All categories

3 years ago

Using the given points and line, determine the slope of the line. (-3, 0) and (2, 7) slope = -7/5

Mathematics

2 answers:

Brilliant_brown [7]3 years ago

7 0

Answer:

Slope= 7/5

Step-by-step explanation:

The slope of a line is determined as a ratio of the change in y to the change in x.

Slope=Δy/Δx

Δy=y₂-y₁

Δx=x₂-x₁

Therefore we can use the the x and y coordinates from the points given, (-3,0) and (2,7) to calculate the slope.

m= (7-0)/(2-⁻3)

Slope= 7/5

When a negative number is subtracted from a number it the operation sign changes to addition.

BigorU [14]3 years ago

4 0

Answer: Last option

$slope=\frac{7}{5}$

Step-by-step explanation:

The equation to find the slope m of a line is:

$m=\frac{y_2-y_2}{x_2-x_1}$

Where $(x_1, y_1)$ and $(x_2, y_2)$ are two points through which the line passes

In this case the points are: (-3, 0) and (2, 7)

Therefore the slope is:

$m=\frac{7-0}{2-(-3)}$

$m=\frac{7}{2+3}$

$m=\frac{7}{5}$

The answer is the last option

You might be interested in

Hii please help asap ill give brainliest thanks

GalinKa [24]

Answer:

The answer is ayurveda

8 0

3 years ago

Luca has 15

pav-90 [236]

Answer:10 question 1/2

Step-by-step explanation:

6 0

3 years ago

If it is 8:50 a.m., what will the time be 4 hours and 20 minutes later?

Marina CMI [18]

Answer:

1:10 p.m.

Step-by-step explanation:

First, add 20 minutes to 8:50 a.m. That would be 9:10 a.m. Then, add the remaining 4 hours to the 9:10 a.m. to get your answer- 1:10 p.m.

3 0

1 year ago

Help please it’s a quiz

Sidana [21]

......................

3 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