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DochEvi [55]
2 years ago
5

Can someone help me with this math equation? Can you explain how to work it out also?

Mathematics
2 answers:
leonid [27]2 years ago
5 0

Answer: 50

<h2>Explanation:</h2><h3>What we know:</h3>
  • The numerator is 6
  • The denominator is equal to (3*5)^(-2)
  • In fraction-style equations, the numerators are divided by denominators
  • Negative exponents are equal to (base)^|power|, and then reversing the products place in the fraction

<h3>How to solve:</h3>

    By utilizing our knowledge of the PEMDAS solving method, negative powers, and fraction-style equations, this problem can be solved with a simple step-by-step procedure

<h2>Process:</h2><h3>Exponent</h3>

Set up equation                                                            6 / [3 * 5^(-2)]

Rewrite power                                                              6 / [3 * (5 * 5)]

Simplify power                                                              6 / [3 * (25)]

Invert power                                                                  6(25) / 3

<h3>Simplifying</h3>

Set up equation                                                            6(25) / 3

Simplify                                                                          150 / 3

Simplify (using common denominator)                        50 / 1

Rewrite in simplest form                                               =50

Answer: 50

lawyer [7]2 years ago
3 0
50 I did this one yesterday
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D = sqrt[ (5 - 9)^2 + (-8 - 0)^2 ]

D = sqrt[ (-4)^2 + (-8)^2 ]

D = sqrt[ 16 + 64]

D = sqrt(80)

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Step-by-step explanation:

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Co-ordinate Geometry
saul85 [17]

Answer:

The value of a  = 14

Step-by-step explanation:

Given

(x₁, y₁) = (2, 4)

(x₂, y₂) = (6, a)

(x₃, y₃) = (-1, 1)

A = 9 sq.units

Area of a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) is:

A=\frac{\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|}{2}

substituting the values (x₁, y₁) = (2, 4), (x₂, y₂) = (6, a), (x₃, y₃) = (-1, 1), A = 9 in th formula

A=\frac{\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|}{2}

9=\frac{\left|2\left(a-1\right)+6\left(1-4\right)+-1\left(4-a\right)\right|}{2}

Multiply both sides by 2

\frac{2\left|2\left(a-1\right)+6\left(1-4\right)-1\left(4-a\right)\right|}{2}=9\cdot \:2

simplify

\left|2\left(a-1\right)+6\left(1-4\right)-1\left(4-a\right)\right|=18

As the area is always positive.

so

2\left(a-1\right)+6\left(1-4\right)-1\cdot \left(4-a\right)=18

2\left(a-1\right)-18-\left(4-a\right)=18

Add 18 to both sides

2\left(a-1\right)-18-\left(4-a\right)+18=18+18

simplify

2\left(a-1\right)-\left(4-a\right)=36

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Divide both sides by 3

\frac{3a}{3}=\frac{42}{3}

a=14

Thus, the value of a  = 14

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alexandr402 [8]

At the point -4


Since 4-8 = -4


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