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Dahasolnce [82]
3 years ago
11

Rearrange the formula to make x the subject : 5( a - 2x) = 9(x + 1)

Mathematics
1 answer:
katovenus [111]3 years ago
6 0

Answer:

x = \frac{5a-9}{19}

Step-by-step explanation:

Given

5(a - 2x) = 9(x + 1) ← distribute parenthesis on both sides

5a - 10x = 9x + 9 ( add 10x to both sides )

5a = 19x + 9 ( subtract 9 from both sides )

5a - 9 = 19x ( divide both sides by 19 )

\frac{5a-9}{19} = x

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The segments in each figure are tangent to the circle. Find each length.
zubka84 [21]
The tangents are equal length, so
.. 8x = x^2
We presume their length is not zero, so we can divide by x to get
.. 8 = 2x
.. 4 = x

Then the length of each tangent is 8*4 = 32 units.
3 0
3 years ago
Root 3 cosec140° - sec140°=4<br>prove that<br><br>​
Lorico [155]

Answer:

Step-by-step explanation:

We are to show that \sqrt{3} cosec140^{0} - sec140^{0} = 4\\

<u>Proof:</u>

From trigonometry identity;

cosec \theta = \frac{1}{sin\theta} \\sec\theta = \frac{1}{cos\theta}

\sqrt{3} cosec140^{0} - sec140^{0} \\= \frac{\sqrt{3} }{sin140} - \frac{1}{cos140} \\= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\

From trigonometry, 2sinAcosA = Sin2A

= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\\\=  \frac{\sqrt{3}cos140-sin140 }{sin280/2}\\=  \frac{4(\sqrt{3}/2cos140-1/2sin140) }{2sin280}\\\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{sin280}\\since sin420 = \sqrt{3}/2 \ and \ cos420 = 1/2  \\ then\\\frac{4(sin420cos140-cos420sin140) }{sin280}

Also note that sin(B-C) = sinBcosC - cosBsinC

sin420cos140 - cos420sin140 = sin(420-140)

The resulting equation becomes;

\frac{4(sin(420-140)) }{sin280}

= \frac{4sin280}{sin280}\\ = 4 \ Proved!

3 0
3 years ago
HELP!!!! I need help
ahrayia [7]

Answer:

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10.74 - 10.60 = 0.14

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Thus, you need 10 ones, 6 tenths and 14 hundredths to make 10.74.

3 0
2 years ago
What is 50% of 30? anyone
tensa zangetsu [6.8K]

Answer:

15

Step-by-step explanation:

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3 0
3 years ago
Read 2 more answers
Use the number line to find the difference of 3 - (-1.5)
ololo11 [35]

your answer is 4.5 you're welcome :-)

5 0
2 years ago
Read 2 more answers
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