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3 years ago

You plan to buy

Mathematics

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uysha [10]3 years ago

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Answer:

2 friends...............

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When the sum of 5 and three times a positive number is subtracted from the square of the number, 0 results. Find the number.

Juliette [100K]

$x-the\ number\\\\x^2-(5+3x)=0\\\\x^2-5-3x=0\\\\x^2-3x-5=0$

$Use\ the\ quadratic\ formula:\\a^2x+bx+c=0\\\\\Delta=b^2-4ac\\\\if\ \Delta > 0\ then\ x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\if\ \Delta=0\ then\ x_0=\dfrac{-b}{2a}\\\\if\ \Delta < 0\ then\ no\ solution$

$x^2-3x-5=0\\\\a=1;\ b=-3;\ c=-5\\\\\Delta=(-3)^2-4\cdot1\cdot(-5)+3+20=29 \ \textgreater \ 0\\\\\sqrt\Delta=\sqrt{29}\\\\x_1=\dfrac{3-\sqrt{29}}{2\cdot1}=\dfrac{3-\sqrt{29}}{2} \ \textless \ 0\\\\x_2=\dfrac{3+\sqrt{29}}{2\cdot1}=\dfrac{3+\sqrt{29}}{2} \ \textgreater \ 0$

$Answer:\ \boxed{x=\dfrac{3+\sqrt{29}}{2}}$

6 0

3 years ago

Can anyone help me with this

marin [14]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

The volume of a prism is 48 cubic centimeters. If the length is 2 centimeters and the width is 4 centimeters. What is the height

Pie

Answer:

Step-by-step explanation:

volume is (lwh) since youre trying to find the height just multiply the 2 and 4 together to get 8. Then divide 8 by 48 to get your height.

4 0

3 years ago

PLZ Hurry!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!The sum of two numbers is 19 and their difference is 7, What are the two num

yuradex [85]

Answer:

13 and 6

Step-by-step explanation:

4 0

3 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

3 years ago