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Alchen [17]
4 years ago
12

Last year, liliana earned

Mathematics
1 answer:
olga_2 [115]4 years ago
3 0

Answer:

$376,528

Explanation: You have to ÷ $4,518,336 by 12 because there is 12 months in 1 year. That equals 376,528. Therefore her average income was $376,528.

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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}}


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3 years ago
A student solved the following problem and made an error. In which line did the student make the first mistake. Line A, Line 2,
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Answer:

line 5. i just took the test and got it correct. oh wait nvm urs is different

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Answer:

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Step-by-step explanation:

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Find the slope and y-intercept of the line in each graph. (Explore Activity 1) Please help me
Anit [1.1K]

Answer:

1. Slope = -2 , y-intercept = 1

2. Slope = 5 , y-intercept = -15

3. Slope = 1.5 , y-intercept = -2

4. Slope = -3 , y-intercept = 9

Step-by-step explanation:

1. The slope is -2 because using the formula y₁-y₂÷x₁-x₂ you can use the points given to you to get -2. It doesn't matter which one you pick as y1 or y2, just make sure you keep it consistent with the x numberings. I did 1--3 (which becomes 1+3) and 0-2. Those equal 4/-2 which gives you -2. Use these same steps for all of them, but with ones that don't have the points given, just pick two that are whole numbers and not decimals. For finding the y-intercept, just find the point that is on the y axis line. It is given on number one as (0,1). So, the y-intercept is 1. Hope this helps!

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