Answer: 9/10 of a mile per hour.
Step-by-step explanation:
Answer:
<h2> </h2><h2>

</h2>
Step-by-step explanation:
<h3>
<u>Question</u><u>:</u><u>-</u></h3>
<h3>
<u>Equation</u><u>:</u><u>-</u></h3>
<h3>
<u>Solution</u><u>:</u><u>-</u></h3>
=> 3x + 4y + 6z = 15
- <em>[</em><em>On</em><em> </em><em>subtracting</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>with</em><em> </em><em>4y</em><em>]</em>
=> 3x + 4y + 6z - 4y = 15 - 4y
- <em>[</em><em>On</em><em> </em><em>Simplification</em><em>]</em>
=> 3x + 6z = 15 - 4y
- <em>[</em><em>On</em><em> </em><em>subtracting</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>with</em><em> </em><em>6z</em><em>]</em>
=> 3x + 6z - 6z = 15 - 4y - 6z
- <em>[</em><em>On</em><em> </em><em>Simplification</em><em>]</em>
=> 3x = 15 - 4y - 6z
- <em>[</em><em>On</em><em> </em><em>dividing</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>with</em><em> </em><em>3</em><em>]</em>

- <em>[</em><em>On</em><em> </em><em>Simplification</em><em>]</em>

Answer:
24 square units
Step-by-step explanation:
Use the formula for area of a parallelogram to solve. The base is 6 units, and the height is 4 units.
A = bh
A = (6)(4)
A = 24 square units
The area of the parallelogram is 24 square units.
Answer:
k = 7
Step-by-step explanation:
The given figures are lines f(x) and g(x)
For the line f(x), we have the y-intercept at (0, -3) and slope = (-1 - (-3))/(-3 - 0) = -2/3
Therefore, line f(x) = y - (-3) = -2/3·(x - 0) which gives f(x) = y = -3 - 2·x/3
For the line g(x), the y-intercept is (0, 4), and the slope is (4 - 2)/(0 - 3) = -2/3
The equation of the line g(x) is therefore, g(x) = y - 4 = -2/3·x, which simplifies to the slope and intercept form as g(x) = y = 4 - 2/3·x
Therefore, given that the transformation of f(x) to g(x) is given as g(x) = f(x) + k, we have;
k = g(x) - f(x) = 4 - 2/3·x - (-3 - 2·x/3) = 4 - 2/3·x + 3 + 2·x/3 = 7
∴ k = 7
The fundamental theorem of algebra states that a polynomial with degree n has at most n solutions. The "at most" depends on the fact that the solutions might not all be real number.
In fact, if you use complex number, then a polynomial with degree n has exactly n roots.
So, in particular, a third-degree polynomial can have at most 3 roots.
In fact, in general, if the polynomial
has solutions
, then you can factor it as

So, a third-degree polynomial can't have 4 (or more) solutions, because otherwise you could write it as

But this is a fourth-degree polynomial.