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3 years ago

A line goes through points (1, 8) and (5, 6). What is the slope of the line?

Mathematics

2 answers:

kherson [118]3 years ago

7 0

-0.5

Step-by-step explanation:

$\frac{6 - 8}{5 - 1?} = - 0.5$

slope=

$\frac{y2 - y1}{x2 - x1}$

IrinaVladis [17]3 years ago

7 0

Answer:

-2/4

Explanation:

You label the first coordinate points as X(1) and Y(1) and the second coordinate points as X(2) and Y(2) and then apply the formula

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Solve 6(x + 1) > 2x - 1 x > -1/2 x > -7/4 x > 0

Bond [772]

Answer:

x > - $\frac{7}{4}$

Step-by-step explanation:

Given

6(x + 1) > 2x - 1 ← distribute parenthesis on left side

6x + 6 > 2x - 1 ( subtract 2x from both sides )

4x + 6 > - 1 ( subtract 6 from both sides )

4x > - 7 ( divide both sides by 4 )

x > - $\frac{7}{4}$

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3 years ago

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MatroZZZ [7]

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3 years ago

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A line has slope and yintercept 5. which answer is the equation of the line

ioda

Answer: if the y-intercept is 5, it's c. -3/4x + 5. Since it's ax+b a = slope b = y intercept

Step-by-step explanation:

6 0

3 years ago

Consider F and C below. F(x, y, z) = 2xz + y2 i + 2xy j + x2 + 9z2 k C: x = t2, y = t + 3, z = 3t − 1, 0 ≤ t ≤ 1 (a) Find a func

Ray Of Light [21]

Answer:

a)∇f = 2y + 2x + 18z

b) $\int\limits^._C {F} \, dr$ =108

Step-by-step explanation:

Given:

f (x,y,z ) = $(2xz+ y^{2})i + (2xy) j +(x^{2} + 9z^{2})k$

The curve C :

$x=t^{2} ,\\y= t+3\\z= 3t-1$

where 0 ≤ t ≤ 1

Required:

(a) F = ∇f =? (F is a vector here)

(b) $\int\limits^._C {F} \, dr$ =?

Solution

First we will find the directional derivative F = ∇f

for that , we will use the formula :

∇f = $F_{x}i+ F_{y} j+F_{z}k$

Fx= δf/δx = δ/δx $(2xz+ y^{2})i$ = 2z i

Fy= δf/δy=δ/δy (2xy)j = 2x j

Fz= δf/δz=δ/δz $(x^{2} + 9z^{2})k$ = 18z k

∇f = (2z) i .i + (2x) j.j + (18z) k.k

∇f = 2z + 2x + 18z

For part b):

we will use line integral formula:

$\int\limits^._C {F} \, dr$

to calculate dr, we will need the curve C:

r = x(t)+y(t)+z(t)

r= $(t^{2})i + (t+3) j +(3t-1) k$

$\frac{dr}{dt}=\frac{dx}{dt} +\frac{dy}{dt} + \frac{dz}{dt}$

$\frac{dx}{dt} = 2t$

$\frac{dy}{dt} = 1$

$\frac{dz}{dt} = 3$

$\int\limits^._C {F} \, dr$ = $\int\limits^1_0 {F_{x} } \, dx+ F_{y} dy +F_{z} dz$

= $\int\limits^1_0 {(2z(2t) + 2x(1) + 18z (3)} \, )$

put values of y, x and z

= $\int\limits^1_0 {2(3t-1) + 2(t^{2}) +18 (3) (3t-1)} \,$

= ${2t^{2} + 6t+162t -54-2}\, |^1_0$

= ${ 2t^{2}+ 168t - 56} \,|^1_0$ (Note : f(1)-f(0))

=2(1)+162(1)+2(0)+162(0)-56

= 2+162 -56

$\int\limits^._C {F} \, dr$ =108

3 0

3 years ago

43, 45, 2, 5, 13, 11 Median - Q1 - Q3 - IQR -

Gnesinka [82]

Answer:

Median- 12

Q1- 5

Q3- 43

IQR- 38

Step-by-step explanation:

2, 5, 11, 13, 43, 45

Put your numbers in order of least to greatest

Since there is no given median as the amount of numbers is even, find the number in between the 2 middle numbers, in this case, the number would be 12, as it is in between 11 and 13.

To find Q1 and Q3, you need to find the first and third median, which would be 5 and 43. To find the IQR, you need to subtract Q3 from Q1, which would give you 38 in this equation

5 0

3 years ago