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2 years ago

Help me please thank you very much

Mathematics

1 answer:

Butoxors [25]2 years ago

6 0

For the first question the answer is-2,3
For the second question the answer should be 168.83 meters.

Hope I didn't mess up for your sake

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The value of Q is given by the relation Q = 17m. If Q and m are integers with Q greater than 150 and less than 160, then which o

zloy xaker [14]

Answer: Value of m = 9

Step-by-step explanation:

Given that the relationship between Q and m is;

Q = 17m

Make m the subject of formula

M = Q/17

If Q is greater than 150 and less than 160, then, let assume that

Q = 151, then

M = 151/17

M = 8.88

If Q = 159

M = 159/17

M= 9.35

Since m ranges from 8.88 to 9.35, the value of m = 9

6 0

2 years ago

Write an equation in slope intercept form for the line with slope of 1/5 and y-intercept -8

AlladinOne [14]

Answer:

y = (1/5)x - 8

Step-by-step explanation:

It's slope-intercept form y=mx+b and we just substitute m=1/5, b= -8.

8 0

2 years ago

Read 2 more answers

Luis does chores at home to earn extra money. Each week he earns $20. He already has $320 in his savings account, and is trying

Setler [38]

Answer:

30.5 Weeks

Step-by-step explanation:

The phone costs $930, and he already has $320 saved in his bank account so subtract 320 from 930. You will get 610. Then, you take 610 and divide it by 20 because he earns $20 a week. I hope this helped.

8 0

2 years ago

Please help me with these, oh sweet jesus

Lelechka [254]

Answer:

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

Dose anyone know this answer?

LiRa [457]

So I can show you on a piece of paper

It’s reflections
Hope this helps *smiles*

3 0

2 years ago