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d1i1m1o1n [39]
3 years ago
9

(x-5)•(x-5) foil the two binomials and simplify

Mathematics
1 answer:
Anastaziya [24]3 years ago
5 0

Answer:

x^2 - 10x + 25

Step-by-step explanation:

(x - 5). (x - 5)

= x^2 + x*-5 -5 * x - 5*- 5

= x^2 - 5x - 5x + 25

= x^2 - 10x + 25

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Serga [27]
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Length = 21

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Now you’re gonna want to multiply 21 by itself.

21 • 21 = 21 + 420 = ?

Which leaves you with the final sum of...

Area = 441 yds.
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3 years ago
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Goryan [66]

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DBE=50 degrees

ABD=140 degrees

Step-by-step explanation:

CA is a line which means that it equals 180 degrees total. Subtract the other angle measurements to find that DBE is 50 degrees. Then add the angles DBE, EBF, and FBA. Hope this helps!

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He can buy 6 packs of cards and he still gets the free poster.

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A baseball diamond has the shape of a square with side
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4 years ago
Use the following vectors to answer parts​ (a) and​ (b). v1equals=[Start 3 By 1 Matrix 1st Row 1st Column 1 2nd Row 1st Column n
erik [133]

Answer:

(1) No matter what's the value of h, \vec{v}_3 is never in the span of \vec{v}_1 and \vec{v}_2.

(2) The three vectors \vec{v}_1, \vec{v}_2, and \vec{v}_3 are always linearly dependent for all real h.

Step-by-step explanation:

<h3>(a)</h3>

If \vec{v}_3 is in the span of \vec{v}_1 and \vec{v}_2, there need to exist real a and b such that

a\; \vec{v}_1 + b\; \vec{v}_2 = \vec{v}_3.

Assume that such a and b do exist.

In other words,

\displaystyle a \left[\begin{array}{c}{1 \\ -4\\2} \end{array}\right] + b\left[\begin{array}{c}{-4 \\ 16\\-8}\end{array}\right] = \left[\begin{array}{c}5 \\7 \\ h\end{array}\right].

\displaystyle \left[\begin{array}{c}{a \\ -4a\\2a} \end{array}\right] + \left[\begin{array}{c}{-4b \\ 16b\\-8b}\end{array}\right] = \left[\begin{array}{c}5 \\7 \\ h\end{array}\right].

\left\{\begin{array}{rrcr} a & - 4b &=& 5\\ -4a & + 16b &= &7\\2a & -8b & =&h\end{aligned}\right..

Rewrite as an augmented matrix and row-reduce:

\displaystyle \left[ \begin{array}{cc|c} 1 & -4 & 5 \\ -4 & 16 & 7 \\ 2 & -8 & h\end{array}\right].

(Add four times row one to row two and -2 times row one to row three.)

\displaystyle \sim \left[ \begin{array}{cc|c} 1 & -4 & 5 \\ 0 & 0 & 27 \\ 0 & 0 & h - 10\end{array}\right].

Note that in row two,

  • Left-hand side: 0;
  • Right-hand side: 27\neq 0.

In other words, this system is inconsistent. There's no real a and b that would satisfy the condition

a\; \vec{v}_1 + b\; \vec{v}_2 = \vec{v}_3.

Hence \forall h \in \mathbb{R}, \quad \vec{v}_3\not \in \text{Span}\{\vec{v}_1, \vec{v}_2\}.

There's no real h that allows h, \vec{v}_3 to be part of the span of \vec{v}_1 and \vec{v}_2.

<h3>(b)</h3>

If the three vectors are linearly dependent, at least one of them can be expressed as the linear combination of the other two.

Note that

\vec{v}_2 = (-4)\vec{v}_1 + 0 \; \vec{v}_3. In other words, \vec{v}_2 can be written as the linear combination of the other two vectors. Additionally, since the coefficient in front of \vec{v}_3 is zero, neither the exact value of \vec{v}_3 nor the value of h will make a difference. Therefore, for all h \in \mathbb{R}, the three vectors \vec{v}_1, \vec{v}_2, and \vec{v}_3 are linearly dependent.

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4 years ago
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