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Nezavi [6.7K]
3 years ago
14

Question:

Mathematics
2 answers:
andreyandreev [35.5K]3 years ago
5 0
5/2 times x is 5/2x and 5/2 times negative 6 is negative 15 so right now u would have 5/2x - 15 + 1/6 (3-x) then distribute 1/6 into 3 and get 5/2x minus 15 plus 0.5 minus 1/6x then combine all like terms so then it would be 2.666666667x minus 15.5 rounded would be 2.7xminus15.5
OLga [1]3 years ago
3 0

Answer:

2 x /3 − 9/ 2

− y + 36

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Adriel answered 54 questions correctly on his multiple choice history final and earned a grade of 45%. How many total questions
alex41 [277]

Answer:

There were 120 questions on the test.

Step-by-step explanation:

It is given that:

Number of questions answered = 54

Grade percent = 45%

Let,

x be the total number of questions

45% of x = 54

\frac{45}{100}x=54\\0.45x = 54

Dividing both sides by 0.45

\frac{0.45x}{0.45}=\frac{54}{0.45}\\x=120

Therefore,

There were 120 questions on the test.

8 0
3 years ago
Find the value of 9x + 4y when x = 4 and y=-5
omeli [17]

Answer:

16

Step-by-step explanation:

substitute the values into the equation

9(4)+4(-5)

solve...

36-20=16

4 0
3 years ago
Read 2 more answers
How to do this problem? Thank you!
Tcecarenko [31]
(Bc)/(ab)=tan1
(Ef)/(ec)=tan2
Angle 1= angle 2
Tan1=tan2
(Bc)/(ab)=(ef)/(ec)
...
8 0
3 years ago
Is this answer correct?
Aleks [24]
That’s correct!!!!!!
6 0
3 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.

8 0
3 years ago
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