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3 years ago

6

Sachant que le volume de la glace diminue de 11% quand elle fond, donner une valeur approchée, au millimètre près, de la hauteur

de l'eau contenue dans le verre de Samira juste avant d'avoir versé la boisson gazeuse.

1 answer:

igomit [66]3 years ago

8 0

Only if I knew this language

You might be interested in

Can someone please help me?

kvasek [131]

now, let's recall the rational root test, check your textbook on it.

so p = 18 and q = 1

so all possible roots will come from the factors of ±p/q

now, to make it a bit short, the factors are loosely, ±3, ±2, ±9, ±1, ±6.

recall that, a root will give us a remainder of 0.

let us use +3.

$\bf x^4-7x^3+13x^2+3x-18 \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{r|rrrrr} 3&1&-7&13&3&18\\ &&3&-12&3&18\\ \cline{1-6} &1&-4&1&6&0 \end{array}\qquad \implies (x-3)(x^3-4x^2+x+6)$

well, that one worked... now, using the rational root test, our p = 6, q = 1.

so the factors from ±p/q are ±3, ±2, ±1

let's use 3 again

$\bf \begin{array}{r|rrrrr} 3&1&4&1&6\\ &&3&-3&-6\\ \cline{1-5} &1&-1&-2&0 \end{array}\qquad \implies (x-3)(x-3)(x^2-x-2)$

and of course, we can factor x²-x-2 to (x-2)(x+1).

(x-3)(x-3)(x-2)(x+1).

7 0

3 years ago

3 regions are defined in the figure find the volume generated by rotating the given region about the specific line

anastassius [24]

The volume generated by rotating the given region $R_{3}$ about OC is $\frac{4}{g} \pi$

<h3>Washer method</h3>

Because the given region ( $R_{3}$ ) has a look like a washer, we will apply the washer method to find the volume generated by rotating the given region about the specific line.

solution

We first find the value of x and y

$y=2(x)^{\frac{1}{4} }$

$x=(\frac{y}{2} )^{4}$

$y=2x$

$x=\frac{y}{2}$

$\int\limits^a_b {\pi } \, (R_{o^{2} } - R_{i^{2} } ) dy$

$R_{o} = x = \frac{y}{2}$

$R_{i} = x= (\frac{y}{2}) ^{4}$

$a=0, b=2$

$v= \int\limits^2_o {\pi } \, [(\frac{y}{2})^{2} - ((\frac{y}{2}) ^{4} )^{2} ) dy$

$v= \pi \int\limits^2_o= [\frac{y^{2} }{4} - \frac{y^{8} }{2^{8} }} ] dy$

$v= \pi [\int\limits^2_o {\frac{y^{2} }{4} } \, dy - \int\limits^2_o {\frac{y}{2^{8} } ^{8} } \, dy ]$

$v=\pi [\frac{1}{4} \frac{y^{3} }{3} \int\limits^2_0 - \frac{1}{2^{8} } \frac{y^{g} }{g} \int\limits^2_o\\v= \pi [\frac{1}{12} (2^{3} -0)-\frac{1}{2^{8}*9 } (2^{g} -0)]\\v= \pi [\frac{2}{3} -\frac{2}{g} ]\\v= \frac{4}{g} \pi$

A similar question about finding the volume generated by a given region is answered here: brainly.com/question/3455095

6 0

2 years ago

When the hour hand moves from 3 pm to 6 pm, through how many degrees does it move?

taurus [48]

I think it might be 90 degrees

6 0

3 years ago

7. Find the distance between the points (13, 8) and (-12, 6)

Alik [6]

9514 1404 393

Answer:

√629 ≈ 25.08

Step-by-step explanation:

The distance formula is useful for this.

d = √((x2 -x1)² +(y2 -y1)²)

d = √((-12 -13)² +(6 -8)²) = √(625 +4) = √629 ≈ 25.08

The distance between the points is about 25.08 units.

3 0

2 years ago

The price of the product decreased by 10%and then decreased by 20% . calculate total percentage change between the final and ini

aev [14]

What is percentage change?

By quantifying the difference between two numbers and expressing it as an increase or decrease, the Percentage Change Calculator (percent change calculator) may determine the percentage change.

The initial price $V_{1}$ (100%) was decreased by 10%., then this reduced price is reduced more by 20% that becomes the final price $V_{2}$ which is 72.

Calculate percentage change from $V_{1}=100$ to $V_{2}=72$ .

= $& \frac{\left(V_{2}-V_{1}\right)}{\left|V_{1}\right|} \times 100 \\$

$=& \frac{(72-100)}{|100|} \times 100 \\$

$=& \frac{-28}{100} \times 100 \\$

$=&-0.28 \times 100 \\$

$=&-28 \% \text { change } \\$

$=& 28 \% \text { decrease }$

To know more about percentage change visit:

brainly.com/question/7568722

#SPJ4

7 0

2 years ago