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3 years ago

What is the equation of a line with a slope of -4 and a point (-2,5) on the line

Mathematics

1 answer:

Lesechka [4]3 years ago

4 0

Answer:

The equation of the line would be y = -4x - 3

Step-by-step explanation:

To find any equation when you have the slope and a point, start with point-slope form and then solve for y.

y - y1 = m(x - x1)

y - 5 = -4(x + 2)

y - 5 = -4x - 8

y = -4x - 3

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Find the circumference and area of a circle with radius 8 cm. Express your answer in terms of pi

BigorU [14]

Answer:

c=16*pi

a=64*pi

Step-by-step explanation:

7 0

3 years ago

The state of New Jersey makes approximately $1.55 billion annually from tolls of the $1.55 billion about 3/4 comes from tolls on

Rama09 [41]

Percent of turn pike is 75% and percent of parkway is 25%.

Solution:

Given, The state of New Jersey makes approximately $1.55 billion annually from tolls of the $1.55 billion

About $\frac{3}{4}$ comes from tolls on the turnpike

Remaining $\frac{1}{4}$ comes from tolls on the parkway

To find the percent of the total revenue from turnpike:

$\begin{array}{l}{\text { Percent of turnpike }=\frac{\text { amount from parkway }}{\text { total revenue }} \times 100} \\\\ {=\frac{\frac{3}{4} \text { of total revenue }}{\text { tot al revenue }} \times 100} \\\\ {=\frac{\frac{3}{4} \times \text { total revenue }}{\text { total revenue }} \times 100} \\\\ {=\frac{3}{4} \times 100=3 \times 25=75 \%}\end{array}$

To find the percent of the total revenue from parkway:

$\begin{array}{l}{\text { Percent of parkway }=\frac{\text { amount from parkway }}{\text { total revenue }} \times 100} \\\\ {=\frac{\frac{1}{4} \text { of total revenue }}{\text { total revenue }} \times 100} \\\\ {=\frac{\frac{1}{4} \times \text { total revenue}}{\text { total revenue }} \times 100} \\\\ {=\frac{1}{4} \times 100=25 \%}\end{array}$

Summarizing the results:

Percent of turn pike is 75%

Percent of parkway is 25%

3 0

3 years ago

The liquid base of an ice cream has an initial temperature of 86°C before it is placed in a freezer with a constant temperature

Karolina [17]

The temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

From Newton's law of cooling, we have that

$T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$

Where

$(t) = \ time$

$T_{(t)} = \ the \ temperature \ of \ the \ body \ at \ time \ (t)$

$T_{s} = Surrounding \ temperature$

$T_{0} = Initial \ temperature \ of \ the \ body$

$k = constant$

From the question,

$T_{0} = 86 ^{o}C$

$T_{s} = -20 ^{o}C$

∴ $T_{0} - T_{s} = 86^{o}C - -20^{o}C = 86^{o}C +20^{o}C$

$T_{0} - T_{s} = 106^{o} C$

Therefore, the equation $T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$ becomes

$T_{(t)}=-20+106 e^{kt}$

Also, from the question

After 1 hour, the temperature of the ice-cream base has decreased to 58°C.

That is,

At time $t = 1 \ hour$ , $T_{(t)} = 58^{o}C$

Then, we can write that

$T_{(1)}=58 = -20+106 e^{k(1)}$

Then, we get

$58 = -20+106 e^{k(1)}$

Now, solve for $k$

First collect like terms

$58 +20 = 106 e^{k}$

$78 =106 e^{k}$

Then,

$e^{k} = \frac{78}{106}$

$e^{k} = 0.735849$

Now, take the natural log of both sides

$ln(e^{k}) =ln( 0.735849)$

$k = -0.30673$

This is the value of the constant $k$

Now, for the temperature of the ice cream 2 hours after it was placed in the freezer, that is, at $t = 2 \ hours$

From

$T_{(t)}=-20+106 e^{kt}$

Then

$T_{(2)}=-20+106 e^{(-0.30673 \times 2)}$

$T_{(2)}=-20+106 e^{-0.61346}$

$T_{(2)}=-20+106\times 0.5414741237$

$T_{(2)}=-20+57.396257$

$T_{(2)}=37.396257 \ ^{o}C$

$T_{(2)} \approxeq 37.40 \ ^{o}C$

Hence, the temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

Learn more here: brainly.com/question/11689670

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2 years ago

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How do i solve v=pi/3*r (squared)h for h

Colt1911 [192]

$V=\dfrac{\pi}{3}\cdot r^2h\ \ \ \ \ |multiply\ both\ sides\ by\ 3\\\\\pi r^2h=3V\ \ \ \ |\divide\ both\ sides\ by\ \pi r^2\\\\\boxed{h=\dfrac{3V}{\pi r^2}}$

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3 years ago

Someone pls help me with this

Burka [1]

the picture is blurry

6 0

3 years ago