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saveliy_v [14]
3 years ago
5

6. Z1 and 22 are vertical angles. If the measure of Z2 is 105°, find the measure of

Mathematics
1 answer:
Rashid [163]3 years ago
3 0

Answer:

Z1 is also 105°

Step-by-step explanation:

this is because vertical angles are always equal

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To solve this we are going to use the formula for the area of a circle: A= \frac{1}{4} \pi d^2
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Rewrite 32/35 and 9/10
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The question was incorrect. Please find the correct content below.

Compare 32/35 and 9/10.

Comparing 32/35 and 9/10 we have 32/35 is greater than 9/10, that is 32/35 > 9/10.

Fraction is the ratio of two numbers. The upper number is called Numerator and the Lower number is called the Denominator.

We know that if the denominators are the same for two fractions then which has the greatest numerator is a greater fraction than the other.

Given the fractions are 32/35, 9/10

To compare this two fractions we have to make denominators equal first.

LCM of 10,35 = 70

Calculating the fractions,

32/35 = (32*2)/(35*2) = 64/70

9/10 = (9*7)/(10*7) = 63/70

Since 64 > 63

So 64/70 > 63/70

Therefore, 32/35 > 9/10

Hence fraction 32/35 is greater than the other fraction 9/10.

Learn more about Fraction here -

brainly.com/question/78672

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Step-by-step explanation:

6 0
3 years ago
A solution is heated from 0°C to 100°C. Between 0°C and 50°C, the rate of temperature increase is 1.5°C/min. Between 50°C and 10
muminat

Answer:

0.6 °C/min

Step-by-step explanation:

The relationship between rates and movement is ...

time = distance/speed

Here, the "distance" is measured in °C, and the "speed" is the rate of change of temperature.

For the first half of the heating, the time required is ...

(50°C -0°C)/(1.5 °C/min) = 50/(3/2) min = 100/3 min

For the second half of the heating, the time required is ...

(100°C -50°C)/(4/10 °C/min) = 50/(4/10) = 125 min

Then the total time is ...

((100/3) +125) min = (475/3) min

And the average rate of temperature increase is ...

total temperature change / total time

= (100°C -0°C)/(475/3 min) = 300/475 °C/min = 12/19 °C/min ≈ 0.6 °C/min

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