Can someone help me please .. I NEED HELP ASAP .

Mathematics

2 answers:

3241004551 [841]3 years ago

7 0

Answer:

no solution

Step-by-step explanation:

no values of x can make this equation true

(please consider marking brainliest <3 )

leonid [27]3 years ago

5 0

No solution found and here how u would find that out

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38 degrees 2x degrees +6 degrees

ratelena [41]

STEP
1
:
Pulling out like terms

Pull out like factors :

32 - 2x = -2 • (x - 16)

STEP
2
:

Equations which are never true:

2.1 Solve : -2 = 0

This equation has no solution.
A a non-zero constant never equals zero.
Solving a Single Variable Equation:

2.2 Solve : x-16 = 0

Add 16 to both sides of the equation :
x = 16

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Answer:

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3 years ago

1)a chord of length 18cm midway the radius of a circle. calculate the radius of the circle correct to 1d.p. 2)if two parallel ch

kykrilka [37]

Part 1:

Given that the length of the chord is 18 cm and the chord is midway the radius of the circle.

Thus, half the angle formed by the chord at the centre of the circle is given by:

$\cos\theta=\frac{\left( \frac{1}{2} r\right)}{r}= \frac{1}{2} \\ \\ \Rightarrow\theta=\cos^{-1}\left( \frac{1}{2} \right)=60^o$

Now,

$\sin60^o= \frac{9}{r} \\ \\ \Rightarrow r= \frac{9}{\sin60^o} =10.392$

Therefore, the radius of the circle is 10.4 cm to 1 d.p.

Part 2I:

Given that the radius of the circle is 10 cm and the length of chord AB is 8 cm. Thus, half the length of the chord is 4cm. Let the distance of the mid-point O to /AB/ be x and half the angle formed by the chord at the centre of the circle be θ, then

$\sin\theta= \frac{4}{10} = \frac{2}{5} \\ \\ \theta=\sin^{-1}\left( \frac{2}{5} \right)=23.6^o$

Now,

$\cos23.6^o= \frac{x}{10} \\ \\ \Rightarrow x=10\cos23.6^o=9.165\approx9.2cm$

Part 2II:

Given that the radius of the circle is 10cm and the angle distended is 80 degrees. Let half the length of chord CD be y, then:

$\sin40^o= \frac{y}{10} \\ \\ \\ \Rightarrow y=10\sin40^o=6.428$

Thus, the length of chord CD = 2(6.428) = 12.856 which is approximately 12.9 cm.

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