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Komok [63]
4 years ago
8

9x-6=8 ??¿????c?????????????????????????????????????????

Mathematics
2 answers:
Goshia [24]4 years ago
6 0
Solve : 9x-14 = 0

Add 14 to both sides of the equation
9x = 14
Divide both sides of the equation by 9:
x = 14/9


Annette [7]4 years ago
5 0
9x = 8 + 6;
9x = 14;
x = 14 ÷ 9;
x = 1.55 ;
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(3*x)*7<br><br>Simplify the expression. Identify the property used
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5 0
4 years ago
Based on the diagram below, what is the measure of arc DC ?
emmainna [20.7K]

Given:

A diagram of a circle O, arc(AB)=y^\circ, arc(CD)=(80+y)^\circ.

To find:

The measure of arc DC.

Solution:

According to the angle of intersecting chords theorem:

Angle at intersection of chords = Half of the sum of intercepted arcs.

Let \theta be the angle between the intersection of chords AC and BD.

\theta +112^\circ=180^\circ      (Linear pair)

\theta =180^\circ-112^\circ

\theta =68^\circ

Using the angle of intersecting chords theorem, we get

\theta=\dfrac{arc(AB)+arc(CD)}{2}

68^\circ =\dfrac{y^\circ+(80+y)^\circ}{2}

136^\circ =(80+2y)^\circ

On simplification, we get

136=80+2y

136-80=2y

56=2y

Divide both sides by 2.

28=y

Now,

arc(DC)=(80+y)^\circ

arc(DC)=(80+28)^\circ

arc(DC)=108^\circ

Therefore, the correct option is C.

5 0
3 years ago
Prove that: (Sec A- cosec A)(1+ tan A+cot A) = tan A× sec A - cot A × cosec A
mash [69]

Answer:

Step-by-step explanation:

(sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA

we take the LHS so here goes,

(sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA\\(\frac{1}{cosA} -\frac{1}{sinA})(1+\frac{sinA}{cosA}+\frac{cosA}{sinA})\\\\(\frac{sinA-cosA}{sinAcosA})(\frac{sinAcosA+sin^2A+cos^2A}{sinAcosA})\\

since , sin^2A+cos^2A=1

the identity becomes,

(\frac{sinA-cosA}{sinAcosA})(\frac{1+sinAcosA}{sinAcosA})\\\\(\frac{sinA+sin^2AcosA-cosA-cos^2AsinA}{sin^2Acos^2A})\\\\

now, we know,

sin^2A=1-cos^2A and cos^2A=1-sin^2A

the identity becomes,

(\frac{sinA+(1-cos^2A)cosA-cosA-(1-sin^2A)sinA}{sin^2Acos^2A} )\\\\

(\frac{sinA+cosA-cos^3A-cosA-sinA+sin^3A}{sin^2Acos^2A})

sin A and cos A cancel out it becomes zero

\frac{sin^3A-cos^3A}{sin^2Acos^2A} \\\\

Splitting the denominator the identity becomes

\frac{sin^3A}{sin^2Acos^2A}-\frac{cos^3A}{sin^2Acos^2A}  \\\\\frac{sinA}{cos^2A} - \frac{cosA}{sin^2A} \\\\\frac{sinA}{cosA}(\frac{1}{cosA})-\frac{cosA}{sinA}(\frac{1}{sinA})\\\\

Hence,

tanAsecA-cotAcosecA

3 0
3 years ago
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