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Ray Of Light [21]
3 years ago
12

On a given morning the temperature was 71°F. The temperature dropped 5°F and then rose 8°F. How much does the temperature need t

o rise or fall to return to the initial temperature of 71°F?
A) Fall 8°F

B) Fall 3°F

C) Rise 8°F

D) Rise 5°F
Mathematics
2 answers:
Gwar [14]3 years ago
6 0

Answer is B- Fall 3 bacause fall 5 plus rise 8 means its 8 over the original.

Step-by-step explanation:


balandron [24]3 years ago
3 0

Answer:

Im going to have to say B

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I need to figure out the answer
ANEK [815]
Hello there,

If c=-2 just plug in -2 for every c you see.

(-2)²-7(-2)-8
This simplifies to
4+14-8
18-8
10

Hope this helps!
3 0
3 years ago
Read 2 more answers
Shishir bought 4000 orange at 70 paisa each. But 400 of them were rotten. He sold 2000 oranges at 90 paisa each.If he plans to m
Ipatiy [6.2K]

Answer:

He needs to sell the rest of the oranges at 75 paisa each.

Step-by-step explanation:

Consider the given information that, Shishir bought 4000 orange at 70 paisa each.

Note: 1 rupees = 100 paisa

Thus, 70 paisa = 70/100 rupees = 0.70 rupees

Therefore, the cost price of 4000 oranges is:

4000×0.70 rupees = 2800 rupees

The selling price of 2000 oranges is:

2000×0.90 rupees = 1800 rupees

The number of oranges now Shishir have:

4000 - 2000 - 400 = 1600

He wants to make a profit of RS 200. Thus the selling price of 4000 oranges should be:

2800 rupees + 200 rupees = 3000 rupees

He earned 1800 rupees by selling 2000 oranges at 90 paisa. So, the remaining amount that he needs to make with 1600 oranges is:

3000 rupees - 1800 rupees = 1200 rupees

Therefore, the cost of one orange is:

1600 oranges = 1200 rupees

1 orange = 1200/1600 rupees

1 orange = 0.75 rupees

Hence, he needs to sell the rest of the oranges at 75 paisa each.

4 0
3 years ago
To identify high-paying jobs for people that do not like stress, the following data were collected showing the average annual sa
charle [14.2K]

Answer:

Y = -0.09594X + 74.35629

Step-by-step explanation:

Given the data :

Job average annual salary (X) :

81

96

70

70

70

92

92

100

98

102

Stress tolerance (Y) :

69

62

67.5

71.3

63.3

69.5

62.8

65.5

60.1

69

Using technology, the linear model Obtian by fitting the given data is :

Y = -0.09594X + 74.35629 ;

Where ;

Y = stress tolerance (dependent variable)

X= Average annual salary (Independent variable)

Slope = 0.09594

Intercept = 74.35629

7 0
3 years ago
Solve for x. 25 + 15x = 0
MA_775_DIABLO [31]

Answer:

x=−

3

5

​

=−1.667

Step-by-step explanation:

6 0
3 years ago
Integrate <img src="https://tex.z-dn.net/?f=e%5E%7B4x%7D%5Csqrt%7B1%2Be%5E%7B2x%7D%20%7D%20dx" id="TexFormula1" title="e^{4x}\sq
AnnyKZ [126]

Answer:

(\frac{(1+e^{2x}) ^{\frac{5}{2} } }{{5}} + \frac{(1+e^{2x} )^{\frac{3}{2} } }{{3}} )+C

Step-by-step explanation:

<u><em> Step(i):-</em></u>

Given that the function

                    f(x) = e^{4x} \sqrt{1+e^{2x} }

Now integrating on both sides, we get

                 \int\limits{f(x)} \, dx = \int\limits{e^{4x} \sqrt{1+e^{2x} } dx

                               =    \int\limits{e^{2x} e^{2x} \sqrt{1+e^{2x} } dx

                         

<u><em>Step(ii):-</em></u>

  Let  1 + e^{2x}  = t

           e^{2x}  = t -1  

          2e^{2x}dx = d t

          e^{2x}dx = \frac{1}{2} d t

                = \int\limits{( \sqrt{1+e^{2x} }) e^{2x} e^{2x} dx

                  = \int\limits {\sqrt{t}(t-1)\frac{1}{2} dt }

                 = \frac{1}{2} \int\limits {\sqrt{t} (t) -\sqrt{t} ) dt }

                = \frac{1}{2} \int\limits {(t^{\frac{1}{2}  } t^{1} +t^{\frac{1}{2} } ) } \, dx

                = \frac{1}{2} \int\limits {(t^{\frac{3}{2}  } +t^{\frac{1}{2} } ) } \, dx

               = \frac{1}{2} (\frac{t^{\frac{3}{2} +1} }{\frac{3}{2}+1 } + \frac{t^{\frac{1}{2} +1} }{\frac{1}{2}+1 } )+C

              =  \frac{1}{2} (\frac{t^{\frac{3}{2} +1} }{\frac{5}{2} } + \frac{t^{\frac{1}{2} +1} }{\frac{3}{2} } )+C

             = \frac{1}{2} (\frac{t^{\frac{5}{2} } }{\frac{5}{2} } + \frac{t^{\frac{3}{2} } }{\frac{3}{2} } )+C

            = (\frac{(1+e^{2x}) ^{\frac{5}{2} } }{{5}} + \frac{(1+e^{2x} )^{\frac{3}{2} } }{{3}} )+C

<u><em>Final answer:-</em></u>

= (\frac{(1+e^{2x}) ^{\frac{5}{2} } }{{5}} + \frac{(1+e^{2x} )^{\frac{3}{2} } }{{3}} )+C

             

3 0
3 years ago
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