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defon
3 years ago
8

Tres croissants y dos ensaimadas nos ha costado 6,10Є si recordamos que el precio ce una ensaimada es 0,30Є más caro que el de u

n croissant. ¿cual es el precio de una ensaimada y cual es de un croissant?
Mathematics
1 answer:
saveliy_v [14]3 years ago
3 0

Answer:

Step-by-step explanation:

3c + 2e = 6.10

e = c + 0.30

3c + 2(c + 0.30) = 6.10

5c + 0.60 = 6.10

5c = 5.50

c = 1.10 Є

e = c + 0.30 = 1.40 Є

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3 years ago
determine the height in inches of a triangle with an area of 107 square inches and a base of 35 inches.
inna [77]

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2 years ago
I don't know how to solve this helpp
Aleksandr-060686 [28]

neverminding the jumbled lingo, is simply asking for the equation of the tangent line at that point, it says all tangents, well, there's only one passing there.

we can simply get the derivative of f(x) and take it from there.

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since now we know the slope when x = 3/2, then we can just plug that into its point-slope intercept form, along with the coordinates.

(\stackrel{x_1}{\frac{3}{2}}~,~\stackrel{y_1}{7})~\hfill \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{7}=\stackrel{m}{-1}(x-\stackrel{x_1}{\frac{3}{2}}) \\\\\\ y-7=-x+\cfrac{3}{2}\implies y=-x+\cfrac{17}{2}

6 0
3 years ago
Use the shell method to compute the volume of the solids obtained by rotating the region enclosed by the graphs of the functions
Marina CMI [18]

Answer:

44.18 cubic unit

Step-by-step explanation:

The intersection of the curve y = x^2 and y = 8-x^2 is

x^2 = 8 - x^2

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x = \pm 2

Since we are rotating the region where x \geq 0.5 we will use the intersection x = 2 and y = 4

Using the shell method with x ranges from 0.5 to 2. The volume of the shell would be

V = \int\limits^2_{0.5} {2\pi x h} \, dx

where h is the difference between the y-coordinates of the curves, in other words

h = 8 - x^2 - x^2 = 8 - 2x^2

Plug h into the V integral and we have

V = \int\limits^2_{0.5} {2\pi x (8 - 2x^2)} \, dx\\V = 4\pi\int\limits^2_{0.5} {(4x - x^3)} \, dx\\V = 4\pi[2x^2 - x^4/4]^2_{0.5}\\V = 4\pi(2*2^2 - 2^4/4 - 2*0.5^2 + 0.5^4/4)\\V = 4\pi(8 - 4 - 0.5 + 0.015625) \approx 44.18

7 0
3 years ago
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