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3 years ago

-10x - y=-5 write in standard form

Mathematics

1 answer:

ololo11 [35]3 years ago

6 0

Answer:

10 x + y = 5

Step-by-step explanation:

Using the formula, write in standard form.

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Find a unit vector in the direction of the vector left bracket Start 3 By 1 Matrix 1st Row 1st Column negative 8 2nd Row 1st Col

MrRissso [65]

Answer:

$\left[\begin{array}{c}-\frac{8}{\sqrt{117} } \\\frac{7}{\sqrt{117} }\\\frac{2}{\sqrt{117} }\end{array}\right]$

Step-by-step explanation:

We are required to find a unit vector in the direction of:

$\left[\begin{array}{c}-8\\7\\2\end{array}\right]$

Unit Vector, $\hat{a}=\dfrac{\overrightarrow{a}}{|\overrightarrow{a}|}$

The Modulus of $\overrightarrow{a}$ = $\sqrt{(-8)^2+7^2+(-2)^2}=\sqrt{117}$

Therefore, the unit vector of the matrix is given as:

$\left[\begin{array}{c}-\frac{8}{\sqrt{117} } \\\frac{7}{\sqrt{117} }\\\frac{2}{\sqrt{117} }\end{array}\right]$

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3 years ago

Is 3/4 a rational number

Verizon [17]

Hello,

<em>3/4 is a rational number because it can be written as a fraction.</em>

Thanks,

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3 years ago

Evaluate the expression 7^2/ x^2 - 2 for x = 3

vovikov84 [41]

Answer:

Step-by-step explanation:

7²/x²-2 when x=3

49/3²-2

49/9-2

49/7

answer=7

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2 years ago

. Andrew made an error in determining the polynomial equation of smallest degree whose roots are 3, 2+2i

KIM [24]

Answer:

Error of Andrew: Made incorrect factors from the roots

Step-by-step explanation:

Roots of the polynomial are: 3, 2 + 2i, 2 - 2i. According to the factor theorem, if a is a root of the polynomial P(x), then (x - a) is a factor of P(x). According to this definition:

(x - 3) , (x - (2 + 2i)) , (x - (2 - 2i)) are factors of the required polynomial.

Simplifying the brackets, we get:

(x - 3), (x - 2 - 2i), (x - 2 + 2i) are factors of the required polynomial.

This is the step where Andrew made the error. The factors will always be of the form (x - a) , not (x + a). Andrew wrote the complex factors in form of (x + a) which resulted in the wrong answer.

So, the polynomial would be:

$(x - 3)(x - 2 - 2i)(x - 2+2i)=0\\\\ (x-3)(x^{2}-2x+2xi-2x+4-4i-2xi+4i-4i^{2})=0\\\\ (x-3)(x^{2}-4x+4+4)=0\\\\ (x-3)(x^{2}-4x+8)=0\\\\ x^{3}-4x^{2}+8x-3x^{2}+12x-24=0\\\\ x^{3}-7x^{2}+20x-24=0$