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3 years ago

Simplify: −z3+5k6−(−z3+10k6)

Mathematics

2 answers:

natta225 [31]3 years ago

7 0

Answer:

The answer to your question is -5k⁶

Step-by-step explanation:

Polynomial

- z³ + 5k⁶ - (-z³ + 10k⁶)

Use law of signs for the parenthesis

- z³ + 5k⁶ + z³ - 10k⁶

Use associative property for like terms

(-z³ + z³) + (5k⁶ - 10k⁶)

Simplify like terms

(0) + (-5k⁶)

Result

-5k⁶

Likurg_2 [28]3 years ago

4 0

Answer:

-5k ^6 is your answer! Hope this helps!

You might be interested in

F(x) = x2+3x+2 x2+5x+4 Why is there no zero at x = –1?

dmitriy555 [2]

You could rewrite $F(x)$ as

$\dfrac{x^2+3x+2}{x^2+5x+4}=\dfrac{(x+1)(x+2)}{(x+1)(x+4)}$

and be tempted to cancel out the factors of $x+1$ . But this cancellation is only valid when $x\neq-1$ .

When $x=-1$ , you end up with the indeterminate form $\dfrac00$ , which is why $-1$ is not a zero.

7 0

3 years ago

If a = pi +3j - 7k, b = pi - pj +4k and the angle between a and is acute then the possible values for p are given by

PIT_PIT [208]

Answer:

The family of possible values for $p$ are:

$(-\infty, -4) \,\cup \,(7, +\infty)$

Step-by-step explanation:

By Linear Algebra, we can calculate the angle by definition of dot product:

$\cos \theta = \frac{\vec a\,\bullet\,\vec b}{\|\vec a\|\cdot \|\vec b\|}$ (1)

Where:

$\theta$ - Angle between vectors, in sexagesimal degrees.

$\|\vec a\|, \|\vec b \|$ - Norms of vectors $\vec {a}$ and $\vec{b}$

If $\theta$ is acute, then the cosine function is bounded between 0 a 1 and if we know that $\vec {a} = (p, 3, -7)$ and $\vec {b} = (p, -p, 4)$ , then the possible values for $p$ are:

Minimum ( $\cos \theta = 0$ )

$\frac{p^{2}-3\cdot p -28}{\sqrt{p^{2}+58}\cdot \sqrt{2\cdot p^{2}+16}} > 0$

Maximum ( $\cos \theta = 1$ )

$\frac{p^{2}-3\cdot p -28}{\sqrt{p^{2}+58}\cdot \sqrt{2\cdot p^{2}+16}} < 1$

With the help of a graphing tool we get the family of possible values for $p$ are:

$(-\infty, -4) \,\cup \,(7, +\infty)$

7 0

3 years ago

15 points!

Phantasy [73]

Hi there! You are provided with two points, this is all you need to calculate the line's slope.

The equation for slope is:

$\frac{y2 - y1}{x2 - x1}$

$\frac{10 - 8}{3 - 7} = -\frac{2}{4}$

Slope: -2/4

Point slope form, or y-y1 = m(x - x1). All we have to do is take the x and y coordinates of either point, as well as the slope, and plug them into the equation.

y - 10 = -2/4(x - 3)

Write the equation in slope-intercept form (y=mx+b)

$y - 10 = -2/4(x-3)\\y = -2/4(x-3) + 10\\y = -2/4x + 1.5 + 10\\y = -2/4x + 11.5$