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Nataly [62]
3 years ago
15

Which equation has the same solution as x^2 - 6x - 14 = 0?

Mathematics
1 answer:
m_a_m_a [10]3 years ago
5 0

Answer: B

Step-by-step explanation:

x^{2}-6x-14=(x-3)^{2}-14-9=(x-3)^{2}-23

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How many solutions does 6 - 3x = 4 - x - 3 - 2x have?
White raven [17]

Answer:

No Solution

Step-by-step explanation:

6 - 3x = 4 - x - 3 - 2x  (Given)

6 - 3x = -1 -3x (combine like terms)

7 (addition)

The x cancels each other out so no solution.

7 0
3 years ago
which of the following answer choices describes all the transformation found in the cubic function f(x)=-(x+2)3-5
Alona [7]
The cubic function is f(x) = x^3

You need to perform three transformations to the cubic function to obtain
f(x) = - (x + 2)^3 - 5.

Those transfformations are:

1) Shift f(x) = x^3,  2 units leftward to obtain f(x) = (x + 2)^3

2) reflect f(x) = (x + 2)^3 across the x-axis to obtain f(x) = - (x + 2)^3

3) shift f(x) = - ( x + 2) ^3, 5 units downward to obtain f(x) = - (x + 2)^3 - 5
8 0
4 years ago
Need answer within 15 min
nikdorinn [45]

Answer:

10

Step-by-step explanation:

12-6+4

6 0
3 years ago
3. Shannon said she can find the factored form of a trinomial of the form
vovikov84 [41]

Answer:

To make it trinomial it has to be ax²+bx+c

8 0
3 years ago
Od<br> 2<br> (<br> 2 7 1-3<br> 1 3 1-2<br> 6:1-6<br> b =<br> d =<br> IDONE
Lena [83]

Answer:

a = \boxed{1},\: b=\boxed{0}\\\\c=\boxed{0},\: d=\boxed{1}

Step-by-step explanation:

  • \begin{bmatrix} 2 & 7\\\\1 & 3\end{bmatrix}\:\begin{bmatrix} -3 & 7\\\\1 & -2\end{bmatrix}=\begin{bmatrix} a & b\\\\c & d\end{bmatrix}

  • Multiply both the matrices on left side, we find:

  • \begin{bmatrix} 2(-3)+7(1) & 2(7)+7(-2)\\\\1(-3)+3(1) & 1(7)+3(-2)\end{bmatrix}=\begin{bmatrix} a & b\\\\c & d\end{bmatrix}

  • \rightarrow\begin{bmatrix} -6+7 & 14-14\\\\-3+3 & 7-6\end{bmatrix}=\begin{bmatrix} a & b\\\\c & d\end{bmatrix}

  • \rightarrow\begin{bmatrix} 1 & 0\\\\0 & 1\end{bmatrix}=\begin{bmatrix} a & b\\\\c & d\end{bmatrix}

  • Comparing the corresponding elements of both the matrices on both sides, we find:

  • a = \boxed{1},\: b=\boxed{0},\:c=\boxed{0},\: d=\boxed{1}
7 0
3 years ago
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