Answer:
The area of the rectangle is 12 cm² (to two significant figures)
Step-by-step explanation:
To determine the area of a rectangle that is 2.1-cm wide by 5.6-cm long,
The area of a rectangle is given by
Area = l × w
where l is the length
and w is the width
From the question,
l = 5.6 cm
and w = 2.1 cm
Hence, Area of the given rectangle will be
Area = 5.6 cm × 2.1 cm
Area = 11.76 cm²
Now, we will convert the answer to two significant figures
Then,
Area = 12 cm²
Hence, the area of the rectangle is 12 cm² (to two significant figures)
Answer:
12
Step-by-step explanation:
since i helped can i have brainlist please :D
Answer:
Answer is A. 36x-11y
Step-by-step explanation:
34x-12x+14x-18y+7y
=22x+14x-18y+7y
=36x-11y
The simplified expression of 2√8x³(3√10x⁴ - x√5x²) is 24x³√5x - 4x³√10x
<h3>How to determine the simplified product?</h3>
The complete question is added as an attachment
From the attached figure, the product expression is:
2√8x³(3√10x⁴ - x√5x²)
Evaluate the exponents
2√8x³(3√10x⁴ - x√5x²) = 2 *2x√2x(3x²√10 - x²√5)
Evaluate the products
2√8x³(3√10x⁴ - x√5x²) = 4x√2x(3x²√10 - x²√5)
Open the bracket
2√8x³(3√10x⁴ - x√5x²) = 12x³√20x - 4x³√10x
Evaluate the exponents
2√8x³(3√10x⁴ - x√5x²) = 24x³√5x - 4x³√10x
Hence, the simplified expression of 2√8x³(3√10x⁴ - x√5x²) is 24x³√5x - 4x³√10x
Read more about expressions at:
brainly.com/question/12990602
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Answer:
The function for the outside temperature is represented by
, where t is measured in hours.
Step-by-step explanation:
Since outside temperature can be modelled as a sinusoidal function, the period is of 24 hours and amplitude of temperature and average temperature are, respectively:
Amplitude


Mean temperature


Given that average temperature occurs six hours after the lowest temperature is registered. The temperature function is expressed as:
![T(t) = \bar T + A \cdot \sin \left[2\pi\cdot\frac{t-6\,h}{\tau} \right]](https://tex.z-dn.net/?f=T%28t%29%20%3D%20%5Cbar%20T%20%2B%20A%20%5Ccdot%20%5Csin%20%5Cleft%5B2%5Cpi%5Ccdot%5Cfrac%7Bt-6%5C%2Ch%7D%7B%5Ctau%7D%20%5Cright%5D)
Where:
- Mean temperature, measured in degrees.
- Amplitude, measured in degrees.
- Daily period, measured in hours.
- Time, measured in hours. (where t = 0 corresponds with 5 AM).
If
,
and
, the resulting function for the outside temperature is:
![T(t) = 85\º + 15\º \cdot \sin \left[\frac{t-6\,h}{24\,h} \right]](https://tex.z-dn.net/?f=T%28t%29%20%3D%2085%5C%C2%BA%20%2B%2015%5C%C2%BA%20%5Ccdot%20%5Csin%20%5Cleft%5B%5Cfrac%7Bt-6%5C%2Ch%7D%7B24%5C%2Ch%7D%20%5Cright%5D)