Ask question

Ask question

All categories

3 years ago

7

The temperature was 68 F .Later that day the temperature was 82 F .How much does the temperature need to rise or fall to return

to the original temperature of 68F
A.fall 14 F
B.fall 16 F
C. Rise 14 F
D.Rise 16 F

1 answer:

ruslelena [56]3 years ago

4 0

The answer is A because 82-14=68.

You might be interested in

What's the value of <br> The 1s in 5114

MissTica

The one near the five is in the hundreds place and the one near the 4 is in the tenth place.

Feel free to message me if you need help on another problem.

8 0

3 years ago

Read 2 more answers

HELP PLEASEEEEEEEE!!!!!!!!!!!!!!!!!

mariarad [96]

Plus or minus 3. (The 1st answer)

x^2 means (-3*-3) which equals 9

x^2 means (3838) which also equals 9

6*9=54

7 0

3 years ago

Read 2 more answers

Y=arccos(1/x)<br><br> Please help me do them all! I don’t know derivatives :(

Stells [14]

Answer:

$f(x) = {sec}^{ - 1} x \\ let \: y = {sec}^{ - 1} x \rightarrow \: x = sec \: y\\ \frac{dx}{dx} = \frac{d(sec \: y)}{dx} \\ 1 = \frac{d(sec \: y)}{dx} \times \frac{dy}{dy} \\ 1 = \frac{d(sec \: y)}{dy} \times \frac{dy}{dx} \\1 = tan \: y.sec \: y. \frac{dy}{dx} \\ \frac{dy}{dx} = \frac{1}{tan \: y.sec \: y} \\ \frac{dy}{dx} = \frac{1}{ \sqrt{( {sec}^{2} \: y - 1}) .sec \: y} \\ \frac{dy}{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \frac{d( {sec}^{ - 1}5x) }{dx} = \frac{1}{ |5x | \sqrt{25 {x }^{2} - 1} }\\\\y=arccos(\frac{1}{x})\Rightarrow cosy=\frac{1}{x}\\x=secy\Rightarrow y=arcsecx\\\therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} }$

4 0

2 years ago

Calculate the range for both sets of data:

otez555 [7]

Range of Set A=9.1
Range of Set B=5.3

5 0

3 years ago

I need help with #6!! ASAP

Naddika [18.5K]

G) Because 70% * 200 = 140

6 0

3 years ago