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andrezito [222]
3 years ago
8

Solve -2p + 10p = 7p for p.

Mathematics
2 answers:
BartSMP [9]3 years ago
8 0

Answer: P = 0

Step-by-step explanation:

Solve for p by simplifying both sides of the equation, the isolating the variable.

-2p + 10p = 7p, P will equal 0.

***If you found my answer helpful, please give me the brainliest. :) ***

Kobotan [32]3 years ago
6 0

Answer:

Step-by-step explanation:  

-2x7+ 10x7

-14+70

70-14 ( there were more positives than negatives. so subtract the smaller number and you get)

56

no variable because it was replaced with 7.

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Read 2 more answers
[4]
Tanya [424]

Answer:

Angle ACD = 38°

Step-by-step explanation:

The full, correct question is presented the attached image to this solution.

Given

Point O is the centre if the circle

Points A, B, C and D are points on the circle

Angle AOB = 140°

Angle OAC = 14°

Angle AOB = 2 × (Angle ACB) [angle subtended at the centre of the circle is twice the angle subtended at the circumference of the circle)

140° = 2 × (Angle ACB)

Angle ACB = (140°/2) = 70°

(Angle AOB) + (Angle OAB) + (Angle ABO) = 180° [sun of angles in a triangle is 180°]

But Angle OAB = Angle ABO = a [base angles of an iscosceles triangle are equal since OA and OB are both radii for the circle O)

(Angle AOB) + (Angle OAB) + (Angle ABO) = 180°

140° + a + a = 180°

2a = 40°

a = (40/2) = 20°

Angle OAB = Angle ABO = 20°

Angle CAB = (Angle OAC) + (Angle OAB) = 14° + 20° = 34°

Triangle ADC is an iscosceles triangle, hence,

Angle DAC = Angle ACD = x [base angles of an iscosceles triangle are equal]

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(Angle DCB) + (Angle DAB) = 180° [Opposite angles of a cyclic quadilateral (a quadilateral inscribed in a circle) sum up to give 180°]

Angle DCB = (Angle DCA) + (Angle ACB) = (x + 70°)

Angle DAB = (Angle DAC) + (Angle CAB) = (x + 34°)

(Angle DCB) + (Angle DAB) = 180°

(x + 70°) + (x + 34°) = 180°

2x + 104° = 180°

2x = 180° - 104° = 76°

x = (76°/2) = 38°

Angle DAC = Angle ACD = 38°

Angle ACD = 38°

Hope this Helps!!!

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2 years ago
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