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Gemiola [76]
2 years ago
5

Which fraction is closest to 1 7/5 or 5/7

Mathematics
2 answers:
EleoNora [17]2 years ago
3 0

Answer:

5/7

Step-by-step explanation:

17/5 is greater than 5/7

pantera1 [17]2 years ago
3 0
Ujkkkbvfr said the video is a video video that is a key
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I don’t know the answer I just need some points to ask more questions !!! have a good day !!
Scilla [17]

Answer:

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Step-by-step explanation:

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2 years ago
Urgent…. I need help
Firdavs [7]

A.

Step-by-step explanation:

To find out the answer, there is a method that can be used to find out if a line is a function. It is called the vertical line method. And it also has to be per x, One y value has to be there. Analyzing the graphs, the answer should be option A.

8 0
2 years ago
I need help please and thanks
Akimi4 [234]
Perimeter of rectangle = all sides added
= 3+3+7+7 =20


volume of sphere = 4/3 x π x r(cubed)
= 4/3 x π X 4(cubed)
= 268.08257

area of triangle = base x height / 2
= 5 x 6 / 2
=30/2 =15

volume of pyramid = base height x base width x height / 3
= 300

side of triangle ---> use Pythagoras theorem so a(squared) + b (squared) = c(squared)
4(squared) + 3(squared) = c(squared)
16 + 9 = c(squared)
25 = c(squared)
c = 5


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3 years ago
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Zina [86]

Answer:

1/24

Step-by-step explanation:

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4 0
3 years ago
Read 2 more answers
1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves
erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx

\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx

\ln|y| = x - \ln|x+1| + C

When x = 0, we have y = 2, so we solve for the constant C :

\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)

Then the particular solution to the DE is

\ln|y| = x - \ln|x+1| + \ln(2)

We can go on to solve explicitly for y in terms of x :

e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}

(b) The curves y = x² and y = 2x - x² intersect for

x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1

and the bounded region is the set

\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}

The area of this region is

\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}

7 0
2 years ago
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