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3 years ago

The polynomial given below has _____ root(s).

Mathematics

1 answer:

lisov135 [29]3 years ago

8 0

3x²-8x+4=0
(3x-2) (x-2)=0
x=2/3 , x = 2
Hence, the polynomial has two positive roots.
Option A is the correct choice.

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Factorise the expressions and divide them as directed.

Annette [7]

a) 3(x-9)= 3x -9

3x-27=3x -9

3x-3x=-9+27

x=18

b) 2x+3x=5x^2

2x+3x=( 5x)^2

2x+3x=25x

5x=25x

x=25x/5

x=5x

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3 years ago

What is the value of the expression –7 – 5 + 4?

vichka [17]

By solving this problem, I got D.-8.Right?

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4 years ago

Terry earned x dollars selling electronics last month. this month he earned $18 less than twice the amount he earned last month.

ValentinkaMS [17]

Answer:

2x - 18

$3,382

Step-by-step explanation:

x = number of dollars gained last month

this month:

twice the amount = 2x

minus $18 = 2x - 18

x = 1,700

2x - 18

2(1700) - 18

3,400 - 18

$3,382

6 0

3 years ago

Read 2 more answers

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 ! ♡

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4 0

3 years ago

Use the given information to find the exact value of the expression. sin α = 2129 21 29 , α lies in quadrant II, and cos β = 151

beks73 [17]

The correct question is
Use the given information to find the exact value of the expression. sin α = 21/29, α lies in quadrant II, and cos β = 15/17, β lies in quadrant I Find sin (α - β).

we know that
sin(α − β<span>) = </span>sin α cos β − cos α sin β

α lies in quadrant II
so
cos α is negative
sin α is positive

β lies in quadrant I
so
cos β is positive
sin β is positive

step 1
find sin β
cos β=15/17
sin² β+cos² β=1-----------> sin² β=1-cos² β----> sin² β=1-(15/17)²
sin² β=1-225/289-----> 64/289
sin β=8/17

step 2
find cos α
sin α = 21/29
cos² α + sin² α=1----> cos² α=1-sin² α---> cos² α=1-(21/29)²---> 1-441/841
cos² α=400/841------> cos α=-20/29 (remember cos α is negative)

step 3
find sin(α − β)
sin α = 21/29 cos α=-20/29
sin β=8/17 cos β=15/17

sin(α − β) = [21/29]*[15/17] − [-20/29*]*[8/17]
sin(α − β) = [315/493] − [-160/493]
sin(α − β) = 475/493

the answer is
sin(α − β) = 475/493

3 0

4 years ago