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goldenfox [79]
3 years ago
10

I need help? Can anyone help me ?

Mathematics
1 answer:
Rina8888 [55]3 years ago
7 0
Given a coordinate point (x, y), the first value of the point represents the value on the x-axis while the second value represent the value on the y-axis.

1.) To express the values (-4, -1), (-1, 2), (1, -4), (2, -3), (4, 3) as a table, we have:

 x          y
-4         -1
-1          2
 1         -4
 2         -3
 4          3

The values (-4, -1), (-1, 2), (1, -4), (2, -3), (4, 3) expressed as a graph have been attached as graph_1


To express the values (-4, -1), (-1, 2), (1, -4), (2, -3), (4, 3) as a mapping, we have two circles with one labelled x and the other one labelled y.
Inside the circle labelled x are the numbers -4, -1, 1, 2, 4 written vertically and inside the circle labelled y are the numbers -4, -3, -1, 2, 3 written vertically.
There are lines joining from the circle labelled x to the circle labelled y with line joining -4 in circle x to -1 in circle y, -1 in circle x to 2 in circle y, 1 in circle x to -4 in circle y, 2 in circle x to -3 in circle y, 4 in circle x to  3 in circle y.

The domain of the relation is the set of the x-values of the relation, i.e. domain is {-4, -1, 1, 2, 4}.
The range of the relation is the set of the y-values of the relation, i.e. range is {-4, -3, -1, 2, 3}

2.) To express the values (-2, 1), (-1, 0), (1, 2), (2, -4), (4, 3) as a table, we have:

 x          y
-2         1
-1         0
 1          2
 2         -4
 4          3

The values (-2, 1), (-1, 0), (1, 2), (2, -4), (4, 3) expressed as a graph have been attached as graph_2


To express the values (-2, 1), (-1, 0), (1, 2), (2, -4), (4, 3) as a mapping, we have two circles with one labelled x and the other one labelled y.
Inside the circle labelled x are the numbers -2, -1, 1, 2, 4 written vertically and inside the circle labelled y are the numbers -4, 0, 1, 2, 3 written vertically.
There are lines joining from the circle labelled x to the circle labelled y with a line joining -2 in circle x to 1 in circle y, -1 in circle x to 0 in circle y, 1 in circle x to 2 in circle y, 2 in circle x to -4 in circle y, 4 in circle x to  3 in circle y.

The domain of the relation is the set of the x-values of the relation, i.e. domain is {-2, -1, 1, 2, 4}.
The range of the relation is the set of the y-values of the relation, i.e. range is {-4, 0, 1, 2, 3}
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The solution for 7(5x+9)=12-(x+9) and simplified​
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Answer:

x= -5/3

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

7(5x+9)=12−(x+9)

7(5x+9)=12+−1(x+9)(Distribute the Negative Sign)

7(5x+9)=12+−1x+(−1)(9)

7(5x+9)=12+−x+−9

(7)(5x)+(7)(9)=12+−x+−9(Distribute)

35x+63=12+−x+−9

35x+63=(−x)+(12+−9)(Combine Like Terms)

35x+63=−x+3

35x+63=−x+3

Step 2: Add x to both sides.

35x+63+x=−x+3+x

36x+63=3

Step 3: Subtract 63 from both sides.

36x+63−63=3−63

36x=−60

Step 4: Divide both sides by 36.

36x/36 and -60/36

x= -5/3

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What is the cross-sectional area of a cement pipe 2 in. thick with an inner diameter of 5 ft? Use pie 3.14.
Paul [167]

Answer:

1.60 ft^{2}

Step-by-step explanation:

Assuming that this is a cylindrical pipe, the area can be determined by applying the formula for calculating the area of a circle.

i.e Area = \pir^{2}

So that,

for the inner part of the pipe,

r = \frac{diameter}{2}

 = \frac{5}{2}

 = 2.5 feet

Area of the inner part of the pipe = \pir^{2}

                                 = 3.14 x (2.5)^{2}

                                 = 19.625

Are of the inner part of the pipe is 19.63 ft^{2}.

Total diameter of the pipe = 5 + 0.2

                                           = 5.2 feet

r = \frac{5.2}{2}

 = 2.6 feet

Area of the pipe = \pir^{2}

                            = 3.14 x (2.6)^{2}

                            = 21.2264

Area of the pipe is 21.23 ft^{2}.

Thus the cross sectional area = Area of the pipe - Area of the inner part of the pipe

                                           = 21.2264 - 19.625

                                           = 1.6014

The cross sectional area of the cement pipe is 1.60 ft^{2}.

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3 years ago
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