If y represents a child's age, which inequality shows that you must be older than 4 to attend an elementary school?

What is the distance between the points (5, 1) and (-3,-5)?

Reil [10]

Answer: THIRD OPTION.

Step-by-step explanation:

For this exercise you need to use the formula for calculate the distance between two points. This is:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Given the following points:

$(5, 1)$ and $(-3,-5)$

You can identify that:

$x_2=-3\\x_1=5\\\\y_2=-5\\y_1=1$

Knowing these values, the next step is to substitute them into the equation for calculate the distance between two points and then evaluate.

Therefore, the distance between $(5, 1)$ and $(-3,-5)$ is:

$d=\sqrt{(-3-5)^2+(-5-1)^2}\\\\d=10$

8 0

3 years ago

Solve this triangle, by finding each unknown side or angle. Include a triangle summary. Need help please! this is trigonometry b

navik [9.2K]

Use your SOH CAH TOA functions to find the missing legs. Do Sin(53.1) and that is equal to opposite (PU) over hypotenuse (UG) so to solve for the PU leg just set up an equation that says sin(53.1)=PU/36 so then you can just multiply the 36 over and that gives you the leg. To find PG do adjacent over hypotenuse so Cosine, cos(53.1)=PG/36 and again just multiply over the 36 and that gives you PG

8 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

express rental car charges $40 rental fee, $15 for gas and $.25 mile driven. for the same car, reynolds rental car charges $45 f

alexgriva [62]

Here you go! Sorry for the bad handwriting :/