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djverab [1.8K]
3 years ago
8

Plz help with the last 2

Mathematics
1 answer:
juin [17]3 years ago
6 0
I would say the asymptote is at x=2 and the domain is (-infinity, 2]
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24 ÷ (–8 – (–4)) = <br> What is the answer
Dominik [7]
24 ÷ (–8 – (–4)) = -6 hope this helps
8 0
3 years ago
Read 2 more answers
Someone please help I don’t understand this please
sdas [7]

The fraction of the image in blue with respect to the <em>entire</em> hexagon is \frac{1}{3}.

<h3>Estimation of the ratio of the shaded region area to the entire area</h3>

Initially we proceed to create auxiliar constructions , represented by red line segments, to re-define the hexagon as a sum of standard figures. According to the figure, we find two types of triangles (type-1 and type-2), and the following relationship between the two types:

A_{1} = A_{2} (1)

Where:

  • A_{1} - Area of a type-1 triangle.
  • A_{2} - Area of a type-2 triangle.

The formulae for the area of the shaded region (A_{s}) and the <em>entire</em> hexagon (A_{h}) are, respectively:

<h3>Shaded region</h3>

A_{s} = 2\cdot A_{1} + 4\cdot A_{2} (2)

<h3>Entire hexagon</h3>

A_{h} = 8\cdot A_{1} + 10\cdot A_{2} (3)

And the fraction of the image in blue is:

\frac{A_{s}}{A_{h}} =  \frac{2\cdot A_{1}+4\cdot A_{2}}{8\cdot A_{1}+10\cdot A_{2}}

\frac{A_{s}}{A_{h}} = \frac{6\cdot A}{18\cdot A}

\frac{A_{s}}{A_{h}} = \frac{1}{3}

The fraction of the image in blue with respect to the <em>entire</em> hexagon is \frac{1}{3}. \blacksquare

To learn more on hexagons, we kindly invite to check this verified question: brainly.com/question/4083236

6 0
3 years ago
HELP!! will give brainliest!
klemol [59]

The rate of change of function A is greater than the rate of change of function B.

8 0
3 years ago
A person that weighs 179 lbs needs an IV Drip in mL. The drug only comes in powder form but can be converted as 10 mg is equal t
OleMash [197]

Solution:

Weight of person = 179 lbs

As given person needs IV Drip in m L.

Also,→ 10 mg = 2 m L

Now, you must remember , → 1 lbs = 0.453 Kg (approx)

→179 lbs = 179 × 0.453 Kg = 78.47 Kg (approx)

The person needs at least 60 mg for each Kg they weigh.

⇒ 1 Kg → 60 mg

⇒ 78.47 × 1 Kg → 78.47 × 60 mg

⇒ 78.47 Kg →  4708.2 mg

Also, 10 mg → 2 m L

⇒ 1 mg → \frac{1}{5}  m L

⇒ 4708 . 2 mg → \frac{1}{5} \times 4708.2  m L

⇒ 4708 . 2 mg → 941. 64 ml


8 0
3 years ago
Solve for m.<br><br> 3m+7/2=−2m+5/2<br> I will give brainleist
DiKsa [7]
M = -1/2 or m = 0.2.
8 0
3 years ago
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