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3 years ago

What is the range of possible sizes for side x?

Mathematics

2 answers:

svlad2 [7]3 years ago

8 0

D but with 20 more characters respectfully

oksian1 [2.3K]3 years ago

6 0

Answer: 1.48

Step-by-step explanation:

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Find the equation of the line shown in the graph. Write the equation in slope-intercept form.

Marysya12 [62]

Answer:

y=-2x+6

Step-by-step explanation:

y=mx+b

b= y-intercept

b=6

m= slope

m=y2-y1/x2-x1

take 2 points from the graph

(-3,10) and (3,2)

2-10/3-(-3)

-8/6

-4/3

7 0

3 years ago

HELP I WILL GIVE 100 POINTS!!! NO FAKE ANSWERS, PLEASE ANSWER ALL PARTS!!

katrin2010 [14]

option 1 is not linear because it does not change by the same amount every year. It does not have a constant slope.

option 2 increases by 100 per year, constant slope, straight line, linear

-----------------------

a = A y^n

1100 =A y

1210 = Ay^2

1331 = A y^3

---------------

1210/1100 = y = 1.1

check 1331/1210 = y = 1.1 sure enough

1100 = A (1.1)

A = 1000

a = 1000 * 1.1^n

and the linear option 2

a = 1000+100n

You can put the numbers in for part C

10% compounded is better than 10% simple :)

5 0

3 years ago

Help please if you can

Citrus2011 [14]

It goes up each time bc its 2x2 is r 3x3 is 9 4x4 is 16 and 5x5 is 25. its each number time each number and it goes up 1 everytime. what are the options ?

7 0

3 years ago

PLSSS GET THIS RIGHT

Anuta_ua [19.1K]

6 más 8 son 14 es muy fácil suerte

8 0

3 years ago

A contaminant is leaking into a lake at a rate of R(t) = 1700e^0.06t gal/h. Enzymes that neutralize the contaminant have been ad

olasank [31]

Answer:

16,460 gallons

Step-by-step explanation:

This is a differential equation problem, we have a constant flow of contaminant into the lake, but also we know that only a fraction of that quantity of contaminant remains because of the enzymes. For that reason, the differential equation of contaminant's flow into the lake would be:

$\frac{dQ}{dt} =1700exp(0.06t)*exp(-0.32t)\\\frac{dQ}{dt} =1700exp(-0.26t)\\$

Then, we have to integrate in order to find the equation for Q(t), as the quantity of contaminant in the lake, in function of time.

$\int\limits^0_t {dQ}=\int\limits^0_t {1700exp(-0.26t)dt}\\Q(t)=\frac{1700}{-0.26} exp(-0.26t)+C \\$

Now, we use the given conditions to replace them in the equation, in order to solve for $C$

$t_{0} =0\\Q_{0}=10,000\\Q_{0}=-6538exp(-0.26*0)+C\\C=10,000+6538=16538$

Then, we reorganize the equation and we replace t for 17 hours, in order to determine the quantity of contaminant at that time:

$Q_{t} =-6538exp(-0.26t)+16538\\Q_{17} =-6538exp(-0.26*17)+16538\\Q_{17} =16460 gallons$

3 0

3 years ago