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3 years ago

Rewrite without absolute value for the given conditions: y=|x-3|+|x+2|-|x-5|, if x>5

Mathematics

2 answers:

Wittaler [7]3 years ago

5 0

Answer:

y = x + 4

Step-by-step explanation:

Since x > 5, x-3 > 0, x + 2 > 0 and x - 5 > 0

So Ix - 3I = x - 3, Ix + 2I = x + 2 and x - 5 = x - 5

Therefore, y = x - 3 + x + 2 - x + 5

y = x + 4

Svetlanka [38]3 years ago

5 0

Answer:

Step-by-step explanation:

It might interest you to know what this question means without the restriction. See graph 36b. This shows exactly what happens to the absolute value. There are all sorts of turns and twists in it. Notice all the changes in direction and the minimum at -2,-2. The y intercept is at 0,0.

0,0 is also one of the x intercepts as well. The other is -4,0. The whole point of showing you this is to make you aware of what the restriction does.

The graph is reduced to a line starting at (5,9)

You might be interested in

Scooter the clown buys flower hats for $27 each and plain hats for $22 each. If he spends a total of $250 on hats, how many hats

gavmur [86]

Answer:

6 Flower hats

Step-by-step explanation:

If you are asking for whole number answers, then the answer is 6 Flower hats.

check:

6X27=162 and 4 plain hats, 4X22=88, 162+88=250.

hope it helps!

3 0

3 years ago

Solve the matrix and prove that it is equal 0

Art [367]

Step-by-step explanation:

$\underline{ \underline{ \text{Given}}} :$

$\tt{ {A}^{T} = \begin{bmatrix} 2 & - 4 \\ 4 & 3 \\ \end{bmatrix}}$

$\underline{ \underline { \text{To \: Find}}} :$

$\sf{ {A}^{2} - 5A+ 22I= 0}$

$\underline{ \underline{ \text{Solution}}} :$

The new matrix obtained from a given matrix by interchanging it's rows and columns is called the transposition of matrix. It is denoted by $\sf{ {A}^{T}}$ . Again , Interchange it's rows and columns in order to find ' A '.

$\tt{A = \begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix}}$

Now , LEFT HAND SIDE ( L.H.S )

$\tt{ {A}^{2} - 5A+ 22I}$

Here, I refers to identity matrix. A diagonal matrix in which all the elements of leading diagonal are 1 ( unit ) is called unit or identity matrix.

⟼ $\begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix} \times \begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix} - 5 \times \begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix} + 22 \times \begin{bmatrix} 1 & 0 \\ 0 & 1\\ \end{bmatrix}$

⟼ $\begin{bmatrix} 2 \times 2 + 4 \times ( - 4)& 2 \times 4 + 4 \times 3 \\ - 4 \times 2 + 3 \times ( - 4) & - 4 \times 4 + 3 \times 3 \\ \end{bmatrix} - \begin{bmatrix} 10 & 20 \\ - 20& 15 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}$

⟼ $\begin{bmatrix} 4 + ( - 16) & 8 + 12 \\ - 8 + ( - 12) & - 16 + 9 \\ \end{bmatrix} - \begin{bmatrix} 10 & 20 \\ - 20 & 15 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}$

⟼ $\begin{bmatrix} - 12 & 20\\ - 20& - 7 \\ \end{bmatrix} - \begin{bmatrix} 10 & 20 \\ - 20 & 15 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}$

⟼ $\begin{bmatrix} - 22 & 0 \\ 0& - 22 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}$

⟼ $\begin{bmatrix} - 22 + 22 & 0 + 0 \\ 0 + 0 & - 22 + 22 \\ \end{bmatrix}$

⟼ $\begin{bmatrix} 0 & 0\\ 0 & 0 \\ \end{bmatrix}$

⟼ $\sf{0}$

RIGHT HAND SIDE ( R.H.S ) : 0

L.H.S = R.H.S [ Hence , proved ! ]

Hope I helped ! ♡

Have a wonderful day / night ! ツ

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4 0

3 years ago

Please help me in this question

charle [14.2K]

Number defective=4
number of non-defective=total-defective
=20-4
=16
the questions asked for 1 defective
so we choose only 1 defective from 4 defective items
4C1

and for 1 non-defective
we choose only 1 non-defective from 16 non-defective items
16C1

we choose 2 items from 20 items
so 20C2

answer:
probability= [(4C1) * (16C1)] /(20C2)
=32/95