Please help, will mark brainliest

How does solving the equation arithmetically compare to solving an equation algebraically

IRISSAK [1]

In an arithmetic equation, there is no variable in the 'meat' of the equation(example: 5-5=0). In an algebraic equation, there is a variable in the meat of the equation(example: 5-x=0).
Hope this helps and please give brainliest!

4 0

3 years ago

If the Volume of a Cylinder is 287 and the radius is 5, what is the height "h"?

Sidana [21]

Answer:

3.7 is the height

Step-by-step explanation:

so the volume is 287 radius is 5 . The formula Volume = base area x height which we come to equate 287 =3.142x5x5 to get 287 divides by 78.55 answers round off to the nearest tenths 3.653 round off 3.7

8 0

3 years ago

Hiiiiiii im soooooooo tired but i gotta clean my room lol

kumpel [21]

Answer:

lol

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

4x²-y²=171 and 2x-y=9, value of 2x+y

11111nata11111 [884]

So you will need to solve for x and y before evaluating 2x+y....

2x-y=9, y=2x-9 now this will make 4x^2-y^2=171 become:

4x^2-(2x-9)^2=171

4x^2-(4x^2-36x+81)=171

36x-81=171

36x=252

x=7, now we can use 2x-y=9 to solve for y...

2(7)-y=9

14-y=9

-y=-5

y=5

now we know that x=7 and y=5, 2x+y becomes:

2(7)+5

14+5

19

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