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Andrej [43]
2 years ago
12

How do you write 0.050 in scientific notation?

Mathematics
2 answers:
Ulleksa [173]2 years ago
5 0

Answer:

5.0 x 10^-2

Step-by-step explanation:

you need to move the decimal place to the right in order to be 1 or over.

moving the decimal to the right will cause the exponent to be negative. opposite direction is positive.

11111nata11111 [884]2 years ago
5 0
5.0 x 10^-2
please mark brainliest
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Find the value of the variable y, where the sum of the fractions 6/(y+1) and y/(y-2) is equal to their product.
Blizzard [7]

Answer:

The answer is

y = 3

y =  - 4

Step-by-step explanation:

We must find a solution where

\frac{6}{y + 1}  +  \frac{y}{y - 2}  =  \frac{6}{y + 1}  \times  \frac{y}{y - 2}

Consider the Left Side:

First, to add fraction multiply each fraction on the left by it corresponding denomiator and we should get

\frac{6}{y + 1}  \times  \frac{y - 2}{y - 2}  +  \frac{y}{y - 2}  \times  \frac{y + 1}{y + 1}

Which equals

\frac{6y - 12}{(y -2) (y + 1)}  +  \frac{ {y}^{2} + y }{(y - 2)(y + 1)}

Add the fractions

\frac{y {}^{2} + 7y - 12 }{(y - 2)(y + 1)}  =  \frac{6}{y + 1}  \times  \frac{y}{y - 2}

Simplify the right side by multiplying the fraction

\frac{6y}{(y  + 1)(y + 2)}

Set both fractions equal to each other

\frac{6y}{(y + 1)(y - 2)}  =  \frac{ {y}^{2} + 7y - 12}{(y + 1)(y - 2)}

Since the denomiator are equal, we must set the numerator equal to each other

6y =  {y}^{2}  + 7y - 12

=  {y}^{2}  + y - 12

(y  + 4)(y - 3)

y =  - 4

y = 3

6 0
2 years ago
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You measured the width of your front door to be 3 feet and 2 inches. the chair you are trying to get through the door measured t
Citrus2011 [14]
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5 0
3 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
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