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3 years ago

PLEASE HELP! i have a couple minutes left

Mathematics

2 answers:

swat323 years ago

8 0

Answer:

1.8

Step-by-step explanation:

slope formula: (y2-y1)/(x2-x1)

(86-32)/(30-0)

54/30

1.8

Olin [163]3 years ago

6 0

Answer:

Im just guessing here because i havent done this in a while but i think the answer is 1.8

Step-by-step explanation:

I used the slope formula y2-y1 over x2-x1
So placed the corridanats in and i got 32-86 over o-30
I then subtracted 86 from 32 and got -54
I subtracted 30 from 0 and got -30
Then i divided -54 by -30 and got 1.8
aI hope this helps i havent done this in a while
(good luck)

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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:

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

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:

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

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3 years ago

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marshall27 [118]

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Step-by-step explanation:

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Slav-nsk [51]

Complete question:

CHECK ATTACHMENT FOR THE DIAGRAM.

Answer:

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Step-by-step explanation:

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