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denis23 [38]
2 years ago
13

If ∠G is supplementary to ∠H and the measure of ∠G is 65° what is the measure of ∠H?

Mathematics
1 answer:
Elina [12.6K]2 years ago
7 0

Solution:

<u>Given:</u>

  • ∠G = 65°

Supplementary angles are a pair of angles that sum up to 180°.

<u>It should be noted:</u>

  • If ∠G and ∠H are a pair of supplementary angles, they both sum up to 180°.

Equation formed: ∠G + ∠H = 180

<u>Substitute the values into the equation.</u>

  • ∠G + ∠H = 180
  • => 65 + ∠H = 180

<u>Subtract 65 both sides.</u>

  • => 65 - 65 + ∠H = 180 - 65
  • => ∠H = 180 - 65 = 115°

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Step-by-step explanation:

Step 1: Add like terms

4x² - 6x² = -2x²

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Step 2: Rewrite

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Sara hina and Arslan have a total of RS 390in their wallet . Sara has Rs. 7more than Hina .Arslan has 3 times what sara has . Ho
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Answer:Sara hina and Arslan Have RS79.4,RS 72.4 and RS238.2 respectively.

Step-by-step explanation:

Step 1

Let the amount that hina has be x

the amount that sara has be represented as  7+x

and the amount that Arslan have be represented as 3(7+x)

such that the total amount in their wallet which is 390 can be expressed as

x+7+x + 3(7+x)=390

Step 2

Solving

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5x+28=390

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2 years ago
A bar of gold has a volume of 722 cubic centimeters and weighs 13.949 kilograms. What's its density in grams per cubic centimete
Ahat [919]

Answer:

The density of the  bar of gold is  19.32 grams per cubic centimeter .

Step-by-step explanation:

Formula

Density = \frac{Mass}{Volume}

As given

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As 1 kilogram = 1000 gram

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13.949 kilograms = 13.949 × 1000 grams

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Volume = 722 cubic centimeter

As put in the formula

Density = \frac{13949}{722}

Density = 19.32 grams per cubic centimeter (Approx)

Therefore the density of the  bar of gold is  19.32 grams per cubic centimeter .


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Answer: -116

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