Item 9 Evaluate (3−1) 3 +7(6)− 5 2 (3−1)3+7(6)−52 .

HELP!!! How do you do a diophantine equation? Plz help as soon as possible.

Annette [7]

Answer and Step-by-step explanation:

A Diophantine equation is simply an equation that relates whole numbers.

There are various types of Diophantine equations, but probably the most basic is the linear version: ax + by = c.

Different methods also abound for solving these types of equations:

- Coordinate geometry

- Modular arithmetic

- Induction

Hope this helps!

4 0

3 years ago

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If ln(2x=y)=x=1 then dy/dx =

umka21 [38]

N(2x+y) = x +1

Differentiate both sides, using the Chain Rule on the left side.
(1 / (2x + y)) * d(2x + y)/dx = 1
(1 / (2x + y)) * (2 + dy/dx) = 1

Rearrange to isolate dy/dx.

3 0

3 years ago

I have to get this so I pass help plz

pshichka [43]

Answer:

$\boxed{\sf x=-2}$

$\boxed{\sf y=-4}$

Step-by-step explanation:

First, Let's solve for x in -2x+2y=-4:

$\sf -2x+2y=-4$

Subtract 2y from both sides:

$\sf -2x+2y-2y=-4-2y$

$\sf -2x=-4-2y$

Divide both sides by -2:

$\sf \cfrac{-2x}{-2}=-\cfrac{4}{-2}-\cfrac{2y}{-2}$

$\bold{ x=y+2}$

Now, we'll substitute x=y+2 to 3x+3y=-18:

$\sf 3x+3y=-18$

→ let x=2+y

$\sf 3\bold{(2+y)}+3y=-18$

Simplify:

$\sf 6+6y=-18$

Now, let's solve for y in 6+6y=-18

$\sf 6+6y=-18$

Subtract 6 from both sides:

$\sf 6+6y-6=-18-6$

$\sf 6y=-24$

Divide both sides by 6:

$\sf \cfrac{6y}{6}=\cfrac{-24}{6}$

$\bold{ y=-4}$

Now, substitute y=-4 into x=2+y:

$\sf x=2+y$

→ let y = -4

$\sf x=2+\bold{-4}$

$\bold{x=-2}$

Therefore, x=-2 and y=-4.

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5 0

2 years ago

11. John puts a fence around his rectangular backyard. The width of his backyard is 10 feet. He

Lady_Fox [76]

Length = 20 ft

If he uses 60 ft of fencing to line his rectangular backyard, then the perimeter is 60 ft.

Perimeter formula is ((Width × 2) + (Length × 2))

10 × 2 = 20

20 × 2 = 40

40 + 20 = 60

4 0

2 years ago

Solve this for Brainliest! See the attachment [Level - GCSE]

madreJ [45]

Answer:

$x = 3 + 4\sqrt{3}$

Step-by-step explanation:

OPQ is a right angle triangle.

Using Pythagoras

$x^{2} + (x+5)^{2} = (x+8)^{2}\\x^{2} + x^{2} + 10x +25 = x^{2} + 16x + 64\\x^{2} + x^{2} - x^{2} + 10x - 16x + 25 - 64 = 0\\x^{2} - 6x -39 = 0$

Using quadratic formula

$x = \frac{-b \± \sqrt{b^{2}-4ac}}{2a}\\x = \frac{-(-6) \± \sqrt{(-6)^{2}-4(1)(-39)}}{2(1)}\\x = \frac{6}{2} \± \frac{\sqrt{192}}{2}\\x = 3 \± \frac{8\sqrt{3}}{2}\\x = 3 \± 4\sqrt{3}$

$x = 3 + 4\sqrt{3} = 9.928(4s.f.)$

OR

$x = 3 - 4\sqrt{3} = -3.928(4s.f.)$ (Length can't be negative)

∴ $x = 3 + 4\sqrt{3}$

3 0

2 years ago

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