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3 years ago

What is the difference between the mean high and mean low temperature for this 5-day period?

Mathematics

1 answer:

amid [387]3 years ago

3 0

Answer:

the mean high temperature is 71 degrees and the mean low is 31 degrees

Step-by-step explanation:

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Helpppp pleaseeeee e e e e e e e e e!!!!!!

AleksandrR [38]

Answer: 80

Step-by-step explanation:

50+50=100

180-100=80

8 0

3 years ago

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For the expression 34- 34a to have a negative value, what must be true about the value of a ?

MArishka [77]

Answer:

a>0

Step-by-step explanation:

if a is negative then the subtraction sign would switch. therefore, for the expression to have a negartive value a must be greater than 0

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3 years ago

A system of equations consists of a line s of the equation y = x - 5 that is graphed in orange, and a line t that passes through

diamong [38]

Answer: um what grade are you in

Step-by-step explanation:

7 0

3 years ago

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The ratio of the quantities of flour and sugar needed to bake a cake is 1:5. What is the quantity of sugar needed for cake, if 7

Pepsi [2]

You multiply 750 and 5 to get the answer to the sugar. 750×5=3750 so the answer would be 3,750 grams of sugar.

4 0

3 years ago

Integrate the following

enot [183]

I suppose you mean to have the entire numerator under the square root?

$\displaystyle\int_2^4\frac{\sqrt{x^2-4}}{x^2}\,\mathrm dx$

We can use a trigonometric substitution to start:

$x=2\sec t\implies\mathrm dx=2\sec t\tan t\,\mathrm dt$

Then for $x=2$ , $t=\sec^{-1}1=0$ ; for $x=4$ , $t=\sec^{-1}2=\frac\pi3$ . So the integral is equivalent to

$\displaystyle\int_0^{\pi/3}\frac{\sqrt{(2\sec t)^2-4}}{(2\sec t)^2}(2\sec t\tan t)\,\mathrm dt=\int_0^{\pi/3}\frac{\tan^2t}{\sec t}\,\mathrm dt$

We can write

$\dfrac{\tan^2t}{\sec t}=\dfrac{\frac{\sin^2t}{\cos^2t}}{\frac1{\cos t}}=\dfrac{\sin^2t}{\cos t}=\dfrac{1-\cos^2t}{\cos t}=\sec t-\cos t$

so the integral becomes

$\displaystyle\int_0^{\pi/3}(\sec t-\cos t)\,\mathrm dt=\boxed{\ln(2+\sqrt3)-\frac{\sqrt3}2}$

7 0

3 years ago