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ASHA 777 [7]
3 years ago
10

At 6:00 the outside air temperature was 15°F. By 8:00 the temperature had dropped 9°F, and by 11:00 I had dropped another 12°. W

hat was the temperature at 11:00?
Mathematics
1 answer:
laila [671]3 years ago
8 0

Answer:

- 6°F

Step-by-step explanation:

Given :

Temperature at 6:00 = 15°F

Temperature dropped by 9°F by 8:00

Temperature by 8:00 = 15°F - 9°F = 6°F

Temperature dropped another 12°F by 11:00

Hence, Temperature at 11:00 will be :

6°F - 12°F = - 6°F

Temperature at 11:00 = - 6°F

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Quantitative data refers to any information that can be quantified, counted or measured, and given a numerical value. Qualitative data is descriptive in nature, expressed in terms of language rather than numerical values.
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2 years ago
Solve the following system of equations algebraically:<br> 4x - 5y = 18<br> 3x – 2y = 10
fiasKO [112]

Answer:

D

Step-by-step explanation:

Given the 2 equations

4x - 5y = 18 → (1)

3x - 2y = 10 → (2)

Multiplying (1) by 3 and (2) by - 4, then adding will eliminate the x- term

12x - 15y = 54 → (3)

- 12x + 8y = - 40 → (4)

Add (3) and (4) term by term to eliminate x, that is

- 7y = 14 ( divide both sides by - 7 )

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Substitute y = - 2 into either of the 2 equations and solve for x

Substituting into (1)

4x - 5(- 2) = 18

4x + 10 = 18 ( subtract 10 from both sides )

4x = 8 ( divide both sides by 4 )

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solution is (2, - 2 ) → D

8 0
3 years ago
Solve the equation x/8=3
Sidana [21]

Answer:

24

Step-by-step explanation:

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6 0
2 years ago
Find the general solution to 3y′′+12y=0. Give your answer as y=... . In your answer, use c1 and c2 to denote arbitrary constants
lozanna [386]

Answer:

y(x)=c_1e^{2ix}+c_2e^{-2ix}

Step-by-step explanation:

You have the following differential equation:

3y''+12y=0     (1)

In order to find the solution to the equation, you can use the method of the characteristic polynomial.

The characteristic polynomial of the given differential equation is:

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The solution of the differential equation is:

y(x)=c_1e^{m_1x}+c_2e^{m_2x}   (2)

where m1 and m2 are the roots of the characteristic polynomial.

You replace the values obtained for m1 and m2 in the equation (2). Then, the solution to the differential equation is:

y(x)=c_1e^{2ix}+c_2e^{-2ix}

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3 years ago
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