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3 years ago

3. WEATHER On a cold January day,

Mathematics

1 answer:

erma4kov [3.2K]3 years ago

3 0

Answer: -9 + 21 = t t = 12°C.

Step-by-step explanation: For this problem lets have t be equal to the original temperature in the beginning of the day. The equation -9 + 21 = t represents this situation because it states that if we add the degrees that dropped to the current degrees it will give us our original degrees.

Solve.

-9 + 21 = t (Simply add -9 + 21) -9 + 21 = (+12)

12 = t

So, the original degrees in the beginning of the day was 12°C.

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What is the product of 5.06 and 2.1

Oksana_A [137]

Answer:

10.626

Step-by-step explanation:

There's multiple ways to go about it. See how

5.06 * 2.1 = (5 + 0.06)(2 + 0.1) = 5*2 + 0.06*2 + 5*0.1 + 0.06*0.1 =

= 10 + 0.12 + 0.5 + 0.006 = 10.626

Please notice how breaking up the product into smaller parts has let us make all the computations without too much hassle.

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2 years ago

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Sales of 10 000 increase by 65 each year

lorasvet [3.4K]

Answer:

Whats the question

Step-by-step explanation:

also if you are looking for the unit rate its 153.846...

or if you are looking for how many in 2 years then 650,000

Hope this helps

Have a nice day :)

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3 years ago

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Which is an equivalent form of the expression 1800(1.2)^-t

ololo11 [35]

Answer:

1800(5/6)^t

Step-by-step explanation:

I don't see any answer choices, but this would be another form if it is listed. The negative exponent means that you get the reciprocal of the fraction (in this case 1.2). This turns 1.2 into (1/1.2), which is equivalent to 5/6.

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1 year ago

John leaves school to go home.his bus drives 6 kilometers north and then goes 7 kilometers west.how far is John's house from the

otez555 [7]

Answer:

John is 9.21 km form the school.

Step-by-step explanation:

John leaves school to go home. His bus drives 6 kilometres north and then goes 7 kilometres west. It is required to find John's distance from the school. It is equal to the shortest path covered or its displacement. So,

$d=\sqrt{6^2+7^2} \\\\d=9.21\ km$

So, John is 9.21 km form the school.

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3 years ago

What is the coefficient of x2 in the expansion of (x + 2)3?

Vinvika [58]

I think 6 is the answer. Hopefully I answered your question

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2 years ago