15% of 20 gal = 0.15 * 20 gal = 3 gal, so the solution contains 3 gal of salt.
If we add <em>x</em> gal of water to the solution, we end up with (20 + <em>x</em>) gal of solution. We want the new mixture to have a concentration of 10%, or
10% of (20 + <em>x</em>) gal = 0.1 * (20 + <em>x</em>) gal = 2 + 0.1<em>x</em> gal
of salt.
The amount of salt in the tank hasn't changed. Solve for <em>x </em>:
2 + 0.1<em>x</em> = 3
0.1<em>x</em> = 1
<em>x</em> = 10 gal
Answer:
the first one
Step-by-step explanation:
check which graph passes through the point (2,4)
Answer:
dy/dx = -b/a cot α
Step-by-step explanation:
x² / a² + y² / b² = 1
Take derivative with respect to x.
2x / a² + 2y / b² dy/dx = 0
2y / b² dy/dx = -2x / a²
dy/dx = -b²x / (a²y)
Substitute:
dy/dx = -b²a cos α / (a²b sin α)
dy/dx = -b cos α / (a sin α)
dy/dx = -b/a cot α
Answer:
Yes they can all be written in y = mx + b. You just have to move the terms around.
Step-by-step explanation:
y = 2x -3, this is already in slope-intercept form
Now, y - 2 = x + 2: We can add 2 on both sides to cancel out the one on the left side:
y - 2 = x + 2
y - 2 + 2 = x + + 2
y = x + 4 <-- This is in y = mx + b form
Now the last one, 3x = 9 + 3y
We can first divide all terms by 3,
3x = 9 + 3y
/3 /3 /3
x = 3 + y: Then we can subtract 3 from both sides:
x - 3 = 3 + y - 3
x - 3 = y
These are all linear equations because none of the x's have bigger powers than 1. x^2 is a quadratic equation and x^3 is cubic equation.
I think you mean b is not equal to 0.
We have that 11.424242... = 11 + 0.424242...
Now let x = 0.424242...
Then 100x = 42.424242...
Now 99x = 100x - x = 42 so 99x = 42 and therefore x = 42/99
Now we can simplify 42/99 to 14/33
Now combining this we get 11.424242... = 11 + 14/33
= 363/33 + 14/33 = 377/33
