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4 years ago

I NEED HELP PLEASE!! THANKS :)

Mathematics

1 answer:

Artyom0805 [142]4 years ago

3 0

Answer:

Their atomic theory of the elements

Step-by-step explanation:

The ancient Greeks believed that all matter consisted of four elements — earth, air, fire, and water

Everything on Earth consisted of these elements in different combinations.

Robert Boyle is best known for his famous Law — p₁V₁ = p₂V₂.

However, he was also an excellent chemist who did many experiments.

He found it was impossible to combine the four Greek elements into any substance.

Boyle found he could decompose many substances into simpler substances, but none of them was a Greek element.

He therefore rejected the Greek theory.

Almost 150 years before Dalton's atomic theory, Boyle proposed a new definition of an element:

An element is any substance that cannot be broken into a simpler substance.

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E^(xy) + y^2 = sin(x + y) <br><br> Q: Express the variable y in terms of x only.

choli [55]

Y cannot be expressed explicitly.

8 0

4 years ago

please help me, Prove a quadrilateral with vertices G(1,-1), H(5,1), I(4,3) and J(0,1) is a rectangle using the parallelogram me

mestny [16]

Answer:

Step-by-step explanation:

We are given the coordinates of a quadrilateral that is G(1,-1), H(5,1), I(4,3) and J(0,1).

Now, before proving that this quadrilateral is a rectangle, we will prove that it is a parallelogram. For this, we will prove that the mid points of the diagonals of the quadrilateral are equal, thus

Join JH and GI such that they form the diagonals of the quadrilateral.Now,

JH= $\sqrt{(5-0)^{2}+(1-1)^{2}}=5$ and

GI= $\sqrt{(4-1)^{2}+(3+1)^{2}}=5$

Now, mid point of JH= $(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

= $(\frac{5+0}{2},\frac{1+1}{2})=(\frac{5}{2},1)$

Mid point of GI= $(\frac{5}{2},1)$

Since, mid point point of JH and GI are equal, thus GHIJ is a parallelogram.

Now, to prove that it is a rectangle, it is sufficient to prove that it has a right angle by using the Pythagoras theorem.

Thus, From ΔGIJ, we have

$(GI)^{2}=(IJ)^{2}+(JG)^{2}$ (1)

Now, JI= $\sqrt{(4-0)^{2}+(3-1)^{2}}=\sqrt{20}$ and GJ= $\sqrt{(0-1)^{2}+(1+1)^{2}}=\sqrt{5}$

Substituting these values in (1), we get

$5^{2}=(\sqrt{20})^{2}+(\sqrt{5})^{2} }$

$25=20+5$

$25=25$

Thus, GIJ is a right angles triangle.

Hence, GHIJ is a rectangle.

Also, The diagonals GI= $\sqrt{(4-1)^{2}+(3+1)^{2}}=5$ and HJ= $\sqrt{(0-5)^2+(1-1)^2}=5$ are equal, thus, GHIJ is a rectangle.

6 0

3 years ago

I need help ASAP

3241004551 [841]

1/7-9/7n+6/7

1-9/7n

that is the extended form and is simplified

Hope it help! :)

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4 years ago

CAn someone please help me with this? I been getting bots all day

allochka39001 [22]

The answer is C i’m pretty sure because you can divide y by 15 and get x

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3 years ago

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bija089 [108]

Answer:

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3 years ago