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2 years ago

Triangles ABC, EDC, and EFG are similar triangles. The measures of the three interior angles of triangle EFG are °, °, and °.

Mathematics

2 answers:

iren [92.7K]2 years ago

7 0

Answer:

<h2>The interior angles of triangle EFG are 30°, 50° and 100°.</h2>

Step-by-step explanation:

We know by given that $\triangle ABC \sim \triangle EDC \sim \triangle EFG$ .

Similarities refers to proportional sides and congruent angles, so

$\angle BAC \cong \angle DEC \cong \angle FEG$ , by corresponding elements.

Therefore, $\angle BAC = 30\° =\angle DEC = \angle FEG$

By the same reason, $\angle ECD = \angle ACB = \angle EGF = 50 \°$

Then,

$\angle FEG + \angle EGF + \angle EFG = 180\°$ , by internal angles theorem.

Replacing values, we have

$30\° + 50\° + \angle EFG = 180\°\\\angle EFG = 180\° -80\°\\\angle EFG = 100\°$

Therefore, the interior angles of triangle EFG are 30°, 50° and 100°.

Misha Larkins [42]2 years ago

4 0

Because of the vertical angles theorem, BAC becomes 50°. And angle B is 180-(50+30)=100.

If all 3 triangles are similar, the angles of triangle EFG are 30°, 50°, 100°

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How much heat is required to raise the temperature of

Elina [12.6K]

Answer:

The amount of heat required to raise the temperature of liquid water is 9605 kilo joule .

Step-by-step explanation:

Given as :

The mass of liquid water = 50 g

The initial temperature = $T_1$ = 15°c

The final temperature = $T_2$ = 100°c

The latent heat of vaporization of water = 2260.0 J/g

Let The amount of heat required to raise temperature = Q Joule

Now, From method

Heat = mass × latent heat × change in temperature

Or, Q = m × s × ΔT

or, Q = m × s × ( $T_2$ - $T_1$ )

So, Q = 50 g × 2260.0 J/g × ( 100°c - 15°c )

Or, Q = 50 g × 2260.0 J/g × 85°c

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Or, Q = 9,605 × 10³ joule

Or, Q = 9605 kilo joule

Hence The amount of heat required to raise the temperature of liquid water is 9605 kilo joule . Answer

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2 years ago

Without properly working spark plugs, a vehicle will not run. For a specific vehicle, the spark plugs are supposed to have a gap

Salsk061 [2.6K]

Answer:

The probability that the spark plugs are supposed to have a gap between 3.9mm and 4.3mm.

P(3.9≤X≤4.3) = 0.9922

Step-by-step explanation:

Step(i):-

Given that the mean of the Normal distribution = 4.1mm

Given that the standard deviation of the Normal distribution = 0.0075mm

Let 'X' be the random variable in a normal distribution

Given that X₁ = 3.9mm

$Z_{1} = \frac{X_{1} -mean}{S.D}$

$Z_{1} = \frac{3.9 -4.1}{0.075} = -2.666$

Given that X₂ = 4.3mm

$Z_{2} = \frac{X_{2} -mean}{S.D}$

$Z_{1} = \frac{4.3 -4.1}{0.075} = 2.666$

Step(ii):-

The probability that the spark plugs are supposed to have a gap between 3.9mm and 4.3mm.

P(3.9≤X≤4.3) = P(-2.666≤Z≤2.666)

= P(Z≤2.666)-P(Z≤-2.666)

= 0.5 +A(2.666) - (0.5-A(2.666)

= 2 × A(2.666)

= 2×0.4961

= 0.9922

Final answer:-

The probability that the spark plugs are supposed to have a gap between 3.9mm and 4.3mm.

P(3.9≤X≤4.3) = 0.9922

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1 year ago

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Drupady [299]

Answer:

Step-by-step explanation:

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