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2 years ago

13

In Buffalo, New York, the temperature was "14 °F in the morning. If the temperature drops 7°F, and then rose 9°F, what is the te

mperature now?

1 answer:

elena55 [62]2 years ago

7 0

Answer:

16 Degrees

Step-by-step explanation:

14-7=7 7+9=16

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Nora had already read 30% of the book before her teacher assigned it to the class. She is choosing between two options to finish

guapka [62]

Answer:

Let p represent the # of pages in the book. Then, Nora has already read 0.30p pages and has 0.70p pages left to read.

If she reads 25% pages/night, that means reading 0.25(0.70)p pages per night, or 17.5 pages/night. If 28% p/n, that means 0.28(0.70)p pages/night, or 19.6p pages/night.

How many nights will it take Nora to finish the book if she reads 25% of 7/10 of the book per night? Without any calculations, we can answer this by "4 nights, since she reads 1/4 of the unread portion of the book per night."

If she reads 28% of 7/10 of the book per night, that will require fewer nights:

First night: 28%

Second night: 28%

Third night: 28%

Total: 3(28%) = 84%

This leaves 16% to read on the final night.

This is one interpretation of what I think is a poorly worded question.

The author of this question might have meant reading 25% of the remaining unread pages per night, which leads to a different answer.

7 0

3 years ago

Multiply.<br><br> (−2.04)(4.3)<br><br> PLEASE HURRY

Valentin [98]

Answer:

The answer is -8.772

Step-by-step explanation:

3 0

3 years ago

Nadya [2.5K]

Answer:

8+8+9+9.. so the answer is 34 yards.

Step-by-step explanation:

6 0

3 years ago

Suppose that a cyclist began a 476

mojhsa [17]

Well, we know the cyclist left the western part going eastwards, at the same time the car left the eastern part going westwards

the distance between them is 476 miles, and they met 8.5hrs later

let's say after 8.5hrs, the cyclist has travelled "d" miles, whilst the car has travelled the slack, or 476-d, in the same 8.5hrs

we know the rate of the car is faster... so if the cyclist rate is say "r", then the car's rate is r+33.2

thus $\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
\textit{eastbound cyclist}&d&r&8.5\\
\textit{westbound car}&476-d&r+33.2&8.5
\end{array}\\\\
-----------------------------\\\\
\begin{cases}
\boxed{d}=8.5r\\\\
476-d=(r+33.2)8.5\\
----------\\
476-\boxed{8.5r}=(r+33.2)8.5
\end{cases}$

solve for "r", to see how fast the cyclist was going

what about the car? well, the car's rate is r + 33.2

6 0

3 years ago

I will give brainllits

Zina [86]

Answer:

image please

information is missing

4 0

3 years ago