Write an equation for the line that passes through (2,10),(0,4)

1 answer:

8 0

I'm going to go right ahead and assume that we're talking about a linear equation.

The linear equation passing through the points given is y=3x+4, since it satisfies both of the points given.

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Whats is the answer -2/3 (-1 1/6)

grin007 [14]

Answer:

$\frac79$

Step-by-step explanation:

$\frac{-2}{3} \cdot-1\frac{1}{6}$

Let's convert the mixed number to an important fraction.

-1 1/6 = -7/6

Now multiply:

$\frac{-2}{3} \cdot \frac{-7}{6} = \\\\\\\frac{-2\cdot-7}{3\cdot6} = \\\\\\\frac{14}{18} = \frac79$

-Chetan K

7 0

2 years ago

Turn this into a equation the sum of twice r and three timea s is identical to thirteen

Goryan [66]

In equation form it would be the following:

2r + 3s = 13.

I believe.

5 0

3 years ago

The sum of the diagonals of a rhombus is 5√2.

Alex

Greetings!
Let ABCD be a rhombus
AC + BD = 5√2 cm

Area of ABCD = Ar△ABD + Ar △BCD
$= \frac{1}{2} \times BD \times AO + \frac{1}{2} \times BD \times OC$
$= \frac{1}{2} BD (AO + OC)$
$\frac{1}{2} BD \times AC$

$So, \frac{ BD \times AC}{2} = 4 \: cm {}^{2}$
$BD \times AC = 8$
$Now, AC + \: BD = 5 \sqrt{2}$

Squaring both sides, we get
$AC {}^{2} + BD {}^{2} + 2 AC.BD =50$
$AC {}^{2} + BD {}^{2} + 2 \times 8 = 50$
$AC {}^{2} + BD {}^{2} = 50 - 16$
$AC {}^{2} + BD {}^{2} = 34$

In △AOB, we have
$OA {}^{2} + OB {}^{2} =AB {}^{2}$
$( \frac{ AC }{2} ) {}^{2} + ( \frac{ BD}{2} ) {}^{2} = AB {}^{2}$
$\frac{ AC {}^{2} } {4} + \frac{ BD {}^{2} }{4} = AB {}^{2}$
$AC {}^{2} + BD {}^{2} = 4 AB {}^{2}$
$34 = 4 AB {}^{2}$

Square rooting both sides
$\sqrt{34} = 2 AB$
$Perimeter = 4 \: AB \\ = 2 \times 2 \: AB \\ = 2 \: \times \sqrt{34 } \\ = 2 \sqrt{34 \: } units$ .

Hope it helps!