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3 years ago

7

Use the formulas to generate a Pythagorean Triple with x = 5 and y = 2. The three side lengths, from smallest to largest are: a0

, a1, and a2.

2 answers:

stiv31 [10]3 years ago

8 0

<span>Perhaps you mean that for any integers x and y, x > y, xÂ˛ - yÂ˛, 2xy, and xÂ˛ + yÂ˛ form a Pythagorean triple. If GCF(x, y) = 1, the triple is primitive. For x = 5, y = 2, the triple is 21, 20, 29, or from smallest to largest, 20, 21, 29 âś Let's just check 20Â˛ + 21Â˛ = 841= 29Â˛</span>

Sliva [168]3 years ago

7 0

Answer:

Step-by-step explanation:

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3 years ago

What is the difference of the polynomials?

Vilka [71]

Answer:

$8r^6s^3-5r^5s^4+r^4s^5+5r^3s^6$

Step-by-step explanation:

We want to simplify;

$(8r^6s^3-9r^5s^4+3r^4s^5)-(2r^4s^5-5r^3s^6-4r^5s^4)$

We expand the bracket to obtain;

$8r^6s^3-9r^5s^4+3r^4s^5-2r^4s^5+5r^3s^6+4r^5s^4$

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We now simplify to get;

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The correct answer is C

3 0

4 years ago

Read 2 more answers

your automobile insurance semiannual premium is 750 and you have 1050 deductible per incident. In a particularly bad year, you h

nataly862011 [7]

Answer:

$3600

Step-by-step explanation:

In each case, the repair cost exceeded the deductible, so your out-of-pocket cost for 2 repairs is ...

2 × $1050 = $2100

In addition, you had to pay the premium of $750 per 6-month period, or $1500 per year, so your total was ...

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6 0

4 years ago

The length of a 200 square foot rectangular vegetable garden is 4feet less than twice the width. Find the length and width of th

inna [77]

Answer:

Length = 18.099 ft

Width = 11.049 ft

Step-by-step explanation:

let the length of the field be x ft

and the width be y ft

as per the condition given in problem

x=2y-4 -----------(A)

Also the area is given as 200 sqft

Hence

xy=200

Hence from A we get

y(2y-4)=200

taking 2 as GCF out

2y(y-2)=200

Dividing both sides by 2 we get

$y(y-2)=100$

$y^2-2y=100$

subtracting 100 from both sides

$y^2-2y-100=0$

Now we solve the above equation with the help of Quadratic formula which is given in the image attached with this for any equation in form

$ax^2+bx+c=0$

Here in our case

a=1

b=-1

c=-100

Putting those values in the formula and solving them for y

$y=\frac{-(-2)+\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

$y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

Solving first

$y=\frac{2+\sqrt{4+400}{2}$

$y=\frac{2+\sqrt{404}{2}$

$y=\frac{2+20.099}{2}$

$y=\frac{22.099}{2}$

$y=11.049$

Solving second one

$y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

$y=\frac{2-\sqrt{4+400}{2}$

$y=\frac{2-\sqrt{404}{2}$

$y=\frac{2-20.099}{2}$

$y=\frac{-18.99}{2}$

$y=-9.045$

Which is wrong as the width can not be in negative

Our width of the field is

y=11.099

Hence the length will be

x=2y-4

x=2(11.049)-4

x=22.099-4

x=18.099

Hence our length x and width y :

Length = 18.099 ft

Width = 11.049 ft

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3 years ago

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Korvikt [17]

Answer:C

Step-by-step explanation:on edge

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