Ask question

Ask question

All categories

3 years ago

11

Find the slope of the line that passes through the points A(0,2) and B(4,-1)

2 answers:

Amiraneli [1.4K]3 years ago

8 0

The slope would be -3/4 and the y-intercept would be 2 so your equation is y=-3/4x+2

sergiy2304 [10]3 years ago

4 0

The slope is -3/4 for your question

You might be interested in

There a 12 pumpkins at the grocery store and you are going to select 3 to take home. How

Virty [35]

Hello do 9+46=684 there you go to find sum 69

3 0

3 years ago

Just Seabra 5.2 pounds of grapes for $7.75 using the unit rate how much would 3 pounds of grapes cost

lora16 [44]

Answer: 4.47

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

What is the sum of the polynomials

NeX [460]

Answer:

its the first answer choice!

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Given the equation...

Semenov [28]

Should be all real numbers

8 0

3 years ago

the height h(t) of a trianle is increasing at 2.5 cm/min, while it's area A(t) is also increasing at 4.7 cm2/min. at what rate i

nekit [7.7K]

Answer:

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

Step-by-step explanation:

From Geometry we understand that area of triangle is determined by the following expression:

$A = \frac{1}{2}\cdot b\cdot h$ (Eq. 1)

Where:

$A$ - Area of the triangle, measured in square centimeters.

$b$ - Base of the triangle, measured in centimeters.

$h$ - Height of the triangle, measured in centimeters.

By Differential Calculus we deduce an expression for the rate of change of the area in time:

$\frac{dA}{dt} = \frac{1}{2}\cdot \frac{db}{dt}\cdot h + \frac{1}{2}\cdot b \cdot \frac{dh}{dt}$ (Eq. 2)

Where:

$\frac{dA}{dt}$ - Rate of change of area in time, measured in square centimeters per minute.

$\frac{db}{dt}$ - Rate of change of base in time, measured in centimeters per minute.

$\frac{dh}{dt}$ - Rate of change of height in time, measured in centimeters per minute.

Now we clear the rate of change of base in time within (Eq, 2):

$\frac{1}{2}\cdot\frac{db}{dt}\cdot h = \frac{dA}{dt}-\frac{1}{2}\cdot b\cdot \frac{dh}{dt}$

$\frac{db}{dt} = \frac{2}{h}\cdot \frac{dA}{dt} -\frac{b}{h}\cdot \frac{dh}{dt}$ (Eq. 3)

The base of the triangle can be found clearing respective variable within (Eq. 1):

$b = \frac{2\cdot A}{h}$

If we know that $A = 130\,cm^{2}$ , $h = 15\,cm$ , $\frac{dh}{dt} = 2.5\,\frac{cm}{min}$ and $\frac{dA}{dt} = 4.7\,\frac{cm^{2}}{min}$ , the rate of change of the base of the triangle in time is:

$b = \frac{2\cdot (130\,cm^{2})}{15\,cm}$

$b = 17.333\,cm$

$\frac{db}{dt} = \left(\frac{2}{15\,cm}\right)\cdot \left(4.7\,\frac{cm^{2}}{min} \right) -\left(\frac{17.333\,cm}{15\,cm} \right)\cdot \left(2.5\,\frac{cm}{min} \right)$

$\frac{db}{dt} = -2.262\,\frac{cm}{min}$

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

6 0

3 years ago