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3 years ago

Answer the question in the picture pls WORTH 50 POINTS PLUS BRAINLIEST FOR FIRST CORRECT ANSWER

Mathematics

2 answers:

Pachacha [2.7K]3 years ago

6 0

Answer:

Step-by-step explanation:

| 4-y| ≥8

Separate into two inequalities, one positive and one negative (remembering to flip the inequality)

4-y ≥8 or 4-y≤ -8

Solve

Subtract 4 from all sides

4-y -4 ≥8-4 or 4-y-4≤ -8-4

-y ≥4 or -y≤ -12

Divide each side by -1, remembering to flip the inequality

y ≤-4 or y ≥12

The smallest positive number is 12

Alexus [3.1K]3 years ago

5 0

Answer:

Step-by-step explanation:

Step 1: Simplify both sides of the inequality.

−y+4≥8

Step 2: Subtract 4 from both sides.

−y+4−4≥8−4

−y≥4

Step 3: Divide both sides by -1.

−y

−1

≥

−1

y≤−4

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Answer:

Step-by-step explanation:

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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:

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

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3 years ago