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Semenov [28]
3 years ago
15

Given that a = -1, b=2 and c= -3 ,find the value of a(b2+c2)

Mathematics
1 answer:
KiRa [710]3 years ago
7 0

Answer:

a(b^{2}+c^{2}) = = -13

Step-by-step explanation:

Given that a = -1, b=2 and c= -3

find the value of a(b^{2}+c^{2})

Just substitute the values given for each variable

a(b^{2}+c^{2}) = (-1)(2^{2}+(-3)^{2}) ; do the exponents first

a(b^{2}+c^{2}) = (-1)(4+9)=(-1)(13) ; now add inside the parenthesis;

a(b^{2}+c^{2}) = -13 ; and multiply by -1

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please help solve this problem the - is a negative sign. (2b+-7)+5(b+-3)
enyata [817]

Answer:


Step-by-step explanation:

(2b+-7)+5(b+-3)

Multiply 5 throughout the second parentheses

(2b-7)+(5b-15)

Add 2b and 5b

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Subtract -7 and -15

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3 0
3 years ago
Limit as x approaches 9 of x^2 -81/sqrt of x - 3
Ipatiy [6.2K]

Answer:

108

Step-by-step explanation:

Limit as x approaches 9 of x^2 -81/sqrt of x - 3

First substitute x into the expression

= 9²-81/√9 - 3

= 81-81/3-3

= 0/0 (indeterminate)

Apply l'hospital rule

= lim x -> 9 d/dx(x²-81)/√x - 3

= lim x -> 9 2x/1/2√x

Substitute x = 9

= 2(9)/1/2√9

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Hence the limit of the function is 108

7 0
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

<u>1 |      2   -5   3</u>

  | 2   -5    3  0

<u> 1 |      2     -3   </u>

    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

  <u>2  |     2  2     </u>

         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

<u>-2 |      -12    10     -2</u>

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

6 0
2 years ago
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