute

3x - 6 = 4x

Subtract 3x from each side

3x-3x-6 = 4x-3x

-6 =x

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What is the area for 10 and 6 1/2

schepotkina [342]

Answer:

65 units squared

Step-by-step explanation:

Why:area is base length times width. Considering what you gave me, 10 * 6.5 =65

6 0

3 years ago

Please help!! x = 2 - 3 cos t, y = 1 + 4 sin t in rectangular form?<br><br> thank you! :)

sukhopar [10]

Answer:

$\frac{(x-2)^2}{9}+\frac{(y-1)^2}{16}=1$

Step-by-step explanation:

$x=2-3\cos(t)$

$y=1+4\sin(t)$

Let's solve for $\cos(t)$ in the first equation and then solve for $\sin(t)$ in the second equation.

I will then use the following identity to get right of the parameter, $t$ :

$\cos^2(t)+\sin^2(t)=1$ (Pythagorean Identity).

Let's begin with $x=2-3\cos(t)$ .

Subtract 2 on both sides:

$x-2=-3\cos(t)$

Divide both sides by -3:

$\frac{x-2}{-3}=\cos(t)$

Now time for the second equation, $y=1+4\sin(t)$ .

Subtract 1 on both sides:

$y-1=4\sin(t)$

Divide both sides by 4:

$\frac{y-1}{4}=\sin(t)$

Now let's plug it into our Pythagorean Identity:

$\cos^2(t)+\sin^2(t)=1$

$\frac{x-2}{-3})^2+(\frac{y-1}{4})^2=1$

$\frac{(x-2)^2}{(-3)^2}+\frac{(y-1)^2}{4^2}=1$

$\frac{(x-2)^2}{9}+\frac{(y-1)^2}{16}=1$

5 0

2 years ago

Read 2 more answers

-3 (3c + 5) = 2 (10 - c)

Ilya [14]

Answer:-5

Step-by-step explanation:

You need to distribute on both sides so u get

9c-15=20-2c

Move variable to left side and change the sign

-9c+2c-15=20

Move constant to right side and change the sign

-9c+2c=20+15

Connect like terms

-7c=35

Divide

C=-5

5 0

2 years ago

I don't understand how to solve this pls help

Charra [1.4K]

the sine function is a many-to-one function and therefore has no inverse function.

However if the domain is restricted to -90° ≤ x ≤ 90°

Then the function is one-to-one for this domain

Thus, $sin^{-1}$ x is defined as the angle such that - π/2 ≤ x ≤ π/2

5π/6 is therefore outwith the domain

7 0

3 years ago

HELP HELP HELP HELP

scoundrel [369]

Answer:

there is no solutions.

I hope this helps!

7 0

2 years ago