MATHEMATICS HIGH SCHOOL

3(2x + 1 ) = 2(x + 1) + 1

Answers

Answer 1

Answer:

3(2x+1)=2(x+1)+1

Step 1:Simplify both sides of the equation

(3)(2x)+(3)(1)=(2)(x)+(2)(1)+ -(distribute)

6x+3=2x+2+1

6x+3=(2x)+(2+1) -(Combine Like Terms)

6x+3=2x+3

Step 2: Subtract 2x from both sides

6x+3-2x=2x+3-2x

4x+3-3=3-3

4x÷4=0÷4

x=0

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What is the value of 36x - 8y to the power of 2, when x=3 and y= -6

Answers

$36x-8y^2 \hbox{ when } x=3 \hbox{ and } y=-6 \n \n36 * 3 -8 * (-6)^2=3 * 36 - 8 * 36=(3-8) * 36= -5 * 36=\n =\boxed{-180}$

$(36x-8y)^(2)$

$[36(3)-8(-6)]^2$

$(108+48)^2$

$156 ^(2)$

$=24336$

How would I solve these?

(-3xy)³ (-x³)

and

(-9b²a³)² (3³b)²

Answers

$(-9b^2a^3)^2 \cdot (3^3b)^2 = (-9)^2\cdot(b^2)^2 \cdot (a^3)^2 \cdot (3^3)^2\cdot (b)^2 = \n \n= 81\cdot b^4\cdot a^6 \cdot 3^6\cdot b ^2 = 3^4 \cdot a^6 \cdot b^4\cdot 3^6\cdot b ^2 =\n \n= 3^(4+6) \cdota^(6) \cdot b ^(4+2) = 3^(10) \cdot a^(6) \cdot b ^6$

$(-3xy)^3(-x^3)=(-3)^3x^3y^3\cdot(-x^3)=-27x^3y^3\cdot(-x^3)=27x^6y^3\n\n\n(-9b^2a^3)^2\cdot(3^3b^2)^2=(-3^2a^3b^2)^2\cdot(3^3b^2)^2=3^(2\cdot2)a^(3\cdot2)b^(2\cdot2)\cdot3^(3\cdot2)b^(2\cdot2)\n\n=3^4a^6b^4\cdot3^6b^4=3^(4+6)a^6b^(4+4)=3^(10)a^6b^8$

Help plz I’ll mark Brainly!!!!!

Answers

Answer:

SAS

Step-by-step explanation:

What is the common difference between successive terms in the sequence?0.36, 0.26, 0.16, 0.06, –0.04, –0.14,

Answers

the common difference between successive terms

To find the common difference we subtract the first term from second term

0.36, 0.26, 0.16, 0.06, –0.04, –0.14,

0.26 - 0.36 = -0.10

0.16 - 0.26 = -0.10

0.06 - 0.16 = -0.10

-0.04 - 0.06 = -0.10

-0.14 - (-0.04) = -0.10

We can see that the common difference is =0.10

the common difference between successive terms = -0.10

The correct answer is:

-0.10.

Explanation:

The common difference between successive terms in a sequence is the number you add to each term to find the next one.

To find this number, you can subtract the first term from the second, the second from the third, and so on:

0.26-0.36 = -0.10

0.16-0.26 = -0.10

0.06-0.16 = -0.10

-0.04-0.06 = -0.10

-0.14--0.04 = -0.10

The common difference is -0.10.

Write an equation for the line below. Show work

Answers

x=3 is your answer. This is because it only crosses the x-axis and this occurs at positive 3. y=3 on the other hand would be a horizontal line that only touches positive 3 on the y-axis.

The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A, in astronomical units, AU. If planet Y is twice the mean distance from the sun as planet X, by what factor is the orbital period increased? A. 2^1/3. B. 2^1/2. C. 2^2/3. D. 2^3/2

Answers

The given equation for the relationship between a planet's orbital period, T and the planet's mean distance from the sun, A is T^2 = A^3. Let the orbital period of planet X be T(X) and that of planet Y = T(Y) and let the mean distance of planet X from the sun be A(X) and that of planet Y = A(Y), then A(Y) = 2A(X) [T(Y)]^2 = [A(Y)]^3 = [2A(X)]^3 But [T(X)]^2 = [A(X)]^3 Thus [T(Y)]^2 = 2^3[T(X)]^2 [T(Y)]^2 / [T(X)]^2 = 2^3 T(Y) / T(X) = 2^3/2 Therefore, the orbital period increased by a factor of 2^3/2

Answer:

The orbital increased by factor $2^{(3)/(2)}$ .Hence, correctander is option D.