If x is equal to 9 and y is equal to -3, then what is the value of…….? x - 9y - 3x + 6y -13?

Answers

Answer 1

We are given –

x =9 y = -3

Now put the values

[tex]\qquad[/tex] [tex]\twoheadrightarrow\bf x - 9y - 3x + 6y -13[/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf 9 -(9 \times -3) -(3 \times 9) +(6 \times -3) -13 [/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf 9 + 27 -27 -18 -13 [/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf 9 +\pink{\cancel{27}}-\pink{\cancel{27}} -31[/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\bf-22[/tex]

Henceforth, value of x - 9y - 3x + 6y -13 is -22.

________________________________________

Answer 2

Answer:

Value of x - 9y - 3x + 6y -13 is -22 .

Step-by-step explanation:

In the question we have given an expression and values of x and y that is 9 and -3 respectively and ,

We have asked to find the value of x - 9y - 3x + 6y -13 by substituting values of x and y .

[tex] \longmapsto \:x - 9y - 3x + 6y -13[/tex]

So , Substituting the values of x and y ,

[tex] \longmapsto \: 9 - 9 \bold{( - 3)} - 3\bold{(9)} + 6\bold{( - 3)} - 13[/tex]

Now , calculating the values ,

[tex] \longmapsto \: 9 + 27 - 27 - 18 - 13[/tex]

[tex] \longmapsto \: 9 + \cancel{ \red{27}} - \cancel{ \red{27 }}- 18 - 13 [/tex]

[tex] \longmapsto \: 9 - 31[/tex]

[tex] \longmapsto \: \pink{\boxed{- 22}}[/tex]

Therefore , value of given expression that is x - 9y - 3x + 6y -13 is -22 .

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Answers

Step-by-step explanation:

We have that

[tex](x + \frac{1}{x} ) {}^{2} = 3[/tex]

We are trying to find the number value so that we can apply in the later equation.

Qe first simplify.

Remeber that

[tex](a + b) {}^{2} = a {}^{2} + 2ab + {b}^{2} [/tex]

Also remeber that

[tex] \frac{1}{x} = {x}^{ - 1} [/tex]

so

[tex](x + x {}^{ - 1} ) {}^{2} = {x}^{2} + 2x {}^{0} + {x}^{ - 2} = 3[/tex]

We then simply remeber that x^0=1 so

[tex] {x}^{2} + 2 + \frac{1}{ {x}^{2} } = 3[/tex]

Multiply both sides by x^2.

[tex] {x}^{4} + 2 {x}^{2} + 1 = 3 {x}^{2} [/tex]

Subtract both sides by 3x^2

[tex] {x}^{4} - {x}^{2} + 1 = 0[/tex]

Notice that x^4= (x^2)^2.

So our reformed equation is

[tex]( {x}^{2} ) {}^{2} - {x}^{2} + 1 = 0[/tex]

Let a variable , w equal x^2. This means that we subsitute variable, w in for x^2.

[tex]w {}^{2} - w + 1 = 0[/tex]

Now we use the quadratic formula

[tex] w = \frac{ - b + \sqrt{b {}^{2} - 4ac } }{2a} [/tex]

and

[tex]w = - b - \frac { \sqrt{b {}^{2} - 4ac } }{2a} [/tex]

Let a=1 b=-1 and c=1.

[tex]w = \frac{1 + \sqrt{1 - 4(1)} }{2} [/tex]

[tex]w = \frac{1 - \sqrt{1 - 4} }{2} [/tex]

Now, we get

[tex]w = \frac{1}{2} + \frac{i \sqrt{3} }{2} [/tex]

and

[tex]w = \frac{1}{2} - \frac{ i\sqrt{3} }{2} [/tex]

Now since we set both of these to the x^2 we solve for x.

and

[tex] {x}^{2} = \frac{1}{2} + \frac{i \sqrt{3} }{2} [/tex]

and

[tex] {x}^{2} = \frac{1}{2} - \frac{i \sqrt{3} }{2} [/tex]

We can represent both of these as complex number in the form of a+bi. Next we can convert this into trig form which is

[tex] {x}^{2} = 1( \cos(60) + i \: \sin(60) [/tex]

and

[tex] {x}^{2} = 1( \cos(300) + i \: sin(300))[/tex]

Next we take the sqr root of 1 which is 1, and divide the degree by two.

