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iren2701 [21]
3 years ago
7

Find 12 ÷ 3/5. Show or explain your reasoning.

Mathematics
2 answers:
scZoUnD [109]3 years ago
3 0
20 because when you divide 12 by 3/5 you get 20 use a calculator
pychu [463]3 years ago
3 0
I think it’s 20 good luck
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Write the expression: y divided by the quantity x minus 5.
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Find the degree of the polynomial: 7.5x^3yz+x^7z^2+2x^3y^3z−x^4z
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The degree of this polynomial is 9


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Describe the end behavior of a 14th degree polynomial with a positive leading coefficient.
kipiarov [429]
The function will enter the graph graph in the upper left hand region and exit in the upper right hand region and overall the graph will be concave upwards.  
For determining the end behavior of a polynomial, there is just 2 things to take notice of. 
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2. Is the degree of the polynomial odd or even?  
For odd ordered polynomials, the curve starts in either quadrant II or III, and ends in quadrant IV, or I. Basically, if it's positive, the curve enters the graph somewhere in the lower left hand region, and exits the graph in the upper right hand region. If the coefficient is negative, it enters in the upper left hand region, and exits in the lower right hand region.  
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4 0
3 years ago
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Gre4nikov [31]
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6 0
3 years ago
Please help me! I’m failing!!!
choli [55]

a. If G(t) = a (1 + r)ᵗ, then G(0) = a corresponds to the starting area of the glacier, which is given to be 142 acres. So a = 142.

The area of the glacier shrinks by 4.4% each year. This means that after 1 year, the area is reduced to

142 - (4.4% of 142) = 142 (1 - 0.044) ≈ 135.75 acres

After another year,

135.75 - (4.4% of 135.75) = 135.75 (1 - 0.044) ≈ 129.78 acres

or equivalently, 142 (1 - 0.044)² acres.

And so on. After t years, the glacier would have an area of

G(t) = 142 (1 - 0.044)ᵗ   ⇒   G(t) = 142 × 0.956ᵗ

b. If the year 2007 corresponds to t = 0, then 2012 refers to t = 5. Then the area of the glacier is

G(5) = 142 × 0.956⁵   ⇒   G(5) ≈ 113.39 acres

c. Simply take the difference between G(5) and G(0) :

G(5) - G(0) ≈ 142 - 113.39 ≈ 28.61 acres

d. The average rate of change of G(t) from 2007 to 2012 is given by the difference quotient of G(t) over the interval 0 ≤ t ≤ 5 :

ARC_{[2007,2012]} = \dfrac{G(5) - G(0)}{5 - 0} \approx \dfrac{113.39 - 142}5

so that ARC ≈ -5.72 acres/year. This rate tells us that a little less than 6 acres of glacial ice is lost each year, based on the reduction over the first 5 years.

e. The year 2017 refers to t = 10, so now we compute

ARC_{[2012,2017]} = \dfrac{G(10) - G(5)}{10 - 5} \approx \dfrac{90.55-113.39}5

which comes out to about ARC ≈ -4.57 acres/year. Since this rate is smaller than the ARC between 2007 and 2012, this means that the glacial ice is disappearing at a slower rate in later years.

8 0
2 years ago
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