How can you determine the sale price if you are giving the regular price and the percent of markdown?

15x + 50y = _____ (3x + 10y)

Ber [7]

Answer:

15x + 50y = 5 (3x + 10y)

5 is a common factor

15/3x = 3x

50/5y = 10y

the answer is 5

3 0

2 years ago

4 times the sum of 7 and 3 is subtracted from 41

Montano1993 [528]

Answer:

not so sure though but I think thats the answer

7 0

3 years ago

Use the substitution u = tan(x) to evaluate the following. int_0^(pi/6) (text(tan) ^2 x text( sec) ^4 x) text( ) dx

Rudiy27

If we use the substitution $u = \tan x$ , then $du = \sec^2 {x}\ dx$ . If you try substituting just $u$ and $du$ into the integrand, though, you'll notice that there's a $\sec^2x$ left over that we have to deal with.

To get rid of this problem, use the identity $\tan^2 x + 1 = \sec^2 x$ and substitute in the left side of the identity for the extra $\sec^2x$ , as shown:

$\int\limits^{\pi/6}_0 {tan^2 x \ sec^4 x} \, dx$
$\int\limits^{\pi/6}_0 {tan^2 x \ (tan^2 x + 1) \ sec^2 x} \, dx$

From there, we can substitute in $u$ and $du$ , and then evaluate:

$\int\limits^{\pi/6}_0 {tan^2 x \ (tan^2 x + 1) \ sec^2 x} \, dx$
$\int\limits^{\frac{1}{\sqrt{3}}}_0 {u^2(u^2 + 1)} \, du$
$\int\limits^{\frac{1}{\sqrt{3}}}_0 {u^4 + u^2} \, du$
$= \left.\frac{u^5}{5} + \frac{u^3}{3}\right|_0^\frac{1}{\sqrt{3}}$
$= (\frac{(\frac{1}{\sqrt{3}})^5}{5} + \frac{(\frac{1}{\sqrt{3}})^3}{3}) - (\frac{(0)^5}{5} + \frac{(0)^3}{3})$
$= \frac{1}{45\sqrt{3}} + \frac{1}{9\sqrt{3}} = \frac{6}{45\sqrt{3}} = \bf \frac{2}{15\sqrt{3}}$

8 0

3 years ago

What is the value of X

alexdok [17]

Answer:

$x=12$