All categories

3 years ago

What is the y-intercept of the function, represented by the table of values below? x: -2,1,2,4,7 y: 14,8,6,2,-4

Mathematics

1 answer:

anastassius [24]3 years ago

7 0

Answer:

The y-intercept is at y = 10.

Step-by-step explanation:

It will be between y = 14 and y = 7 because the corresponding x values are -2 and 1.

An increase of 3 units of x gives a decrease of 6 units of y fro the above values.

Then an increase of 1 ( 1 to 2) in x gives decrease in y of 2 (8 to 6). The other values show the same pattern.

So very unit increase in x, the y values change by -2.

So from x = -2 to 0 is +2 units for x and this will be -4 units for y so the y-intercept ( when x = 0) will be at y = 14-4 = 10

y-intercept is (0,10).

You might be interested in

FUNNY MEMES THE FUNNIEST GETS BRAINLIEST!!!!!!!!!!!!!!!!

diamong [38]

Answer:

Does February March?.... NO, but APRIL MAY

Step-by-step explanation:

Does February March?.... NO, but APRIL MAY

\Does February March?.... NO, but APRIL MAY

Does February March?.... NO, but APRIL MAY

3 0

2 years ago

Read 2 more answers

Find the solution of 4x2y′′−4x2y′+y=0,x>04x2y″−4x2y′+y=0,x>0 of the form y1=xr(1+c1x+c2x2+c3x3+⋯)

BabaBlast [244]

Answer:

r = 1/2

c1 = 3/4

c2 = 27/64

c3 = 405/1408

Step-by-step explanation:

Find the solution of 4x2y′′−4x2y′+y=0,x>04x2y″−4x2y′+y=0,x>0 of the form y1=xr(1+c1x+c2x2+c3x3+⋯)

r = 1/2

c1 = 3/4

c2 = 27/64

c3 = 405/1408

The solution is attached.

5 0

3 years ago

Find the prime factorization of 2,520

Anit [1.1K]

<span>the prime factorization of 2,520

2 * 2 * 2 * 3 * 3 *5 * 7</span>

5 0

3 years ago

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.

Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Step-by-step explanation:

GIven that:

$f(x) = 5e^{-x^2} cos (4x)$

The Maclaurin series of cos x can be expressed as :

$\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+... \ \ \ (1)}$

$\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0} \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!} -\dfrac{x^6}{3!}+... \ \ \ (2)}$

From equation(1), substituting x with (4x), Then:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}- \dfrac{(4x)^6}{6!}+...}$

The first three terms of cos (4x) is:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}-...}$

$\mathtt{cos (4x) = 1 - \dfrac{16x^2}{2}+ \dfrac{256x^4}{24}-...}$

$\mathtt{cos (4x) = 1 - 8x^2+ \dfrac{32x^4}{3}-... \ \ \ (3)}$

Multiplying equation (2) with (3); we have :

$\mathtt{ e^{-x^2} cos (4x) = ( 1- x^2 + \dfrac{x^4}{2!} ) \times ( 1 - 8x^2 + \dfrac{32 \ x^4}{3} ) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1+ (-8-1)x^2 + (\dfrac{32}{3} + \dfrac{1}{2}+8)x^4 + ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + (\dfrac{64+3+48}{6})x^4+ ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Finally , multiplying 5 with $\mathtt{ e^{-x^2} cos (4x) }$ ; we have:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

7 0

3 years ago

Which segment represents the radius of the circle?

BARSIC [14]

Answer:The answer would be A) blue

Step-by-step explanation: The blue segment represents the measurement of the part of a circle

5 0

3 years ago