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Lostsunrise [7]
3 years ago
12

ПОЖАЛУЙСТА,РЕШИТЕ!!!Решите уравнение:

Mathematics
1 answer:
Anuta_ua [19.1K]3 years ago
6 0

Answer:

Step-by-step explanation:

<em>1).</em> -\frac{5}{12} x = 0 ⇒ <em>x = 0</em>

<em>2).</em> 5,4 ( x + 6,3 ) = 0 ⇒ x + 6,3 = 0 ⇒ <em>x = - 6,3</em>

<em>3).</em> ( x - 3 )( x + 4 ) = 0 ⇒ x - 3 = 0 или x + 4 = 0 ⇒ <em>x = 3</em> <u><em>или</em></u> <em>x = - 4</em>

<em>4).</em> 23,5 | x | = 0 ⇒ | x | = 0 ⇒ <em>x = 0</em>

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If the price of the early bird session is 2x - 3 and the price of the mid-morning session is 5x + 6, find an expression that rep
Travka [436]
Apply distributive property
(5x+6)-(2x-3)
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8 0
3 years ago
I need help in partial fraction!! With simple explaination would be nice!
WITCHER [35]
\dfrac{x^4-7x^2+17x-10}{x(x^2-3)}

The degree of the numerator (4) is larger than the degree of the denominator (3), so first you need to divide. (Added screenshot of long division procedure.)

\dfrac{x^4-7x^2+17x-10}{x(x^2-3)}=x-\dfrac{4x^2-17x+10}{x(x^2-3)}

Now the second term can be decomposed into partial fractions.

\dfrac{4x^2-17x+10}{x(x^2-3)}=\dfrac{r_1}x+\dfrac{r_2x+r_3}{x^2-3}
\dfrac{4x^2-17x+10}{x(x^2-3)}=\dfrac{r_1(x^2-3)+x(r_2x+r_3)}{x(x^2-3)}
4x^2-17x+10=r_1(x^2-3)+x(r_2x+r_3)
4x^2-17x+10=(r_1+r_2)x^2+r_3x-3r_1
\implies\begin{cases}r_1+r_2=4\\r_3=-17\\-3r_1=10\end{cases}\implies r_1=-\dfrac{10}3,r_2=\dfrac{22}3,r_3=-17=-\dfrac{51}3
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So

\dfrac{x^4-7x^2+17x-10}{x(x^2-3)}=x+\dfrac{10}{3x}-\dfrac{22x-51}{x^2-3}

3 0
3 years ago
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7 0
3 years ago
A student performed the following steps to find the solution to the question x^2 - 2x - 15 = 0. Where did the student go wrong?
nataly862011 [7]

Answer:

D  In step 3

Step-by-step explanation:

Given polynomial:  x² - 2x - 15 = 0

Step 1:  factor the polynomial ⇒ (x + 3)(x - 5) = 0

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7 0
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