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2 years ago

9

The formula for converting temperatures in degrees Celsius to degrees Fahrenheit is F = 95C + 32. If the temperature is 20°C, wh

at is the temperature in degrees Fahrenheit?

1 answer:

Sveta_85 [38]2 years ago

4 0

Answer:

68 degrees fahrenheit

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Justin weighted 8lb 12oz when he was born at his two week check-up he had gained 8oz what was his new weight in pounds and ounce

avanturin [10]

The answer is 9 lbs and 4 oz

5 0

3 years ago

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Convert into decimals (use your calculator): a) 6/38 b) 45/55 c) 4/28 d) 35/28 e) 1030/2030

Delicious77 [7]

Answer:

$a.\hspace{3} \frac{6}{38} = 0.15789474\\\\b.\hspace{3} \frac{45}{55} = 0.81818181\\\\c.\hspace{3} \frac{4}{28} = 0.14285714\\\\d.\hspace{3} \frac{35}{28} = 1.25\\\\e.\hspace{3} \frac{1030}{2030} = 0.50738916$

Step-by-step explanation:

Rational numbers, expressed as decimal numbers, are obtained from the operation of division between the integer of the numerator and the integer of the denominator. Then:

$a.\hspace{3} \frac{6}{38} = 6\div38 = 0.15789474\\\\ b.\hspace{3} \frac{45}{55} = 45\div55 = 0.81818181\\\\c.\hspace{3} \frac{4}{28} = 4\div28 = 0.14285714\\\\ d.\hspace{3} \frac{35}{28} = 35\div28 = 1.25\\\\ e.\hspace{3} \frac{1030}{2030} = 1030\div2030 = 0.50738916$

3 0

3 years ago

The perimeter of a square is 44 centimeters. what is its area?

aalyn [17]

Answer:

Perimeter = 44cm

we know that square has 4 sides which are same

so perimeter means sum of all 4 sides

4 × s = 44cm

s = 11cm

Each side will be of 11cm.

Now area of square is Square of its side length.

so area = 11 × 11 = 121 cm^2

8 0

3 years ago

Can someone help please and thank you!

kondor19780726 [428]

Answer:

10 and 40 I think maybe?

Step-by-step explanation:

I hope this is correct if not I am so so so so so so sorry

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2 years ago

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Find the exact length of the curve. x=et+e−t, y=5−2t, 0≤t≤2 For a curve given by parametric equations x=f(t) and y=g(t), arc len

Rama09 [41]

The length of a curve <em>C</em> parameterized by a vector function <em>r</em><em>(t)</em> = <em>x(t)</em> i + <em>y(t)</em> j over an interval <em>a</em> ≤ <em>t</em> ≤ <em>b</em> is

$\displaystyle\int_C\mathrm ds = \int_a^b \sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2} \,\mathrm dt$

In this case, we have

<em>x(t)</em> = exp(<em>t</em> ) + exp(-<em>t</em> ) ==> d<em>x</em>/d<em>t</em> = exp(<em>t</em> ) - exp(-<em>t</em> )

<em>y(t)</em> = 5 - 2<em>t</em> ==> d<em>y</em>/d<em>t</em> = -2

and [<em>a</em>, <em>b</em>] = [0, 2]. The length of the curve is then

$\displaystyle\int_0^2 \sqrt{\left(e^t-e^{-t}\right)^2+(-2)^2} \,\mathrm dt = \int_0^2 \sqrt{e^{2t}-2+e^{-2t}+4}\,\mathrm dt$

$=\displaystyle\int_0^2 \sqrt{e^{2t}+2+e^{-2t}} \,\mathrm dt$

$=\displaystyle\int_0^2\sqrt{\left(e^t+e^{-t}\right)^2} \,\mathrm dt$

$=\displaystyle\int_0^2\left(e^t+e^{-t}\right)\,\mathrm dt$

$=\left(e^t-e^{-t}\right)\bigg|_0^2 = \left(e^2-e^{-2}\right)-\left(e^0-e^{-0}\right) = \boxed{e^2-\frac1{e^2}}$

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2 years ago