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3 years ago

How to find the side of a rhombus with given diagonals?

1 answer:

7 0

The diagonals of a rhombus are perpendicular bisectors of each other. You can use the Pythagorean theorem. If the diagonals are length "a" and "b", the side length of the rhombus (s) is
s = (1/2)√(a²+b²)

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3x 2y = 4 2x - y = 5 this system of equations has no solution. has one solution. is coincident.

juin [17]

Answer:

no solution

Step-by-step explanation: because if it has no variables that are the same it has no solution.

7 0

2 years ago

100 POINTS! PLEASE HELP!

Semmy [17]

Apply angle bisector theorem

$\\ \sf\longmapsto \dfrac{9}{15}=\dfrac{3}{5}=\dfrac{2x-1}{3x}$

$\\ \sf\longmapsto 9x=5(2x-1)$

$\\ \sf\longmapsto 9x=10x-5$

$\\ \sf\longmapsto 10x-9x=5$

$\\ \sf\longmapsto x=5$

$\\ \sf\longmapsto \dfrac{4}{3}=\dfrac{x}{2.25}$

$\\ \sf\longmapsto 4(2.25)=3x$

$\\ \sf\longmapsto 3x=9$

$\\ \sf\longmapsto x=3$

Hope it helps

4 0

3 years ago

Read 2 more answers

1. Two object accumulated a charge of

In-s [12.5K]

Answer:

The magnitude of the electric force between these two objects

will be: 181.274 N.

i.e. F $=181.274$ N

Step-by-step explanation:

Two object accumulated a charge of 4.5 μC and another a charge of 2.8 μC.

q₁ = 4.5 μC = 4.5 × 10⁻⁶ C

q₂ = 2.8 μC = 2.8 × 10⁻⁶ C

separated distance = d = 2.5 cm

Calculating the magnitude of the force between two charged objects using the formula:

$F = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2}$

$=\:\frac{4.5\times \:\:10^{-6}\:\times \:\:2.8\:\times \:\:10^{-6}}{4\:\times \:\left(3.14\right)\:\times \:\left(8.85\times \:10^{-12}\right)\times \left(2.5\times \:\:10^{-2}\right)^2}$

$=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:4\times \:8.85\pi \left(10^{-2}\times \:2.5\right)^2}$ ∵ $4.5\times \:10^{-6}\times \:2.8\times \:10^{-6}=10^{-12}\times \:12.6$

$=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:35.4\pi \left(10^{-2}\times \:2.5\right)^2}$ ∵ $\mathrm{Multiply\:the\:numbers:}\:4\times \:8.85=35.4$