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skad [1K]
3 years ago
13

Which fraction is greater than 2/3

Mathematics
1 answer:
Firlakuza [10]3 years ago
5 0
4/5, 5/6, 3/4.... these are all greater than 2/3. 

I hope this helps!
~kaikers
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What is the value of y in the system of equations shown below?
vlabodo [156]
just multiple the numbers.
for example 7×7 is - 49y
and 3×16 is - 48y
6 0
4 years ago
Please help
ratelena [41]

Answer:

rom the relation for linear regression formula, Y = b·X + a, we have ;

For car 1

ΣXY = 102780

(ΣX)² = 36

ΣX = 6

ΣY = 52390

ΣX² = 14

N = 3

b = -1000

a = 19463

Y = -1000×X + 19463

For Car 2

ΣXY = 102539.8

(ΣX)² = 36

ΣX = 6

ΣY = 52346.6

ΣX² = 14

N = 3 .

b = -1076.7

a = 19602

Y = -1076.7×X + 19602

Step-by-step explanation:

part B ^

8 0
3 years ago
A Ferrari can travel up to 217 miles per hour. How many feet per second is this equivalent to?
kari74 [83]

Answer:

318.267 feet per second

19096 feet per minute

Step-by-step explanation:

1) 217*5280 to get feet

2) 1145760/60 to get feet per minute

3) 19096/60 to get feet per second

7 0
3 years ago
Read 2 more answers
What is 5 1/4 in decimal form​
Norma-Jean [14]

5 1/4 in decimal form is 5.25 because 25 is 1/4 of 100.

8 0
4 years ago
Find the volume v of the described solid s. the base of s is an elliptical region with boundary curve 4x2 + 9y2 = 36. cross-sect
Tasya [4]
4x^2+9y^2=36\iff\dfrac{x^2}9+\dfrac{y^2}4=1

defines an ellipse centered at (0,0) with semi-major axis length 3 and semi-minor axis length 2. The semi-major axis lies on the x-axis. So if cross sections are taken perpendicular to the x-axis, any such triangular section will have a base that is determined by the vertical distance between the lower and upper halves of the ellipse. That is, any cross section taken at x=x_0 will have a base of length

\dfrac{x^2}9+\dfrac{y^2}4=1\implies y=\pm\dfrac23\sqrt{9-x^2}
\implies \text{base}=\dfrac23\sqrt{9-{x_0}^2}-\left(-\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac43\sqrt{9-{x_0}^2}

I've attached a graphic of what a sample section would look like.

Any such isosceles triangle will have a hypotenuse that occurs in a \sqrt2:1 ratio with either of the remaining legs. So if the hypotenuse is \dfrac43\sqrt{9-{x_0}^2}, then either leg will have length \dfrac4{3\sqrt2}\sqrt{9-{x_0}^2}.

Now the legs form a similar triangle with the height of the triangle, where the legs of the larger triangle section are the hypotenuses and the height is one of the legs. This means the height of the triangular section is \dfrac4{3(\sqrt2)^2}\sqrt{9-{x_0}^2}=\dfrac23\sqrt{9-{x_0}^2}.

Finally, x_0 can be chosen from any value in -3\le x_0\le3. We're now ready to set up the integral to find the volume of the solid. The volume is the sum of the infinitely many triangular sections' areas, which are

\dfrac12\left(\dfrac43\sqrt{9-{x_0}^2}\right)\left(\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac49(9-{x_0}^2)

and so the volume would be

\displaystyle\int_{x=-3}^{x=3}\frac49(9-x^2)\,\mathrm dx
=\left(4x-\dfrac4{27}x^3\right)\bigg|_{x=-3}^{x=3}
=16

6 0
3 years ago
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