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pishuonlain [190]
2 years ago
5

Which lists contain only rational numbers? Select all that apply.

Mathematics
2 answers:
taurus [48]2 years ago
7 0
0.865, 0.4444, -6.37, -11.11
Rational are decimals / fractions
solong [7]2 years ago
7 0
O will be your correct answer.
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Evaluate the determination for the following matrix [144] [522] [155]
Anettt [7]
Matrix :

1    4    4

5    2    2

1    5    5

=> Determinant =

1 * (2*5 - 2*5) - 4 * (5*5 - 1*2) + 4 * ( 5*5 - 2*1)

= 1*(10 - 10) - 4*(25 - 2) + 4*(25 -2) = 1*0 + 4* 23 + 4*(25 - 2) = 0 - 4*23 + 4*23 = 0

Answer: 0
8 0
3 years ago
Find the area of the shape
Tema [17]

Answer:

  • A. 104 cm²

Step-by-step explanation:

<u>The area is:</u>

  • 22*4 + 2*(1/2)*4*4 = 88 + 16 = 104 cm²

Correct choice is A

5 0
2 years ago
Which expression is equivalent to the expression below? 4 x 2/3 A.
Tema [17]
Bro just go on Symbolab and use the calculator
8 0
2 years ago
Which equation does the graph below represent?<br> y=1/3+x<br> y=1/3x<br> y=3+x<br> y=3x
barxatty [35]
Lets find 2 points, one obvious one is (0,0), another point is (12,4)

Slope = \frac{4-0}{12-0} =\frac{4}{12} = \frac{4\times 1}{4\times 3} = \frac{1}{3}

y = mx + b

m is the slope

we got m = 1/3

We can eliminate answer choices C and D.

y = 1/3x + b

we need to solve for b

Plug in the point (0,0) and solve for b

0 = 1/3(0) + b

0 = 0 + b

b = 0

The equation of the line is y = 1/3x + 0, or y = 1/3x

Your answer is B.
3 0
3 years ago
Field book of an land is given in the figure. It is divided into 4 plots . Plot I is a right triangle , plot II is an equilatera
Kazeer [188]

Answer:

Total area = 237.09 cm²

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)

From the figure attached,

Area of the right triangle I = \frac{1}{2}(\text{Base})\times (\text{Height})

Area of ΔADC = \frac{1}{2}(\text{CD})(\text{AD})

                        = \frac{1}{2}(\sqrt{(AC)^2-(AD)^2})(\text{AD})

                        = \frac{1}{2}(\sqrt{(13)^2-(19-7)^2} )(19-7)

                        = \frac{1}{2}(\sqrt{169-144})(12)

                        = \frac{1}{2}(5)(12)

                        = 30 cm²

Area of equilateral triangle II = \frac{\sqrt{3} }{4}(\text{Side})^2

Area of equilateral triangle II = \frac{\sqrt{3}}{4}(13)^2

                                                = \frac{(1.73)(169)}{4}

                                                = 73.0925

                                                ≈ 73.09 cm²

Area of rectangle III = Length × width

                                 = CF × CD

                                 = 7 × 5

                                 = 35 cm²

Area of trapezium EFGH = \frac{1}{2}(\text{EF}+\text{GH})(\text{FJ})

Since, GH = GJ + JK + KH

17 = \sqrt{9^{2}-x^{2}}+5+\sqrt{(15)^2-x^{2}}

12 = \sqrt{81-x^2}+\sqrt{225-x^2}

144 = (81 - x²) + (225 - x²) + 2\sqrt{(81-x^2)(225-x^2)}

144 - 306 = -2x² + 2\sqrt{(81-x^2)(225-x^2)}

-81 = -x² + \sqrt{(81-x^2)(225-x^2)}

(x² - 81)² = (81 - x²)(225 - x²)

x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴

144x² - 11664 = 0

x² = 81

x = 9 cm

Now area of plot IV = \frac{1}{2}(5+17)(9)

                                = 99 cm²

Total Area of the land = 30 + 73.09 + 35 + 99

                                    = 237.09 cm²

7 0
3 years ago
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