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2 years ago

There are a total of 12 bicycles and tricycles in a parking lot. If there are 31 wheels in all, how many of them are bicycles?

Mathematics

1 answer:

leva [86]2 years ago

6 0

Answer:

5 bicycles

Step-by-step explanation:

5 * 2 = 10 wheels

7 * 3 = 21 wheels

10 + 21 = 31

Hope this helped! Brainliest :)

You might be interested in

What are the odd numbers

Ivan

Answer:

1,3,5,7,9....up to soon are odd numbers

Step-by-step explanation:

they are not divided by 2

4 0

3 years ago

Read 2 more answers

Can someone plz help me i will thank u so much

klemol [59]

Answer:

2 hours and 29 minutes

Step-by-step explanation:

if you change 9 hours to minutes you get 540 plus 17 equals 557. if you turn 6 hours to minutes it's 360 plus 48 equals 408. 557 minus 408 equals 149 minutes. Now if you convert it to hours and minutes you need to divide by 60. So if you divide it you get 2 hours and 29 minutes. hope this helped! :)

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

In a class of 40 students, everyone has either a pierced nose or a pierced ear. The professor asks everyone with a pierced nose

ladessa [460]

Answer:3

Step-by-step explanation:

Given

There are total 40 students in a class

No of students with Pierced Nose n(N)=7

No of students with Pierced ear n(E)=36

and we know using sets

$n\left ( N\cup E\right )=n\left ( E\right )+n\left ( N\right )-n\left ( N\cap E\right )$