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2 years ago

Do you know how to find rate of change? X&Y intercepts? Zeros?

Mathematics

1 answer:

Rzqust [24]2 years ago

3 0

#Rate of change

Rate of change is slope

If two points be (x1,y-1) and (X2,Y2) then

Slope=m=y_2-y_1/x_2-x_1

If a line be ax+by+c=0

m=-a/b

#X and y inetercept

If a equation given y=mx+b

To find x intercept put y=0
To find y inetercept put x=0

#Zeros

To find the zeros spot out the x inetercepts

The x values of the x intercepts are the zeros

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How do i do this, my son is struggling and he needs help

expeople1 [14]

Answer:

Step-by-step explanation:

angle a is the outside angle. It creates a circle that will always equal 360 degrees. Since the outside angle is 302, the inside angle must be 58 degrees (360-302). The top square angle will be 90 degrees (a square angle is always 90 degrees). Since the angles in a triangle always add up to 180 degrees, we need to add 58 and 90, and then subtract it from 180, which equals 32. So, angle x equals 32.

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2 years ago

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Carter draws one side of equilateral △PQR on the coordinate plane at points P(-3,2) and Q(5,2). Which ordered pair is a possible

Law Incorporation [45]

Step-by-step explanation:

Hey, there!!!

Let me simply explain you about it.

We generally use the distance formula to get the points.

let the point R be (x,y)

As it an equilateral triangle it must have equal distance.

now,

let's find the distance of PQ,

we have, distance formulae is;

$pq = \sqrt{( {x2 - x1)}^{2} + ( {y2 - y1)}^{2} }$

$or \: \sqrt{( {5 + 3)}^{2} + ( {2 - 2)}^{2} }$

By simplifying it we get,

$8$

Now,

again finding the distance between PR,

$pr = \sqrt{( {x2 - x1}^{2} + ( {y2 - y1)}^{2} }$

or,

$\sqrt{( {x + 3)}^{2} + ( {y - 2)}^{2} }$

By simplifying it we get,

$= \sqrt{ {x}^{2} + {y}^{2} + 6x - 4y + 13 }$

now, finding the distance of QR,

$qr = \sqrt{( {x - 5)}^{2} + ( {y - 2)}^{2} }$

or, by simplification we get,

$\sqrt{ {x}^{2} + {y}^{2} - 10x - 4y + 29 }$

now, equating PR and QR,

$\sqrt{ {x}^{2} + {y}^{2} + 6x - 4y + 13} = \sqrt{ {x}^{2} + {y}^{2} - 10x - 4y + 29 }$

we cancelled the root ,

${x}^{2} + {y}^{2} + 6x - 4y + 13 = {x}^{2} + {y}^{2} -10x - 4y + 29$

or, cancelling all like terms, we get,

6x+13= -10x+29

16x=16

x=16/16

Therefore, x= 1.

now,

equating, PR and PQ,

$\sqrt{ {x}^{2} + {y}^{2} + 6x - 4y + 13 } = 8}$

cancel the roots,

${x}^{2} + {y}^{2} + 6x - 4y + 13 = 8$

now,

(1)^2+ y^2+6×1-4y+13=8

or, 1+y^2+6-4y+13=8

y^2-4y+13+6+1=8

or, y(y-4)+20=8

or, y(y-4)= -12

either, or,

y= -12 y=8

Therefore, y= (8,-12)

by rounding off both values, we get,

x= 1

y=(8,-12)

So, i think it's (1,8) is your answer..

Hope it helps...

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3 years ago

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Answer:

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State the sufficient conditions for existence of Laplace

Yuki888 [10]

What is the question

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3 years ago

1. for what constant k must f(k) always equal the constant term of f(x) for any polynomial f(x) 2. If we multiply a polynomial b

DIA [1.3K]

Answer:

1. k=0

2. yes, result is still a polynomial.

3. yes, f and g must have the same degree to have deg(f+g) < deg(f) or deg(g)

Step-by-step explanation:

1. for what constant k must f(k) always equal the constant term of f(x) for any polynomial f(x)

for k=0 any polynomial f(x) will reduce f(k) to the constant term.

2. If we multiply a polynomial by a constant, is the result a polynomial?

Yes, If we multiply a polynomial by a constant, the result is always a polynomial.

3. if deg(f+g) is less than both deg f and deg g, then must f and g have the same degree?

Yes.

deg(f+g) < deg(f) and

deg(f+g) < deg(g)

then it means that the two leading terms cancel out, which can happen only if f and g have the same degree.

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