Why is 8(s-3) considered a single term when (s-3) is the difference of two terms?

Answers

Answer 1

Answer: When you have 8(s-3) you are find a product of 8 times whatever s-3 is however s-3 you are find the difference between whatever s is and 3.

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In the number 913,256 what is the period that has the digits 913

Answers

The name of the period is the thousands period.

Thousands is the period

For which value of O is sin O= -1

Answers

Answer:

O = 2 π n - π/2 for n element Z

Step-by-step explanation:

Solve for O:

sin(O) = -1

Hint: | Eliminate the sine from the left hand side.

Take the inverse sine of both sides:

Answer: O = 2 π n - π/2 for n element Z

assuming you meant sin(θ) = -1, Check the picture below.

angles of a quadrilateral 80 degrees 100 degrees 100 degrees and 80 degrees in this order what kind of quadrilateral shapes

Answers

Most likely a parrelogram

I Need Help PLease!Ashton estimates the square root of 72 in the following way:
√72=
2√36=
2×6=12
Explain why his reasoning is incorrect.
Estimate the square root of 72 to the nearest tenth without using your calculator. Show your work.
Answer:

Answers

Answer:

$\sqrt {72} = 6\sqrt 2$

Step-by-step explanation:

Ashton estimates the square root of 72 in the following way:

√72=

2√36=

2×6=12

a) Explain why his reasoning is incorrect

This is incorrect because $\sqrt2 * 6$ is the correct answer but he has missed the square root term on 2.

$=\sqrt {2^2X2X3^2 }$

$=\sqrt {2^2} \sqrt 2 \sqrt {3^2 }$

$=2 X \sqrt 2 X 3$

$= \sqrt 2 X 6$

or can be written as:

$= 6\sqrt 2$

b) Estimate the square root of 72 to the nearest tenth without using your calculator. Show your work.

Find the factors of 72 we get $\sqrt {2X2X2X3X3 }$

Now, we will combine 2 terms if the are same so,

$=\sqrt {2^2X2X3^2 }$

$=\sqrt {2^2} \sqrt 2 \sqrt {3^2 }$

$=2 X \sqrt 2 X 3$

$= \sqrt 2 X 6$

$= 6\sqrt 2$

$\sqrt {72} = 6\sqrt 2$

$(a)/(a - x) + (b)/(b - x) = ((a + b)^(2) )/(ab)$

Answers

Answer:

$x_1=(a(a^2+ab+b^2))/((a+b)^2)\n \nx_2=(b(a^2+ab+b^2))/((a+b)^2)$

Step-by-step explanation:

First, simplify the left part:

$(a)/(a-x)+(b)/(b-x)=(a(b-x)+b(a-x))/((a-x)(b-x))=(ab-ax+ab-bx)/(ab-ax-bx+x^2)$

Now, cross multiply:

$ab\cdot (ab-ax+ab-bx)=(a+b)^2\cdot (ab-ax-bx+x^2)\n \n2(ab)^2-a^2bx-ab^2x=(a^2+2ab+b^2)(ab-ax-bx+x^2)\n \n2(ab)^2-a^2bx-ab^2x=a^3b-a^3x-a^2bx+a^2x^2+2(ab)^2-2a^2bx-2ab^2x+2abx^2+ab^3-ab^2x-b^3x+b^2x^2\n \na^3b-a^3x+a^2x^2-2a^2bx-2ab^2x+2abx^2+ab^3-b^3x+b^2x^2=0\n \nx^2(a^2+2ab+b^2)+x(-a^3-2a^2b-2ab^2-b^3)+a^3b+ab^3=0\n \n(a+b)^2x^2-x((a+b)(a^2-ab+b^2)+2ab(a+b))+ab(a^2+b^2)=0\n \n(a+b)^2x^2-x((a+b)(a^2+ab+b^2))+ab(a^2+b^2)=0\n \n(a+b)^2x^2-x(a+b)(a^2+ab+b^2)+ab(a^2+b^2)=0$

$D=(a+b)^2(a^2+ab+b^2)^2-4(a+b)^2ab(a^2+b^2)\n \n=(a+b)^2(a^4+(ab)^2+b^4+2a^3b+2a^2b^2+2ab^3-4a^3b-4ab^3)\n \n=(a+b)^2(a^2-ab+b^2)^2\n \n√(D)=(a+b)(a^2-ab+b^2)=a^3-b^3$

So,

$x_(1,2)=((a+b)(a^2+ab+b^2)\pm (a^3-b^3))/(2(a+b)^2)\n \nx_1=(a^3+a^2b+ab^2+ba^2+ab^2+b^3+a^3-b^3)/(2(a+b)^2)=(2a^3+2a^2b+2ab^2)/(2(a+b)^2)=(2a(a^2+ab+b^2))/(2(a+b)^2)=(a(a^2+ab+b^2))/((a+b)^2)\n \nx_2=(a^3+a^2b+ab^2+ba^2+ab^2+b^3-a^3+b^3)/(2(a+b)^2)=(2b^3+2a^2b+2ab^2)/(2(a+b)^2)=(2b(a^2+ab+b^2))/(2(a+b)^2)=(b(a^2+ab+b^2))/((a+b)^2)$

I really need help on #21 a-c thanks. :)

Answers

assuming that the order is
ok so
distance from greenvile to parker=29
distance from greenvile to hadley=?

distance from hadley to greenvile is 12 and 1/2 less than distance between parker and greenvile means
?=-12 and 1/2+29
add
-12 and 1/2+29=6 and 1/2

?=6 and 1/2 miles

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