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alukav5142 [94]

3 years ago

14

Please help Algebra . ASAP

1 answer:

Norma-Jean [14]3 years ago

8 0

4 different salads, 7 different main courses, and 6 different desserts

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I will give brainlest if it’s right Which expression is equivalent to (24 + 9)?

MrRissso [65]

Step-by-step explanation:

the answer is D

this is because, 24 + 9 = 33

and (3×8 + 3×3 =33)

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3 years ago

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Leeza is making labels in the shape of parallelograms Each label has an area of 18 square centimeters and a base of 6 centimeter

stich3 [128]

I think it's 3

It should be 3 because if you have a base of 6 then your height must be 3 to get 18.

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3 years ago

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I. Write the equation of a parabola, in standard form, that goes through these points: (0, 3) (1, 4) (-1, -6) II. Graph the para

Anna11 [10]

Vertex = (1,4)
Axis of symmetry is x = 1, (1,0) or just 1

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7 0

3 years ago

Please help. Write as a single logarithm, a) log(13)+log(4) b) log(small 5)(14)-log(small 5)(2)

dsp73

Answer:

a) $\log 52$

b) $\log_57$

Step-by-step explanation:

Use logarithm properties:

$\log_ab+\log_ac=\log_a(b\cdot c)\\ \\\log_ab-\log_ac=\log_a\dfrac{b}{c}$

Then

a) $\log 13+\log 4=\log 13\cdot 4=\log 52$

b) $\log_514-\log _52=\log _5\dfrac{14}{2}=\log _57$

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