[tex] {x} = 1( \cos(30) + i \: sin \: 30)[/tex]

and

[tex]x = 1( \cos(150) + i \: sin(150)[/tex]

We are asked for the 2nd root so just add 180 degrees to this and we have

[tex]x = 1 \cos(210) + i \: sin \: 210)[/tex]

and

[tex]x = 1( \cos(330) + i \: sin(330)[/tex]

which both simplified to

[tex]x = - \frac{ \sqrt{3} }{2} - \frac{1}{2} i[/tex]

and

[tex]x = \frac{ \sqrt{3} }{2} - \frac{1}{2} i[/tex]

Now we must find

x^18+x^12+x^6+1.

We just use demovire Theorem. Which is a complex number raised to the nth root is

[tex] {r}^{n} (cos(nx) + i \: sin(nx)[/tex]

So let plug in our first root.

[tex]1( \cos(330 \times 18)) + i \: sin \: (330 \times 18))) + 1( \cos(12 \times 330)) + i \: sin(12 \times 330) + 1( \cos(6 \times 330) + i \: sin(6 \times 330))) + 1[/tex]

To save time we multiply the angle and use rules of terminals angle and we get

[tex]1( \cos(180) + i \sin(180) ) + 1( \cos(0) + i \: sin \:( 0) + 1( \cos(180) + i \: sin(180) + 1[/tex]

And we get

[tex] - 1 + 1 + - 1 + 1 = 0[/tex]

So one of the answer is x=0

And the other, let see

[tex]1 \cos(210 \times 18) + i \: \sin(210 \times 18) + 1 \: cos(210 \times 12) + i \: sin(210 \times 12) + 1 \cos(210 \times 6) + \:i sin(210 \times 6) + 1[/tex]

[tex] \cos(180) + i \: sin(180) + 1 \cos(0) + i\sin(0) +1( \cos(0) + i \sin(0) + 1[/tex]

We get

[tex] - 1 + 1 + 1 + 1 = 2[/tex]

So our answer are 2.

So the answer to the second part is

0 and 2.

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What are the vertices of r(180, 0) (AXYZ)?

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The formula for the Area of a triangle is A = (1/2)bh where b is the base an h is the height. Since the triangle is a right triangle, the base and height are the two shorter sides

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Step-by-step explanation:

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Answers

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Answer:

4x+9y-27=0

compare and fill em

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Answer: the y intercept form is y= 4x-1

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Y=3x+7
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Consider a big number 3,400,000. To convert this number into scientific notation:

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If the points (2,7),(-3,3) and (5,1) are the vertices of a triangle ,find the length of the median drawn through the first vertex.​

Answers

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[tex]\sqrt{26}[/tex] units

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= sqrt(104)/2 = sqrt(104/4) = sqrt(26) =

= 5.099019514...

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Parmi les rectangles de périmètre 100cm, quelles sont les dimensions du rectangle d'aire maximale?

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Answers

La solution:

625 cm^2.

Explication étape par étape:

Si la forme est rectangulaire, elle aura la plus grande superficie possible quand la longueur équivaut à la largeur. Pour avoir un périmètre de 100 cm, cela signifie que chaque côté doit faire 25 cm.

La superficie serait alors de 25 cm x 25 cm = 625 cm^2.

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y=3x +4

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Answer:

9√10

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Answers

Answer:

The answer would be -12, I'm not to sure what simplify means

When adding integers that have different signs you have to subtract and the solution will have the sign of the bigger number

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enter an algebraic expression for the word expression.
2 decreased by n​

Answers

Answer:

2 - n

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2 decreased by n means n less than 2.

Answer:

Write each phrase as an algebraic expression. Phrase, Expression. nine increased by a number x, 9 + x. fourteen decreased by a number p, 14

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2-n       Plz mark brainliest if correct

hello how are you doing today

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I am doing fine thanks bro.

