Write an equation of the line passing through the point (-6,-7)with slope=4.

2 answers:

6 0

The slope is 4, therefore
m = 4
The line passes through the point
(x₁,y₁) = (-6,-7)

Use the following formula to define the equation
y - y₁ = m(x - x₁)

Plug ini the numbers
y - (-7) = 4(x - (-6))
y + 7 = 4(x + 6)
y + 7 = 4x + 24
y = 4x + 17
This is the equation

MAVERICK [17]3 years ago

5 0

And how would I do that via computer? If this is homework, do it your self, it's not that hard. Draw it and scan it (man computer doesn't have the ability to scan things)

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The football team has 2 large water containers. Each container holds 345π in3 of water. Each player is given a cylindrical water

raketka [301]

Answer:

They'll be able to get 34 bottles from the containers.

Step-by-step explanation:

Since the bottles are cylindrical we can calculate their volume by using the following formula:

V = base_area*h

V = \pi*(r^2)*h

r = d/2 = 4/2 = 2 inches

V = 3.14*(2^2)*5 = 3.14*4*5

V = 3.14*20 = 62.8 inches^3

In order to know how many full bottles the players will get we need to divide the total volume of the containers, which is given by the sum of the volume of each container, and divide it by the volume of each bottle. We have:

bottles = (345*pi + 345*pi)/62.8 = 690*pi/62.8 = 2,166.6/62.8 = 34.5

Since the problem wants the amount of full bottles we only take the integer part, so they will be able to get 34 bottles from the containers.

4 0

3 years ago

Find the integral using substitution or a formula.

Nadusha1986 [10]

$\rm \int \dfrac{x^2+7}{x^2+2x+5}~dx$

Derivative of the denominator:
$\rm (x^2+2x+5)'=2x+2$

Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.

$\rm \int \dfrac{x^2+2+5}{x^2+2x+5}~dx$

and then split the fraction,

$\rm \int \dfrac{x^2+2}{x^2+2x+5}~dx+\int\dfrac{5}{x^2+2x+5}~dx$

Based on our previous test, we know that a simple substitution will work for the first integral: $\rm \quad u=x^2+2x+5\qquad\to\qquad du=2x+2~dx$

So the first integral changes,

$\rm \int \dfrac{1}{u}~du+\int\dfrac{5}{x^2+2x+5}~dx$

integrating to a log,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{x^2+2x+5}~dx$

Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.

$\rm x^2+2x+5=(x^2+2x+1)+4=(x+1)^2+2^2$

So we have this going on,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{(x+1)^2+2^2}~dx$

Let's factor the 5 out of the intergral,
and the 4 from the denominator,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\frac{(x+1)^2}{2^2}+1}~dx$

Bringing all that stuff together as a single square,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(\dfrac{x+1}{2}\right)^2+1}~dx$

Making the substitution: $\rm \quad u=\dfrac{x+1}{2}\qquad\to\qquad 2du=dx$

giving us,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(u\right)^2+1}~2du$

simplying a lil bit,

$\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du$

and hopefully from this point you recognize your arctangent integral,

$\rm ln|x^2+2x+5|+\frac52arctan(u)$

undo your substitution as a final step,
and include a constant of integration,

$\rm ln|x^2+2x+5|+\frac52arctan\left(\frac{x+1}{2}\right)+c$

Hope that helps!
Lemme know if any steps were too confusing.

8 0

3 years ago

Help help help math math

Arada [10]

Answer:

$x > - 14$