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Inessa05 [86]
3 years ago
14

Bonjour tout le monde j'ai des exercices de maths que je n'arrive pas pouvez vous m'aidez svp c'est urgent c'est pour demain mer

ci d'avance (c'est les ex 74 et 76)

Mathematics
1 answer:
dedylja [7]3 years ago
4 0
Je regrette je ne comprends pas très bien
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Use multiplying by 1 to find an expression equivalent to 3/14 with a denominator of 14 y.
Soloha48 [4]
As a matter of simplification, same/same = 1.

now, "same" could be anything whatsoever, you name it, a whole polynomial, a whole anything whatsoever.

\bf \cfrac{\sqrt{2}}{\sqrt{2}}=\cfrac{\sqrt{3}}{\sqrt{3}}=\cfrac{100000000}{100000000}=\cfrac{eggs}{eggs}=\cfrac{miles}{miles}=\cfrac{\frac{\quad \sqrt{3}}{\sqrt[5]{17}}\quad }{\frac{\sqrt{3}}{\sqrt[5]{17}}}=\cfrac{whatever}{whatever}=1

now, that said, let us use 1 in this case then ... hmmmm we simply need a "y" in the denominator, so we simply multiply the denominator AND numerator by "y", y/y  = 1 btw.

\bf \cfrac{3}{14}\cdot \cfrac{y}{y}\implies \cfrac{3y}{14y}
6 0
4 years ago
The common minimum multiple between <br><br> A) 2,3 and 6<br> B) 8,9 and 12<br> C) -4,2 and 8
Elan Coil [88]
A 2,3 and 6. There you go. :)
8 0
3 years ago
Which of the following statements justifies why the triangle shown below is
kirill [66]

Answer: B

Step-by-step explanation:

B is basically showing the Pythagorean theorem does not work on the triangle's side lengths.

The pythagorean theorem is supposed to work for all right triangles

6 0
3 years ago
What does $25/1day represent?
Veseljchak [2.6K]

Answer:

it represents how much money someone gets in a day

Step-by-step explanation:

4 0
3 years ago
Determine whether the three points are collinear. (3, -10), (-2, -7), (0, -5)
Mumz [18]

The points are not collinear.

Solution:

Let A, B and C be (3, -10), (-2, -7) and (0, -5).

<em>If slopes of any two points are same, then the points are collinear.</em>

Slope formula:

$m=\frac{y_2-y1}{x_2-x_1}

<u>Slope of AB:</u>

$m_1=\frac{-7-(-10)}{-2-3}

$m_1=\frac{-7+10}{-5}

$m_1=-\frac{3}{5}

<u>Slope of BC:</u>

$m_2=\frac{-5-(-7)}{0-(-2)}

$m_2=\frac{-5+7}{2}

$m_2=\frac{2}{2}

m_2=1

<u>Slope of CA:</u>

$m_3=\frac{-10-(-5)}{3-0}

$m_3=\frac{-10+5}{3}

$m_3=-\frac{5}{3}

m_1\neq m_2 \neq m_3

Slope of AB ≠ Slope of BC ≠ Slope of AC

Therefore the points are not collinear.

6 0
3 years ago
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