<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D%20-12%3D-4x" id="TexFormula1" title="x^{2} -12=-4x" alt="x^{2} -12=-4x" align="abs

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natulia [17]

3 years ago

middle" class="latex-formula">
I know it's easy but I keep getting an answer different from the one in the book.

Mathematics

1 answer:

Anit [1.1K]3 years ago

3 0

Answer:

The answer for Factorization is (x+6)(x-2) = 0 or Solving for x is 2 and -6.

Step-by-step explanation:

x² - 12 = -4x

Factorisation :

x² + 4x - 12 = 0

x² - 2x + 6x - 12 = 0

x(x-2)x + 6(x-2) = 0

(x+6)(x-2) = 0

Solve for x :

(x+6)(x-2) = 0

x + 6 = 0

x = -6

x - 2 = 0

x = 2

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A is an m×n matrix.Check the true statements below:A. The kernel of a linear transformation is a vector space.B. If the equation

Bess [88]

Answer:

Results are (1) True. (2) False. (3) False. (4) True. (5) True. (6) True.

Step-by-step explanation:

Given A is an $m\times n$ matrix. Let $T :U\to V$ be the corresponding linear transformationover the field F and $\theta$ be identity vector in V. Now if $x\in Ker( T)\implies T(x)=\theta$ .

(1) The kernel of a linear transformation is a vector space : True.

Let $x,y\in Ker( T)$ , then,

$T(x+y)=T(x)+T(y)=\theta+\theta=\theta\impies x+y\in Ker( T)$

hence the kernel is closed under addition.

Let $\lambda\in F, x\in Ker( T)$ , then

$T(\lambda x)=\lambda T(x)=\lambda\times \theta=\theta$

$\lambda x\in Ker(T)$ and thus Ket(T) is closed under multiplication

Finally, fore all vectors $u\in U$ ,

$T(\theta)=T(\theta+(-\theta))=T(\theta)+T(-\theta)=T(\theta)-T(\theta)=\theta$

$\implies \theta\in Ker(T)$

Thus Ker(T) is a subspace.

(2) If the equation Ax=b is consistent, then Col(A) is $\mathbb R^m$ : False

if the equation Ax=b is consistent, then Col(A) must be consistent for all b.

(3) The null space of an mxn matrix is in $\mathbb R^m$

: False

The null space that is dimension of solution space of an m x n matrix is always in $\mathbb R^n$ .

(4) The column space of A is the range of the mapping $x\to Ax$

: True.

(5) Col(A) is the set of all vectors that can be written as Ax for some x. : True.

Here Ax will give a linear combination of column of A as a weights of x.

(6) The null space of A is the solution set of the equation Ax=0.

: True

5 0

3 years ago

Solve each equation by completing the square <br> 6) m² + 16m – 8 = 0

Natali5045456 [20]

Answer:

m = - 8 ± 6 $\sqrt{2}$

Step-by-step explanation:

Given

m² + 16m - 8 = 0 ( add 8 to both sides )

m² + 16m = 8

To complete the square

add ( half the coefficient of the m- term )² to both sides

m² + 2(8)m + 64 = 8 + 64

(m + 8)² = 72 ( take the square root of both sides )

m + 8 = ± $\sqrt{72}$ = ± $\sqrt{36(2)}$ = ± 6 $\sqrt{2}$

Subtract 8 from both sides

m = - 8 ± 6 $\sqrt{2}$

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3 years ago

What is the area of the given circle in terms of pi

ser-zykov [4K]

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3.13(10.89)

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3 years ago

Someone help me with my math homework pleaseee. Find the volumes of the pyramids and the height is 7cm for the first one.

gregori [183]

$\bf \textit{volume of a pyramid}\\\\
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now, the first one, on the far-left.... can't see the height.. but I gather you do, now as far as its Base area, well, the bottom is just a 12x12 square, so the area of its base is just 12*12

now, the middle pyramid, has a height of 6, the base is also a square, 8x8, so the Base area is just 8*8

now the last one on the far-right

has a height of 8, the Base is a Hexagon, with sides of 6

$\bf \textit{area of a regular polygon}\\\\
A=\cfrac{1}{4}ns^2cot\left( \frac{180}{n} \right)\qquad 
\begin{cases}
n=\textit{number of sides}\\
s=\textit{length of one side}\\
\frac{180}{n}=\textit{angle in degrees}\\
----------\\
n=6\\
s=6
\end{cases}\\\\\\ A=\cfrac{1}{4}\cdot 6\cdot 6^2\cdot cot\left( \frac{180}{6} \right)$

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3 years ago

Triangles G H L and K H J are connected at point H. Angles Angles L G H and H K J are congruent. Sides G H and H K are congruent

stira [4]

The congruence theorem that can be used is: B. ASA

<h3>What is the ASA Congruence Theorem?</h3>

If we have two triangles that have two pairs of corresponding congruent angles (e.g. ∠LGH ≅ ∠HKJ and ∠LHG ≅ ∠KHJ), and a pair of corresponding congruent sides (e.g. GH ≅ HK), the triangles are said to be congruent triangles by the ASA congruence theorem.

Therefore, triangles GHL and KHL in the image given are congruent triangles by the ASA congruence theorem.

Learn more about the ASA congruence theorem on:

brainly.com/question/2398724

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3 0

2 years ago