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Serhud [2]
2 years ago
9

HELP ASAP

Mathematics
2 answers:
erma4kov [3.2K]2 years ago
6 0
Imposter is sus (okay i’m sorry i just want points for my test or i’ll fail)
STALIN [3.7K]2 years ago
4 0

Answer and Step-by-step explanation:

First, you would subtract \frac{1}{2} (or 0.5) to both sides.

\frac{1}{x} = 1.5

Now multiply x to both sides, then divide both sides by 1.5.

\frac{2}{3}  = x

<u>Two thirds is equal to x.</u>

<u><em>I hope this helps!</em></u>

<u><em></em></u>

<u><em>#teamtrees #WAP (Water And Plant)</em></u>

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5

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A. (4, 5)

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Consider the equation 30÷x=6.
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Which graph represents this equation?
antiseptic1488 [7]

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The third option. I’m not 100% sure but I do think that’s the answer

Step-by-step explanation:

6 0
3 years ago
If f(x) = x/2 -3 and g(x) = 3x² + x - 6, find (f + g)(x). O A. 3x O B. 3x² + 3x-9 3x O C. 3.x2 + x - 3 O D. 4x2-9​
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8 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
2 years ago
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