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3 years ago

What are the square numbers between 99 and 199

Mathematics

2 answers:

Inessa05 [86]3 years ago

7 0

100 (10)
121 (11)
144 (12)
169 (13)
196 (14)

BabaBlast [244]3 years ago

4 0

Squared are:

196(14)

144(12)

100(10)

121(11)

169(13)

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an algebra teacher drove by a farmyard full of chickens and pigs.the teacher happened to notice that there were a totals of 100

LiRa [457]

I'll be honest, this is a pretty interesting question. I've never seen anything like it before!

We can start off by making equations with the given. If there are 100 heads, that means that the total number of chickens and pigs combined is 100. Thus, we can form our first equation:

p + c = 100
('p' represents number of pigs, and 'c' represents number of chickens)

We also know that there are 270 legs total. Since chickens have 2 and pigs have 4, we can make the following equation:
2c + 4p = 270

I suppose the quickest way to solve for this equation is to use the subtraction method. Put the equations on top of each other:

2c + 4p = 270
c + p = 100
___________
Multiply the bottom equation by -2;

2c + 4p = 270
-2c -2p = 100
____________
"Add" these two equations together:

2c + 4p = 270
-2c -2p = -200
___________
2p = 70

Divide both sides by 2:
p = 35

Ah, finally. Now that we know that there are 35 pigs, input this variable into the first equation to find the number of chickens:

c + 35 = 100
Subtract:
c = 65

There are 35 pigs and 65 chickens
-T.B.

3 0

3 years ago

Consider the curve defined by the equation y=6x2+14x. Set up an integral that represents the length of curve from the point (−2,

torisob [31]

Answer:

32.66 units

Step-by-step explanation:

We are given that

$y=6x^2+14x$

Point A=(-2,-4) and point B=(1,20)

Differentiate w.r. t x

$\frac{dy}{dx}=12x+14$

We know that length of curve

$s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx$

We have a=-2 and b=1

Using the formula

Length of curve= $s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx$

Using substitution method

Substitute t=12x+14

Differentiate w.r t. x

$dt=12dx$

$dx=\frac{1}{12}dt$

Length of curve= $s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt$

We know that

$\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C$

By using the formula

Length of curve= $s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})$