All categories

3 years ago

write a point-slope equation for the line that passes through the point (6,8) and is parallel to the line given by y=-5x+4

Mathematics

1 answer:

VashaNatasha [74]3 years ago

3 0

Answer:

y -8 = -5(x -6)

Step-by-step explanation:

The point-slope form of the equation for a line is generally written ...

y -k = m(x -h)

for slope m and point (h, k).

The slope of your parallel line is the same as the slope of the reference line, -5. So your equation is ...

y -8 = -5(x -6)

You might be interested in

Find the value of x.

eduard

Answer:

∠43=1/2(62+x)

43×2=62+x

86-62=x

24=x

-------------------------

hope it helps...

have a great day!!

4 0

3 years ago

Solve the quadratic equation using factoring x^2 + 3x + 2 = 0.

Sedaia [141]

Answer:

Step-by-step explanation:

Given

x² + 3x + 2 = 0

To factor the quadratic

Consider the factors of the constant term (+ 2) which sum to give the coefficient of the x- term (+ 3)

The factors are 2 and 1, since

2 × 1 = 2 and 2 + 1 = 3, thus

(x + 2)(x + 1) = 0

Equate each factor to zero and solve for x

x + 2 = 0 ⇒ x = - 2

x + 1 = 0 ⇒ x = - 1

5 0

3 years ago

Nana has a water purifier that filters \dfrac13 3 1 start fraction, 1, divided by, 3, end fraction of the contaminants each ho

Fiesta28 [93]

Given:

Nana has a water purifier that filters $\dfrac{1}{3}$ of the contaminants each hour.

Water has contaminants = $\dfrac{1}{2}$

To find:

The function that gives the remaining amount of contaminants in kilograms, C(t), t hours after Nana started purifying the water.

Solution:

Let C(t) be the remaining amount of contaminants in kilograms after t hours.

Initial amount of contaminants = $\dfrac{1}{2}$

Decreasing rate is $\dfrac{1}{3}$ .

Using the exponential decay model:

$C(t)=C_0(1-r)^t$

where, $C_0$ is initial amount of contaminants, r is the decreasing rate and t is time in hours.

Substituting the values, we get

$C(t)=\dfrac{1}{2}(1-\dfrac{1}{3})^t$

$C(t)=\dfrac{1}{2}(\dfrac{2}{3})^t$

Therefore, the required function is $C(t)=\dfrac{1}{2}(\dfrac{2}{3})^t$ .

4 0

3 years ago

A delivery man is told to go 60 miles per hour north to make his delivery on time. These instructions describe the delivery man'

sattari [20]

Answer: B: Velocity

Step-by-step explanation: because velocity is the speed of something in a given direction.

3 0

3 years ago

Differentiate y = x5√x. SOLUTION Since both the base and the exponent are variable, we use logarithmic differentiation: ln y = l

AnnZ [28]

Answer:

$y'=x^{5\sqrt{x} } (\frac{5}{\sqrt{x} }+\frac{5}{2\sqrt{x} }lnx)$

Step-by-step explanation:

Given: $y=x^{5\sqrt{x} }$

Take ln of both sides

$lny=lnx^{5\sqrt{x} } = 5\sqrt{x}lnx\\\\lny=5\sqrt{x}lnx$

Differentiate with respect to x

$\frac{d}{dx}(lny)=\frac{d}{dx}(5\sqrt{x}lnx)\\\\\frac{d}{dy}(lny)*\frac{dy}{dx}=5\sqrt{x}\frac{d}{dx}lnx+\frac{d}{dx}5\sqrt{x}*lnx)\\\\\frac{1}{y}*\frac{dy}{dx}=(5\sqrt{x}*\frac{1}{x})+(5*\frac{1}{2}x^{-\frac{1}{2}}lnx)\\ \\\frac{1}{y}*y^{'}=\frac{5}{\sqrt{x} }+\frac{5}{2\sqrt{x} }lnx \\\\y'=y(\frac{5}{\sqrt{x} }+\frac{5}{2\sqrt{x} }lnx)\\\\y'=x^{5\sqrt{x} } (\frac{5}{\sqrt{x} }+\frac{5}{2\sqrt{x} }lnx)$

8 0

3 years ago