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arlik [135]
2 years ago
5

PLS HELP ASAP. QUESTION IN PICTURE

Mathematics
2 answers:
NARA [144]2 years ago
7 0
Answer is
40+60= 100 + x = 180
so x = 80

180-80= 100
____ [38]2 years ago
5 0

Answer:

80

Step-by-step explanation:

180 - 100 = 80

(Meeting character limit)

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Answer:

Step-by-step explanation:

a(20ft)

b(46ft)

c(72.6mm)

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wheelchair ramps for access to public buildings are allowed a maximum of one inch of vertical increase for every one foot of hor
AveGali [126]
Length x width = area
<span>but
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3 years ago
Help me with this question please!
BigorU [14]

Answer:

c = 40d+20

Step-by-step explanation:

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6 0
3 years ago
Read 2 more answers
Round this “0.7660444431”into 4 decimals places!!!!!
ANTONII [103]

Answer:

.7660

Step-by-step explanation:

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5 0
3 years ago
Someone please help me with this lol… have no idea what I’m doing
Sholpan [36]

Given:

\cos \theta =\dfrac{3}{5}

\sin \theta

To find:

The quadrant of the terminal side of \theta and find the value of \sin\theta.

Solution:

We know that,

In Quadrant I, all trigonometric ratios are positive.

In Quadrant II: Only sin and cosec are positive.

In Quadrant III: Only tan and cot are positive.

In Quadrant IV: Only cos and sec are positive.

It is given that,

\cos \theta =\dfrac{3}{5}

\sin \theta

Here cos is positive and sine is negative. So, \theta must be lies in Quadrant IV.

We know that,

\sin^2\theta +\cos^2\theta =1

\sin^2\theta=1-\cos^2\theta

\sin \theta=\pm \sqrt{1-\cos^2\theta}

It is only negative because \theta lies in Quadrant IV. So,

\sin \theta=-\sqrt{1-\cos^2\theta}

After substituting \cos \theta =\dfrac{3}{5}, we get

\sin \theta=-\sqrt{1-(\dfrac{3}{5})^2}

\sin \theta=-\sqrt{1-\dfrac{9}{25}}

\sin \theta=-\sqrt{\dfrac{25-9}{25}}

\sin \theta=-\sqrt{\dfrac{16}{25}}

\sin \theta=-\dfrac{4}{5}

Therefore, the correct option is B.

6 0
2 years ago
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