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svlad2 [7]
3 years ago
14

S is inversely proportional to t.When s0.4, t = 4Work out s when t = 0.8​

Mathematics
1 answer:
Yuliya22 [10]3 years ago
3 0
S = 2

0.4x4 = 1.6
1.6/0.8 = 2
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4. Identify the graph of the solution set of 9 > 3 + 2x.
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Answer:

A

Step-by-step explanation:

9 > 3 + 2x

6 > 2x

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or

x < 3

and that means 3 is NOT included (otherwise there must be a smaller or equal sign). hence A is the right answer.

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Find the equation of this line​
Snowcat [4.5K]
Y=-1/2+1
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Answer:

Statements A and B are both true.

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agasfer [191]
Ok im not sure the answer yet so ima work on it at the same time while im explaining it. (The answer will probably be near the end)

We can use the elimination method to eliminate y out. To do that we multiply the first equation by 3.

6x+3y=-12

Now just subtract it from the other equation.

6x+3y=-12
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Usually after doing the elimination method you will have to solve for x but in this case its already solved for you. If you want to find y now you just take the first equation and fill x with -6 and solve for y.

2(-6)+y=-4
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Brainliest my answer if it helps you out?
4 0
4 years ago
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Find the volume of the wedge with vertices at points (0,0,0), (1,0,0), (0,1,0), (0,0,1) by integrating the area of cross-section
Angelina_Jolie [31]

Answer:

V = 1/6 cubic units

Step-by-step explanation:

Applying the concept of integrals for volume calculation:

V = \int\limits^b_a {S(x)} \, dx          (1)

V = volume of the solid bounded by x = a and x = b

S(x) = cross section area of the solid, perpendicular to the x axis

From the figure we have that S is the area of a triangle that has base Z and height Y

Area of the triangle = S(x)=\frac{y(x)*z(x)}{2}          (2)

Calculation of y(x) and z(x)

We apply the equation of the point-slope line (plane xy):

Slope = m = \frac{y_{2} - y_{1} }{x_{2} - x_{1}}          (3)

Equation of the line = y - y_{1} =m(x-x_{1} )          (4)

Replacing the points (1,0) and (0,1) in (3):

m=\frac{1-0}{0-1} =-1

Replacing the point (1,0) and m = -1 in (4):

y-0=(-1)(x-1)

y(x) = -x + 1 (Line A-B)          (5)

We apply the equation of the point-slope line (plane xz):

Slope = m = \frac{z_{2} - z_{1} }{x_{2} - x_{1}}          (6)

Equation of the line = z - z_{1} =m(x-x_{1} )          (7)

Replacing the points (1,0) and (0,1) in (6):

m=\frac{1-0}{0-1} =-1

Replacing the point (1,0) and m = -1 in (7):

z-0=(-1)(x-1)

z(x) = -x + 1 (Line A-C)        (8)

Replacing (5) and (8) in (2)

S(x) = \frac{(-x + 1) * (-x + 1)}{2} =\frac{(-x + 1)^{2} }{2}          (9)

Replacing (9) in (1) and knowing that a = 0 and b = 1:

V = \int\limits^1_0 {\frac{(-x + 1)^{2} }{2}} \, dx = \int\limits^1_0 {\frac{x^{2}-2x+1 }{2}} \, dx

V =\frac{1}{2} (\frac{x^{3} }{3} -2\frac{x^{2} }{2} +x)  evaluated from x=0 to x=1

V= \frac{1}{2} (\frac{1}{3} -1 +1) = \frac{1}{6}

3 0
4 years ago
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