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1 year ago

Is m = 1 a solution to the inequality below? 55 m > 26 yes no

Mathematics

1 answer:

Bas_tet [7]1 year ago

3 0

Answer: no

Step-by-step explanation:

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Subtract (r - 3q + 5p) - (-4r - 3q - 8p)

Alex787 [66]

Answer: ok so Let's simplify step-by-step.

r−3q+5p−(−4r−3q−8p)

Distribute the Negative Sign:

=r−3q+5p+−1(−4r−3q−8p)

=r+−3q+5p+−1(−4r)+−1(−3q)+−1(−8p)

=r+−3q+5p+4r+3q+8p

Combine Like Terms:

=r+−3q+5p+4r+3q+8p

=(5p+8p)+(−3q+3q)+(r+4r)

=13p+5r

Step-by-step explanation:

4 0

2 years ago

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If tan theta =3, find the value of tan theta + tan (theta+pi) + tan (theta +2pi)

sweet-ann [11.9K]

Answer:

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi) = 9$

Step-by-step explanation:

Given

$tan\theta = 3$

Required

Find

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi)$

Calculate $tan(\theta + \pi) \ and\ tan(\theta+2\pi)$

Using tan rule

$tan(\theta + \pi) = \frac{tan\theta + tan\pi}{1 - tan\theta tan\pi}$

So:

$tan(\theta + \pi) = \frac{tan\theta + 0}{1 - tan\theta *0}$

$tan(\theta + \pi) = \frac{tan\theta}{1 }$

$tan(\theta + \pi) = \tan\theta$

$tan(\theta+2\pi)$

$tan(\theta + 2\pi) = \frac{tan\theta + tan2\pi}{1 - tan\theta tan2\pi}$

$tan(\theta + 2\pi) = \frac{tan\theta + 0}{1 - tan\theta *0}$

$tan(\theta + 2\pi) = \tan\theta$ '

'So:

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi)$ $= tan\theta +tan\theta +tan\theta$

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi) = 3+3+3$

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi) = 9$

4 0

2 years ago

Plz help me out with these problems.

Elanso [62]

I could only do b

3(2x + 4) = 2+6x+10

6x+12 = 2+6x+10

6x- 6x = 2+10-12

0x/0 = 0/0

x = 0

I think

4 0

3 years ago

Is 267 the largest number in the sum of 3 integers evaluated from 269?

Anni [7]

Yes. it is the largest number

4 0

3 years ago

What is the product of 2p + q and –3q – 6p + 1?

Umnica [9.8K]

-12p^2 - 12pq + 2p -3q^2 + q

3 0

3 years ago

Read 2 more answers