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Amiraneli [1.4K]
3 years ago
5

Create a real life example using either temperature or money to compare -3 and -5

Mathematics
2 answers:
lana66690 [7]3 years ago
7 0
The temperature when I went to the bus stop in the morning felt like it was -3°, sadly it was only -5° only two degrees warmer, which means it is still freezing
Olegator [25]3 years ago
4 0

Example with temperature:

The temperature -3 degrees is warmer than -5 degrees. This applies to celsius or fahrenheit. You might also write "negative 3 degrees" as "3 degrees below zero"; the same applies for -5. The absolute difference between the two temperatures is 2 degrees.

=================================================

Example with money:

If your bank balance is -3 dollars and your friend's bank balance is -5 dollars, then your friend owes more money to the bank (s/he owes $2 more than you do).

Negative balances mean you are in debt. Another way to look at this is to consider you want to buy an item worth $10 but you only have $7. If the store owner lends you the difference, which you pay back later, then 7-10 = -3 is the amount of money on your balance sheet indicating you are in debt $3. The order of subtraction is important. Saying 10-7 = 3 shows a positive bank balance and you wouldn't be in debt (and still have $3 in your pocket).

=================================================

In either example, the value -3 is to the right of -5 on the number line so we would write the inequality -3 > -5 to say that -3 is larger than -5. Have a look at the image attached below.

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A student is given that point P(a, b) lies on the terminal ray of angle Theta, which is between StartFraction 3 pi Over 2 EndFra
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Answer:

<em>A.</em>

<em>The student made an error in step 3 because a is positive in Quadrant IV; therefore, </em>

<em />cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}

Step-by-step explanation:

Given

P\ (a,b)

r = \± \sqrt{(a)^2 + (b)^2}

cos\theta = \frac{-a}{\sqrt{a^2 + b^2}} = -\frac{\sqrt{a^2 + b^2}}{a^2 + b^2}

Required

Where and which error did the student make

Given that the angle is in the 4th quadrant;

The value of r is positive, a is positive but b is negative;

Hence;

r = \sqrt{(a)^2 + (b)^2}

Since a belongs to the x axis and b belongs to the y axis;

cos\theta is calculated as thus

cos\theta = \frac{a}{r}

Substitute r = \sqrt{(a)^2 + (b)^2}

cos\theta = \frac{a}{\sqrt{(a)^2 + (b)^2}}

cos\theta = \frac{a}{\sqrt{a^2 + b^2}}

Rationalize the denominator

cos\theta = \frac{a}{\sqrt{a^2 + b^2}} * \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 + b^2}}

cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}

So, from the list of given options;

<em>The student's mistake is that a is positive in quadrant iv and his error is in step 3</em>

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