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2 years ago

12

Find the value of +c15 when =c4.

1 answer:

Montano1993 [528]2 years ago

6 0

It equals 60 u do 15* 4

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What is the range of the function y=2sin x

statuscvo [17]

take the absolute value of the coefficient (2)

then the lower bound is the negative of that (-2)

so the range is [-2,2]

7 0

2 years ago

Read 2 more answers

Please help me with these, oh sweet jesus

Lelechka [254]

Answer:

77. $\cot^{6} x = \cot^{4} x \csc^{2}x - \cot^{4} x$ Proved

78. $\sec^{4}x \tan^{2} x = \sec^{2}x [\tan^{2}x + \tan^{4}x ]$ Proved

79. $\cos^{3} x\sin^{2} x = [\sin^{2}x - \sin^{4}x] \cos x$ Proved.

80. $\sin^{4}x - \cos^{4}x = 1 - 2\cos^{2}x + 2 \cos^{4} x$ Proved.

Step-by-step explanation:

77. Left hand side

= $\cot^{6} x$

= $\cot^{4} x \times \cot^{2} x$

= $\cot^{4}x [\csc^{2}x - 1]$

{Since we know, $\csc^{2} x - \cot^{2}x = 1$ }

= $\cot^{4} x \csc^{2}x - \cot^{4} x$

= Right hand side (Proved)

78. Left hand side

= $\sec^{4}x \tan^{2} x$

= $\sec^{2} x [1 + \tan^{2}x] \tan^{2} x$

{Since $\sec^{2}x - \tan^{2}x = 1$ }

= $\sec^{2}x [\tan^{2}x + \tan^{4}x ]$

= Right hand side (Proved)

79. Left hand side

= $\cos^{3} x\sin^{2} x$

= $\cos x[1 - \sin^{2} x] \sin^{2} x$

{Since $\sin^{2}x + \cos^{2} x = 1$ }

= $[\sin^{2}x - \sin^{4}x] \cos x$

= Right hand side

80. Left hand side

= $\sin^{4}x - \cos^{4}x$

= $[\sin^{2}x + \cos^{2}x]^{2} - 2\sin^{2} x \cos^{2}x$

{Since $\sin^{2}x + \cos^{2} x = 1$ }

= $1 - 2\cos^{2} x[1 - \cos^{2}x ]$

= $1 - 2\cos^{2}x + 2 \cos^{4} x$

= Right hand side. (Proved)

7 0

2 years ago

The question says 5k+6k+4k=90. What does k equal?

alina1380 [7]

$5k+6k+4k=90 \\ \\ 15k = 90 \ / \ simplify \\ \\ k = \frac{90}{15} \ / \ divide \ each \ side \ by \ 15 \\ \\ k = 6 \ / \ simplify \\ \\$

The final result is, k = 6.

5 0

2 years ago

Expand each expression

matrenka [14]

Answer:

Option B - $\ln(\frac{4y^5}{x^2})=\ln 4+5\ln y-2\ln x$

Step-by-step explanation:

Given : Expression $\ln(\frac{4y^5}{x^2})$

To find : Expand each expression ?

Solution :

Using logarithmic properties,

$\ln (\frac{A}{B})=\frac{\ln A}{\ln B}=\ln A-\ln B$

and $\ln (AB)=\ln A+\ln B$

Here, A=4y^5 and B=x^2

$\ln(\frac{4y^5}{x^2})=\frac{\ln 4y^5}{\ln x^2}$

$\ln(\frac{4y^5}{x^2})=\ln 4y^5-\ln x^2$

$\ln(\frac{4y^5}{x^2})=\ln 4+\ln y^5-\ln x^2$

Using logarithmic property, $\logx^a=a\log x$

$\ln(\frac{4y^5}{x^2})=\ln 4+5\ln y-2\ln x$

Therefore, option B is correct.

3 0

2 years ago

Read 2 more answers

Apply the distributive property t create an equivalent expression

earnstyle [38]

Answer:The answer is 24m-28

Step-by-step explanation:

Hope this helps

7 0

2 years ago