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Ira Lisetskai [31]
3 years ago
13

Translate into equation

Mathematics
1 answer:
bearhunter [10]3 years ago
7 0
Let x= smaller#
Let 2x-5=larger#

X+2x-5
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Please help you have to write each expression in two other ways
galina1969 [7]
11. a) .19 b) 19/100
12. a) 23% b) .23
13. a) 7% b) 7/100
14. a) .11 b) 11/100
15. a) .2 b) 2/100
16. a) .9 b) 9%
17. a) 13% b) 13/100
18. a) ¼ b) .25

Hoped it helped. ☺
3 0
3 years ago
Cody has $7. He wants to
Temka [501]
<h3>Answer:    (1,3)</h3>

=====================================================

Explanation:

Note that the two shaded regions overlap to form the darker shaded region. The point (1,1) is only in the green region, but not in the blue region. It needs to be in both regions for it to be a solution to the system. In contrast, the point (1,3) is on the boundary of the blue region and in the green region as well. It's a solution point because of this.

Points on the boundary are included because of the "or equal to" as part of the inequality sign. Hence the solid boundary lines rather than dashed boundary lines.

The solution (1,3) means that x = 1 and y = 3 pair up together. It says that Cody can buy x = 1 hotdog and y = 3 packs of peanuts so that he has at least 4 snacks, and he stays within the $7 budget.

Going back to (x,y) = (1,1) for a moment, this isn't a solution because x+y = 1+1 = 2 is not 4 or larger. In other words, he only buys 2 snacks here instead of 4 or more.

5 0
3 years ago
(-3.5t - 4s + 4.5) + (-7.1 - 0.3s + 4.1t)
Kazeer [188]

Answer:

0.6t - 4.3s -2.6

Step-by-step explanation:

Given the expression

(-3.5t - 4s + 4.5) + (-7.1 - 0.3s + 4.1t)

Collect the like terms

= -3.5t+4.1t-4s-0.3s+4.5-7.1

= 0.6t - 4.3s -2.6

Hence the required expression is 0.6t - 4.3s -2.6

6 0
3 years ago
PLEASE ANWSER FAST IM TAKING A TEST!!<br><br> 20 points!!!!
DochEvi [55]

Answer:

the correct awnser is A

Step-by-step explanation:

HAVE A GOOD DAY!!!!!!

3 0
3 years ago
Read 2 more answers
Use the Divergence Theorem to calculate the surface integral S F · dS; that is, calculate the flux of F across S. F(x, y, z) = x
devlian [24]

Answer:

-14 / 3

Step-by-step explanation:

- Divergence theorem, expresses an explicit way to determine the flux of a force field ( F ) through a surface ( S ) with the help of "del" operator ( D ) which is the sum of spatial partial derivatives of the force field ( F ).

- The given force field as such:

                      F = (x^2y) i + (xy^2) j + (3xyz) k

Where,

         i, j, k are unit vectors along the x, y and z coordinate axes, respectively.

- The surface ( S ) is described as a tetrahedron bounded by the planes:

                      x = 0 \\y = 0\\x + 2y + z = 2

                      z = 0\\

- The divergence theorem gives us the following formulation:

                      _S\int\int {F} \,. dS = _V\int\int\int {D [F]} \,. dV

- We will first apply the del operator on the force field as follows:

                      D [ F ] = 2xy + 2xy + 3xy = 7xy

- Now, we will define the boundaries of the solid surface ( Tetrahedron ).

- The surface ( S ) is bounded in the z - direction by plane z = 0 and the plane [ z = 2 - x - 2y ]. The limits of integration for " dz " are as follows:

                      dz: [ z = 0 - > 2 - x - 2y ]

- Now we will project the surface ( S ) onto the ( x-y ) plane. The projection is a triangle bounded by the axes x = y = 0 and the line: x = 2 - 2y. We will set up our limits in the x- direction bounded by x = 0 and x = 2 - 2y. The limits of integration for " dx " are as follows:

                     dx: [ x = 0 - > 2 - 2y ]

- The limits of "dy" are constants defined by the axis y = 0 and y = -2 / -2 = 1. Hence,

                    dy: [ y = 0 - > 1 ]

- Next we will evaluate the triple integral as follows:

                   \int\int\int ({D [ F ] }) \, dz.dx.dy = \int\int\int (7xy) \, dz.dx.dy\\\\\int\int (7xyz) \, | \limits_0^2^-^x^-^2^ydx.dy\\\\\int\int (7xy[ 2 - x - 2y ] ) dx.dy = \int\int (14xy -7x^2y -14 xy^2 ) dx.dy\\\\\int (7x^2y -\frac{7}{3} x^3y -7 x^2y^2 )| \limits_0^2^-^2^y.dy  \\\\\int (7(2-2y)^2y -\frac{7}{3} (2-2y)^3y -7 (2-2y)^2y^2 ).dy  \\\\

                 7 (-\frac{(2-2y)^3}{6} + (2-2y)^2 ) -\frac{7}{3} ( -\frac{(2-2y)^4}{8} + (2-2y)^3) -7 ( -\frac{(2-2y)^3}{6}y^2 + 2y.(2-2y)^2 )| \limits^1_0\\\\ 0 - [ 7 (-\frac{8}{6} + 4 ) -\frac{7}{3} ( -\frac{16}{8} + 8 ) -7 ( 0 ) ] \\\\- [ \frac{56}{3} - 14 ] \\\\\int\int {F} \, dS  = -\frac{14}{3}

3 0
3 years ago
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