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stira [4]
3 years ago
11

Kyle crawled 49.8 feet in 6 minutes, how far did he crawl every minute…

Mathematics
1 answer:
Blizzard [7]3 years ago
8 0

Answer:

assuming that kyle crawled at the same pace for six minute then it would be 8.3 feet each minute

Step-by-step explanation:

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What is the value of u and v?
Novay_Z [31]

Answer and Explanation:

Since this is an isosceles right triangle, you can use the ratio of side lengths specific to this type of triangle where both legs are the hypotenuse over root 2:

u = v = 164√2 / √2 = 164

so:

u = 164

v = 164

3 0
2 years ago
Read 2 more answers
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
The distribution of the time it takes for different people to solve a certain crossword puzzle is strongly skewed to the right,
dezoksy [38]
<span>Answer: Skewed to right with mean 0 and S.D. 1 Explanation: Let's assume the z-scores as a linear transformation. then z=(x-mean)/S.D. = (x-30)/15 z(mean) = (30-30)/15= 0 z(sigma) = 15/15 = 1 Note: Shape does not change.</span>
5 0
3 years ago
A recipe calls for 5 8 pound of cheese per batch. Assuming you have enough of the other ingredients, how many batches of the rec
aliya0001 [1]
I assume you mean 5/8 of a pound not 58 pounds. In that case convert your total number of pounds into eights to make it easier to calculate. 5 pounds is equal to 40/8 pounds of cheese. Now divide this by 5/8. That equals 8. SO you can make 8 batches.
3 0
3 years ago
Permutations: 5!·4! is equivalent to what (5!)(4!)^2, (5)(4!)^2 or 20! ?
solniwko [45]

Answer:

5!*4! is equivalent to 5*(4!)^2

Step-by-step explanation:

5!*4! = 2880

5*(4!)^2 = 2880

4 0
3 years ago
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