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2 years ago

After i have found my LCD in adding fractions how do I find my new NUMERATIOR

Mathematics

1 answer:

Montano1993 [528]2 years ago

6 0

Can you show a picture on that

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1/3 of a number T is less than or equal to five

ZanzabumX [31]

Answer: x ≤ 15

Step-by-step explanation: All we need to do is divide both sides by 1/3, so1/3x / 1/3 and 5/ 1/3 = 15. So x ≤ 15.

Hope this helps!

8 0

2 years ago

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Write the equation 6x + 2y = 4 in slope-intercept form.

Mademuasel [1]

The slope of 6x + 2y = 4 is
y = -3x + 2

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3 years ago

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Create a word problem that can be represented by the expression below. Then, represent each product or quotient as a single rati

Ber [7]

I have eight coupons that are all worth -.25

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2 years ago

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Blababa [14]

Answer:

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Step-by-step explanation:

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2 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