MATHEMATICS HIGH SCHOOL

Ax+by=1
bx-ay=a+b
solve in linear equation in 2 variables

Answers

Answer 1
Answer: \left\{\begin{array}{ccc}ax+by=1&/\cdot a\nbx-ay=a+b&/\cdot b\end{array}\right\n\n+\left\{\begin{array}{ccc}a^2x+aby=a\nb^2x-aby=ab+b^2\end{array}\right\n------------\n.\ \ \ \ \ a^2x+b^2x=a+ab+b^2\n.\ \ \ \ \ \ (a^2+b^2)x=a+ab+b^2\n.\ \ \ \ \ \ \ \ \ \ \ \ x=(a+ab+b^2)/(a^2+b^2)\n\na\cdot(a+ab+b^2)/(a^2+b^2)+by=1\n\n(a^2+a^2b+ab^2)/(a^2+b^2)+by=1\n\nby=1-(a^2+a^2b+ab^2)/(a^2+b^2)

by=(a^2+b^2)/(a^2+b^2)-(a^2+a^2b+ab^2)/(a^2+b^2)\n\nby=(a^2+b^2-a^2-a^2b-ab^2)/(a^2+b^2)\n\nby=(b^2-a^2b-ab^2)/(a^2+b^2)\n\ny=(b^2-a^2b-ab^2)/(a^2b+b^3)

y=(b(b-a^2-ab))/(b(a^2+b^2))\n\ny=(b-a^2-ab)/(a^2+b^2)\n\nAnswer:\n\nx=(a+ab+b^2)/(a^2+b^2)\ and\ y=(b-a^2-ab)/(a^2+b^2)
Answer 2
Answer: ax+by=1
bx-ay=a+b
The solution in attached file

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If two angles are both obtuse, the two angles are equal. What is the converse??

Answers

Answer:

If the angles are equal, then two angles are both obtuse

Step-by-step explanation:

converse flips the conclusion and hypothesis

Y=2x-1 Graph the equation? (steps please thanks!)

Answers

y=2x-1\n\ny-intercept\ (x=0)\n\ny=2\cdot0-1=-1\to(0;-1)\n\nx-intercept\ (y=0)\n\n2x-1=0\n2x=1\nx=0.5\to(0.5;\ 0)
There are many ways to graph this equation.
Here's one that you should be able to manage:

-- Pick any number at all for 'x'. 
[  y = 2x - 1  ], so double your number and subtract 1 to find 'y'.
Mark a point on your graph at (x, y) .

-- Pick any different number at all for 'x'.
[  y = 2x - 1  ], so double your number and subtract 1 to find 'y'.
Mark a point on your graph at (x, y) .

-- Using your pencil and ruler, connect the two points that you have marked
on your graph.  Without moving the ruler, you may extend the line as far as you
want in each direction.

Find the product. y^5·y^3

Answers

The product of the exponential expressions are A = y^(8).

Given data:

The first expression is represented as p.

The second expression is represented as q.

The value of p=y^(5)

The value of q=y^(3)

So, the product is A = p * q

Substituting the values in the expression:

A=y^(5) *y^(3)

From the laws of exponents:

m^(a)*m^(b) = m^(a+b)

So, A = y^(5+3)

A = y^(8)

Hence, the product is y^(8).

To learn more about exponents:

brainly.com/question/28966438

#SPJ6
It would be y^8 you add if its the same variable in front basically

Find all polar coordinates of point P where P = ordered pair 6 comma negative pi divided by 5.

Answers

Answer:

All polar coordinates of point P are (6,-(\pi)/(5)+2n\pi) and (-6,-(\pi)/(5)+(2n+1)\pi), where n is an integer.

Step-by-step explanation:

The given point is

P=(6,-(\pi)/(5))

If a point is P=(r,\theta), then all polar coordinates of point P are defined as

(r,\theta)=(r,\theta+2n\pi)

(r,\theta)=(-r,\theta-(2n+1)\pi)

where n is an integer.

In the given point r=6 and \theta=-(\pi)/(5). So all polar coordinates of point P are defined as

(6,-(\pi)/(5))=(6,-(\pi)/(5)+2n\pi)

(6,-(\pi)/(5))=(-6,-(\pi)/(5)+(2n+1)\pi)

Therefore all polar coordinates of point P are (6,-(\pi)/(5)+2n\pi) and (-6,-(\pi)/(5)+(2n+1)\pi), where n is an integer.

Answer:

P(6,-π/6) = (6, -π/6 + 2nπ) and P(-6,-π/6) = (-6, -π/6 + (2n+1)π)

Step-by-step explanation:

We need to find all polar coordinates of

P = (6,(-\pi )/(6))

The polar coordinates of any point can be reresented by (r,Ф)

where

(r,Ф) = (r, Ф+2nπ) where n is any integer and r is positive

and

(r,Ф) = (-r, Ф+(2n+1)π) where n is any integer and r is negative.

So, in the question given, r = 6 and Ф = -π/6

So, Polar coordinates will be:

P(6,-π/6) = (6, -π/6 + 2nπ) where where n is any integer and r is positive

and

P(-6,-π/6) = (-6, -π/6 + (2n+1)π) where n is any integer and r is negative.

Find the difference.

(3−2x+2x^2)−(4x−5+3x^2)

Answers

Answer:

the difference is that there is (3-2x) at the start instead of (4x-5)

step explanation:

Which values represent the first two terms of the sequence a1=2 and an=-2(an-1)^2?2, -8
-2, 16
-8,-128
16, 1024

Answers

Hello,

a_(1)=2
a_(n+1)=-2*a_(n)^2

a_(1)=2
a_(2)=-2*2^(2)=-8
a_(3)=-2*(-8)^(2)=-2*64=-128


Answer C
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