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icang [17]
3 years ago
11

Plz help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
snow_lady [41]3 years ago
8 0

Answer:

option 9 answer

Step-by-step explanation:

keep j Merrick

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Triangle (-2,-2) (-6,-8) (-8,-8) over the x-axis. What are the new vertices?
ivanzaharov [21]
(-2,2) (-6,8) (-8,80
5 0
3 years ago
Please help me with these, oh sweet jesus
Lelechka [254]

Answer:

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

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
3 years ago
Steven has some money if he spends nine dollars then he will have 3/5 of the amount he started with
Vesna [10]
Total money * fraction = part
x* \frac{3}{5} =9 \\ ( \frac{5}{3}) * \frac{3x}{5} = 9* \frac{5}{3} \\ x= \frac{45}{3} =15
5 0
3 years ago
Read 2 more answers
The graph below shows the line of best fit for data collected on the number of cell phones and cell phone cases sold at a local
Dovator [93]

Answer:

C

Step-by-step explanation:

The equation of a line is given by y-y_1=m(x-x_1)

and

slope (m) is given by:

m=\frac{y_2-y_1}{x_2-x_1}

Where (x_1,y_1) are the first set of point in the line and (x_2,y_2) is the second set of point

<em>Let's take 2 points arbitrarily. (0,0) & (25,20)</em>

<em />

<em>Let's plug it and find the equation:</em>

<em>m=\frac{20-0}{25-0}=0.8</em>

<em>Now</em>

<em>y-y_1=m(x-x_1)\\y-0=0.8(x-0)\\y=0.8x</em>

<em />

<em>C is the correct answer.</em>

5 0
3 years ago
Read 2 more answers
HELP 100 POINTS PLEASE
Ymorist [56]

w o w 100 points! t h a n k s :D

3 0
3 years ago
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