The face of a clock is divided into 12 equal parts. The radius of the clock face is 10 inches. Assume the hands of the clock will form a central angle. The face of a clock is divided into 12 equal parts. Which statements about the clock are accurate? Select three options. The central angle formed when one hand points at 1 and the other hand points at 3 is 30°. The circumference of the clock is approximately 62. 8 inches. The minor arc measure when one hand points at 12 and the other hand points at 4 is 120°. The length of the major arc between 3 and 10 is approximately 31. 4 inches. The length of the minor arc between 6 and 7 is approximately 5. 2 inches.

Answers

The accurate statements are:

b. The circumference of the clock is approximately 62.8 inches. c. The minor arc measure when one hand points at 12 and the other hand points at 4 is 120°.  e. The length of the minor arc between 6 and 7 is approximately 5.2 inches.

The given parameters are:

[tex]n = 12[/tex] --- number of parts

[tex]r = 10[/tex] --- the radius

(a) The central angle

Between points 1 and 3, there are 2 sections, each of which has a measure of 30 degrees.

So, the measure of the two sections is:

[tex]\theta = 30^o \times 2[/tex]

[tex]\theta = 60^o[/tex]

Hence, (a) is false

(b) The circumference

This is calculated using:

[tex]C = 2\pi r[/tex]

So, we have:

[tex]C = 2 \times 3.14\times 10[/tex]

[tex]C = 62.8[/tex]

Hence, (b) is true

(c) The measure of the minor arc

Between points 12 and 4, there are 4 sections, each of which has a measure of 30 degrees.

So, the measure of the four sections is:

[tex]\theta = 30^o \times 4[/tex]

[tex]\theta = 120^o[/tex]

Hence, (c) is true

(d) The length of the major arc

Between points 3 and 10, there are 7 sections, each of which has a measure of 30 degrees.

So, the measure of the seven sections is:

[tex]\theta = 30^o \times 7[/tex]

[tex]\theta = 210^o[/tex]

The length of the arc is:

[tex]L = \frac{\theta}{360} \times 2\pi r[/tex]

So, we have:

[tex]L = \frac{210}{360} \times 2 \times 3.14 \times 10[/tex]

[tex]L = \frac{13188}{360}[/tex]

[tex]L = 36.3[/tex]

Hence, (d) is false

(e) The length of the minor arc

There is only one section between points 6 and 7

So, the measure of the section is:

[tex]\theta = 30^o[/tex]

The length of the arc is:

[tex]L = \frac{\theta}{360} \times 2\pi r[/tex]

So, we have:

[tex]L = \frac{30}{360} \times 2 \times 3.14 \times 10[/tex]

[tex]L = \frac{1884}{360}[/tex]

[tex]L = 5.2[/tex]

Hence, (e) is true

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Answer:

-B

-C

-E

Step-by-step explanation:

just took the test...

I got them right

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Answers

AB=EF

ABEF=ABF+AEF

NOW CONTINUES THE SOLUTION

if the compound interest on a sum for 2 years at 4% p.a. is ₹408, then the simple interest on the same sum at the same rate and for the same period is (I) ₹400 (ii) ₹398 (iii) ₹200 (iv) ₹204​

Answers

Given that:

CI = ₹408

years = 2 years

Rate of interest = 4%

A = P{1+(R/100)}^

A-P = p{1+(R/100)}^n - P

I = P[1+(R/100)}^n - 1]

408 = P[{1+(4/100)²} - 1]

= P[{1+(1/25)²} - 1]

= P[(26/25)² - 1]

= P[(676/625) - 1]

= P[(676-625)/625]

408 = P(51/625)

P = 408*(625/51)

= 8*625 = 5000

Sum = 5000

Simple Interest (I) = (P*R)/100

= 5000*2*(4/100)

= 50*2*4 = 400

From the given above options, option (a) ₹400 is your correct answer.

Find the equation of a line parallel to y = x + 8 that passes through the point
(-3,3).

Answers

Answer:

y=x + 6

Step-by-step explanation:

G1=G2

so G2 =1

y-3 /x+3 =1/1

y-3 = x +3

y=x + 6

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