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aev [14]
2 years ago
15

A(3,5)and C(7,9) are the opposite vertices of a rhombus ABCD.find the equation of diagonal BD.

Mathematics
1 answer:
anzhelika [568]2 years ago
6 0

Answer:

y = 12 - x  

Step-by-step explanation:

The diagonals of a rhombus bisect each other at right angles,

Thus, segment BD is a line perpendicular to AC and passing through the midpoint of AC.

1. Find the midpoint of AC

The midpoint of two points is half-way between their x- and y-values.

For the x-coordinate,

(x₂ + x₁)/2 = (7 + 3)/2 = 10/2 = 5

For the y-coordinate,

(y₂ + y₁)/2 = (9 + 5)/2 = 14/2 = 7

The coordinates of the midpoint are (5,7).

2. Calculate the equation of the diagonal BD

(a) Slope of AC

m₁ = Δy/Δx = (y₂ -y₁)/(x₂ - x₁) = (9 - 5)/(7 - 3) = 4/4 = 1

(b) Slope of BD

The slope m₂ of the perpendicular line BD must be the negative reciprocal of the slope of AC.

m₁ = -1/m₁ = -1

(c) Calculate the y-intercept of BD

The general equation for a straight line is

y = mx + b

Insert point (5,7).

7 = -1×5 + b

7 = -5 + b

b = 12

The y-intercept is (0,12).

(d) Write the equation for the line

y = -x + 12 or y = 12 - x

Points B and D can be any two points on the line that are equidistant from Point O, as shown in the Figure below.

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Answer: Yes ,because it isn't the same the number since the number of digits aren't the same so you make it the same number of digits by putting 0 in the back so 1.7 turns to 1.700 to make it the same digits of 1.700.

Step-by-step explanation:

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3 years ago
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lara31 [8.8K]

Answer:

'The closest distance between Earth and Jupiter is greater than the closest distance between Earth and Mercury by approximately 7.66 times'.

Step-by-step explanation:

We are given that,

Distance between Earth and Jupiter = 5.9 × 10⁸ kilometers

Distance between Earth and Mercury = 7.7 × 10⁷ = 0.77 × 10⁸ kilometers

So, we see that,

5.9 × 10⁸ > 0.77 × 10⁸

Also, Ratio = \frac{5.9\times 10^8}{0.77\times 10^8} =7.66

That is,

'The closest distance between Earth and Jupiter is greater than the closest distance between Earth and Mercury by approximately 7.66 times'.

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3 years ago
The area of a rectangle is 15 square inches. What wil be the area, in square inches, of the rectangle after it is dilated by a
Alex73 [517]

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60 inches²

Step-by-step explanation:

Unit test

5 0
3 years ago
3.) SOLVE FOR X.
gayaneshka [121]

The value of x in the secants intersection is 1 units

The value of NM in the tangent and secant intersection is 51 units

<h3>How to find length when secant and tangent intersect?</h3>

The first question, two secant intersect outside the circle.

Therefore,

(6x + 8x)8x = (9 + 7)7

14x(8x) = 16(7)

112x² = 112

x² = 112 / 112

x = √1

x = 1

The second question, tangent and secant intersect,

Therefore,

(x + 3)² = (x - 3)(16 + x - 3)

(x + 3)² = (x - 3)(x + 13)

(x + 3)(x + 3) = (x - 3)(x + 13)

x² + 3x + 3x + 9 = x² + 13x - 3x - 39

x² + 9x + 9 = x² + 10x - 39

x² - x² + 9x - 10x = -39 - 9

-x = - 48

x = 48

NM = 48 + 3 = 51 units

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6 0
1 year ago
COMPUTE<br><br> 3 ( 2 1/2 - 1 ) + 3/10
Juli2301 [7.4K]

Answer:

<h3>\boxed{ \frac{24}{5} }</h3>

Step-by-step explanation:

\mathsf{3(2 \frac{1}{2}  - 1) +  \frac{3}{10} }

Convert mixed number to improper fraction

\mathrm{3( \frac{5}{2}  - 1) +  \frac{3}{10} }

Calculate the difference

⇒\mathrm{3( \frac{5 \times 1}{2 \times 1} -  \frac{1 \times 2}{1 \times 2}  }) +  \frac{3}{10}

⇒\mathrm{ 3 \times( \frac{5}{2}  -  \frac{2}{2}) } +  \frac{3}{10}

⇒\mathrm{3 \times ( \frac{5 - 2}{2} ) +  \frac{3}{10} }

⇒\mathrm{3 \times  \frac{3}{2}  +  \frac{3}{10} }

Calculate the product

⇒\mathrm{ \frac{3 \times 3}{1 \times 2}  +  \frac{3}{10} }

⇒\mathrm{ \frac{9}{2}  +  \frac{3}{10}}

Add the fractions

⇒\mathsf{ \frac{9  \times 5}{2 \times 5}  +  \frac{3 \times 1}{10 \times 1} }

⇒\mathrm{ \frac{45}{10}  +  \frac{3}{10} }

⇒\mathrm{ \frac{45 + 3}{10 } }

⇒\mathrm{ \frac{48}{10} }

Reduce the numerator and denominator by 2

⇒\mathrm{ \frac{24}{5} }

Further more explanation:

<u>Addition </u><u>and </u><u>Subtraction</u><u> </u><u>of </u><u>like </u><u>fractions</u>

While performing the addition and subtraction of like fractions, you just have to add or subtract the numerator respectively in which the denominator is retained same.

For example :

Add : \mathsf{ \frac{1}{5}  +  \frac{3}{5}  =  \frac{1 + 3}{5} } =  \frac{4}{5}

Subtract : \mathsf{ \frac{5}{7}  -  \frac{4}{7}  =  \frac{5 - 4}{7}  =  \frac{3}{7} }

So, sum of like fractions : \mathsf{ =  \frac{sum \: of \: their \: number}{common \: denominator} }

Difference of like fractions : \mathsf{ \frac{difference \: of \: their \: numerator}{common \: denominator} }

<u>Addition </u><u>and </u><u>subtraction</u><u> </u><u>of </u><u>unlike </u><u>fractions</u>

While performing the addition and subtraction of unlike fractions, you have to express the given fractions into equivalent fractions of common denominator and add or subtract as we do with like fractions. Thus, obtained fractions should be reduced into lowest terms if there are any common on numerator and denominator.

For example:

\mathsf{add \:  \frac{1}{2}  \: and \:  \frac{1}{3} }

L.C.M of 2 and 3 = 6

So, ⇒\mathsf{ \frac{1 \times 3}{2 \times 3}  +  \frac{1 \times 2}{3 \times 2} }

⇒\mathsf{ \frac{3}{6}  +  \frac{2}{6} }

⇒\frac{5}{6}

Multiplication of fractions

To multiply one fraction by another, multiply the numerators for the numerator and multiply the denominators for its denominator and reduce the fraction obtained after multiplication into lowest term.

When any number or fraction is divided by a fraction, we multiply the dividend by reciprocal of the divisor. Let's consider a multiplication of a whole number by a fraction:

\mathsf{4 \times  \frac{3}{2}  =  \frac{4 \times 3}{2}  =  \frac{12}{2}  = 6}

Multiplication for \mathsf{ \frac{6}{5}  \: and \:  \frac{25}{3} } is done by the similar process

\mathsf{ =  \frac{6}{5}  \times  \frac{25}{3}  = 2 \times 5 \times 10}

Hope I helped!

Best regards!

5 0
3 years ago
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