Pls help me do this

A survey was taken of students in math classes to find out how many hours per day students spend on social media. The survey res

Fiesta28 [93]

median because there is and outlier

7 0

3 years ago

Find pb, if pj=14 and jb= 28

anyanavicka [17]

Answer:

42

Step-by-step explanation:

Based on the information given, point j is the midpoint, as it is shared by both line segments. Set the equation:

pb = pj + jb

Note:

pj = 14

jb = 28

Plug in the corresponding numbers to the corresponding variables:

pb = pj + jb

pb = 14 + 28

pb = 42

pb = 42 is your answer.

~

3 0

3 years ago

Read 2 more answers

A 1L IV contains 60meq of kcal. The IV was discontinued after 400ml has infused. How much lvl did the patient receive

Leya [2.2K]

If a 1L IV contains 60meq of kcal and the IV was discontinued after 400ml has infused, then the amount of IV received by the patient is 24meq of kcal.

Calculation for the Amount of IV

It is given that,

1L IV contains 60meq of kcal

⇒ 1000 mL of IV contains 60meq of kcal

⇒ 1 mL of IV will contain 60 / 1000 meq of kcal

As per the question,

The amount of IV infused in the patient = 400 mL

⇒ The amount of IV received by the patient in meq of kcal = 400 × (60 / 1000)

= 4 × 6

= 24 meq of kcal

Hence, the patient receives 24meq of kcal amount of IV.

Learn more about amount here:

brainly.com/question/4567186

#SPJ1

6 0

1 year ago

If x+1/x = 5 find the value of x^3 +1/x^3. pls solve

Korolek [52]

Answer:

1) solve x+1/x = 5

$\frac{x + 1}{x} = 5$

$x + 1 = 5x$

$x - 5x = - 1$

$- 4x = - 1$

$x = \frac{ -1}{ - 4}$

$x = \frac{1}{4}$

2) solve x³+1/x³

$x {}^{3} + \frac{1}{x {}^{3} }$

substitute x = 1/4 into the expression

$( \frac{1}{4} ) {}^{3} + \frac{1}{( \frac{1}{4}) {}^{3} }$

$\frac{1}{64} + \frac{1}{ \frac{1}{64} }$

$\frac{1}{64} + \frac{64 \times 1}{1}$

$\frac{1}{64} + 64$

$\frac{4097}{64} \: or \: 64.02$

5 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