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3 years ago

Can someone please answer. There is one question. There's a picture thank you!

Mathematics

1 answer:

soldier1979 [14.2K]3 years ago

8 0

<span>The range of Cos(x) is [-1,1]. Therefore the range of ln(Cos(x)) will be the image of [-1,1] using the natural log function. However, the domain of ln(x) is (0,infinity) and the log function is strictly increasing with vertical asymptote at "x=0". Therefore, the range of ln(Cos(x)) will be:

(- infinity, ln(1) ] = (-infinity, 0] !</span><span>so its true </span>

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WHICH OF THE FOLLOWING IS THE SLOPE AND Y INTERCEPT?

vivado [14]

Answer:

the correct answer is Slope = 45 and y intercept = 175

Step-by-step explanation:

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3 years ago

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Phoenix [80]

Answer:

1/49

Step-by-step explanation:

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3 years ago

Read 2 more answers

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vovangra [49]

Answer:

1. $27^{\frac{2}{3} } =9$

2. $\sqrt{36^{3} } =216$

3. $(-243)^{\frac{3}{5} } =-27$

4. $40^{\frac{2}{3}}=4\sqrt[3]{25} =4325$

5. Step 4: $(\frac{343}{27}) ^{-1} =\frac{27}{343}$

6. $D. -72cd^{7}$

Step-by-step explanation:

Use the following properties:

$a^{\frac{x}{y} } =\sqrt[x]{a^{y} }$

$\sqrt[n]{ab} =\sqrt[n]{a} \sqrt[n]{b}$

$a^{-n} =\frac{1}{a^{n} }$

$(xy)^{z} =x^{z} y^{z} \\\\$

$(x^{y}) ^{z} =x^{yz}$

$x^{y} x^{z} =x^{y+z}$

So:

1. $27^{\frac{2}{3} } =\sqrt[3]{27^{2}} =\sqrt[3]{729} }=9$

2. $\sqrt{36^{3} } =\sqrt{36*36*36} =\sqrt{36} \sqrt{36} \sqrt{36} =6*6*6=216$

3. $(-243)^{\frac{3}{5} } =\sqrt[5]{-243^{3} } =\sqrt[5]{-14348907} =-27$

4. $40^{\frac{2}{3}}=\sqrt[3]{40^{2} } =\sqrt[3]{2^{6} 5^{2} } =\sqrt[3]{2^{6} } \sqrt[3]{5^{2} } =2^{\frac{6}{3} } 5^{\frac{2}{3} } =4 *5^{\frac{2}{3} } =4\sqrt[3]{5^{2} } =4\sqrt[3]{25}=4325$