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White raven [17]

1 year ago

7

Help please thanks yoi

2 answers:

Naddika [18.5K]1 year ago

6 0

Answer The answer is 60
Explaintion

dusya [7]1 year ago

5 0

Answer:

The answer is 60

Step-by-step explanation:

This is because they are corresponding angles.

Hope this helps!!

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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

3 years ago

Serhud [2]

Answer: c

Step-by-step explanation:

5 0

2 years ago

50+x(greater than or equal to)300

nikdorinn [45]

Answer:

x ≥ 250

Step-by-step explanation:

Note that "greater than or equal to" looks like ≥

50 + x ≥ 300

Isolate the variable. What you do to one side, you do to the other. Subtract 50 from both sides.

x + 50 (-50) ≥ 300 (-50)

x ≥ 300 - 50

x ≥ 250

x ≥ 250 is your answer.

~

4 0

3 years ago

What is the value of g(0)<br> a. 1<br> b. 0<br> c. 8<br> d. 4

uysha [10]

Answer:

c. 8

Step-by-step explanation:

1) $g(x)=8(\frac{1}{2} )^{x}$

2) $g(0)=8(\frac{1}{2} )^{0}$

3) $g(0)=8*1$

4) $g(0)=8$

4 0

2 years ago

Show that the equation x^3+6x-5=0 has a solution between x=0 and x=1

Mnenie [13.5K]

Answer:

Since the value of f(0) is negative and the value of f(1) is positive, then there is at least one value of x between 0 and 1 for which f(x) =0.

Step-by-step explanation:

The equation f(x) given is:

$f(x) = x^3+6x-5$

For x = 0. the value of the expression is:

$f(0) = 0^3+0-5\\f(0) = -5$

For x = 1, the value of the expression is:

$f(1) = 1^3+6-5\\f(1)=2$

Since the value of f(0) is negative and the value of f(1) is positive, then there is at least one value of x between 0 and 1 for which f(x) =0.

In other words, there is at least one solution for the equation between x=0 and x=1.

6 0

2 years ago