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3 years ago

Help me it’s urgent

Mathematics

1 answer:

Ket [755]3 years ago

3 0

Answer:

$a_1 = -\dfrac{3}{4}$

Step-by-step explanation:

$a_1 \times r = a_2$

$a_1 \times r^2 = a_3$

$a_1 \times r^3 = a_4$

...

$a_1 \times r^8 = a_9$

$\dfrac{a_9}{a_4} = \dfrac{a_1 \times r^8}{a_1 \times r^3} = \dfrac{-196}{6}$

$r^5 = -32$

$r = -2$

$a_1 \times r^3 = a_4$

$a_1 \times (-2)^3 = 6$

$a_1 = \dfrac{6}{-8}$

$a_1 = -\dfrac{3}{4}$

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Step-by-step explanation:

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2 years ago

A store manager sets up a cardboard display to advertise a new brand of perfume the display is a square peer amid whose base is

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3 years ago

Read 2 more answers

Tony has 7/8 of a candy bar he wants to split the bar into servings that are 3/10 of the bar how many servings can he make

Irina-Kira [14]

Answer:

35/12 servings

Step-by-step explanation:

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2 years ago

92% of what number is 115 ?

Andrej [43]

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3 years ago

Find the roots of:

Elena-2011 [213]

$\large\displaystyle\text{$\begin{gathered}\sf \pmb{1) \ 2x^3-7x^2+8x-3=0 } \end{gathered}$}$

Synthetic division is used since the equation is of the third degree. The divisors of -3 are 1, -1, 3, +3. So:

| 2 -7 8 -3

1 | 2 -5 3

| 2 -5 3 0

1 | 2 -3

2 -3 0

So the factorization is (x-1)² (2x-3)=0. So:

$\bf{ x_1=x_2=1 \qquad x_2=\dfrac{3}{2} }$

$\large\displaystyle\text{$\begin{gathered}\sf \pmb{2) \ x^3-x^2-4=0 } \end{gathered}$}$

Synthetic division is used since the equation is of the third degree. The divisors of -4 are 1, -1, 2, -2, 4, -4. So:

| 1 -1 0 -4

2 | 2 2

1 2 2 0

So the factorization is (x-2)(x²+x+2)=0 . When calculating the discriminant of the trinomial, it is concluded that it has no roots since the result is negative. So you only have one solution.

$\bf{ 1^2-4(2)(2)=1-16=-15 < 0 \quad \Longrightarrow \quad x=2 }$

$\large\displaystyle\text{$\begin{gathered}\sf \pmb{3) \ 6x^3+7x^2-9x+2=0 } \end{gathered}$}$

Synthetic division is used since the equation is of the third degree. The divisors of 2 are 1, -1, 2, -2. So:

| 6 7 9 2

-2 | -12 10 -2

6 -5 1 0

So the factorization is (x+2)(6x²-5x+1)=0 . The quadratic equation is solved by the general formula:

$\bf{ x_{2, 3}&=\dfrac{5\pm \sqrt{(5)^2-4(6)(1)}}{2(6)}=\dfrac{5\pm \sqrt{25-24}}{12}=\dfrac{5\pm 1}{12} }}$

$\large\displaystyle\text{$\begin{gathered}\sf \begin{matrix} x_1=-2&\ \ \ \ \ \ x_{2}=\dfrac{6}{12} \qquad &\ \ \ x_3=\dfrac{4}{12}\\ &\ \ \ x_2=\dfrac{1}{2} \qquad &x_3=\dfrac{1}{3} \end{matrix} \end{gathered}$}$

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2 years ago