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adell [148]
3 years ago
8

If m∠1 = 3x, m∠3 = 2x + 20 and m∠5 = 30, what is the value of x?

Mathematics
1 answer:
iogann1982 [59]3 years ago
6 0
M∠2 = 180 - m∠1 = 180 - 3x   [supplementary angles<span>]

</span>In a triangle, the three interior angles always add to 180° ⇒
m∠2 + m∠3 + m∠5 = 180 <span>
180-3x + 2x+20 + 30 = 180
230 - x = 180
x = 230 - 180
x = 50</span>
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What is the 8th term of this sequence 2,8,18,36,66, 112,178
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3 years ago
Can you please solve this please and thank you.
Lilit [14]

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If (ax+2)(bx+7)=15x2+cx+14 for all values of x, and a+b=8, what are the 2 possible values fo c
dolphi86 [110]

Given:

(ax+2)(bx+7)=15x^2+cx+14

And

a+b=8

Required:

To find the two possible values of c.

Explanation:

Consider

\begin{gathered} (ax+2)(bx+7)=15x^2+cx+14 \\ abx^2+7ax+2bx+14=15x^2+cx+14 \end{gathered}

So

\begin{gathered} ab=15-----(1) \\ 7a+2b=c \end{gathered}

And also given

a+b=8---(2)

Now from (1) and (2), we get

\begin{gathered} a+\frac{15}{a}=8 \\  \\ a^2+15=8a \\  \\ a^2-8a+15=0 \end{gathered}a=3,5

Now put a in (1) we get

\begin{gathered} (3)b=15 \\ b=\frac{15}{3} \\ b=5 \\ OR \\ b=\frac{15}{5} \\ b=3 \end{gathered}

We can interpret that either of a or b are equal to 3 or 5.

When a=3 and b=5, we have

\begin{gathered} c=7(3)+2(5) \\ =21+10 \\ =31 \end{gathered}

When a=5 and b=3, we have

\begin{gathered} c=7(5)+2(3) \\ =35+6 \\ =41 \end{gathered}

Final Answer:

The option D is correct.

31 and 41

8 0
1 year ago
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