Mr young has 30 times as many pencils as jack. The whole class has 200 times as many pencils as jack. If jack has 2pencils, how many pencils does mr young have ? And the whole school have?

Answers

Answer 1

Answer: well you would do 30*2= 60
and that would be your answer

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P + 4 Solve for p
--------- = 5
6

Answers

Answer:

p = 26

Step-by-step explanation:

(p+4)/6 = 5

Multiply each side by 6

(p+4)/6 *6 = 5*6

p+4 = 30

Subtract 4 from each side

p+4-4 = 30-4

p = 26

What is the product of 2x + y and 5x – y + 3?

Answers

Answer: The required product is $10x^2-y^2+3xy+6x+3y.$

Step-by-step explanation: We are to find the product f the following two algebraic expressions:

$E_1=2x+y,\n\nE_2=5x-y+3.$

To find the product of the above two expressions, we must multiply each term of the first expression with each term of the second expression.

The multiplication is as follows:

$M\n\n=E_1 * E_2\n\n=(2x+y)*(5x-y+3)\n\n=10x^2-2xy+6x+5xy-y^2+3y\n\n=10x^2-y^2+3xy+6x+3y.$

Thus, the required product is $10x^2-y^2+3xy+6x+3y.$

Answer: $10x^2-y^2+3xy+6x+3y$

Step-by-step explanation:

Given expressions are 2x+ y and 5x -y +3

Product of 2x+y and 5x-y+3 is

$(2x+y)* (5x-y+3) = 2x* (5x-y+3) + y* (5x-y+3)$ ( by applying distributive property under multiplication over addition)

⇒ $(2x+y)* (5x-y+3) = 10x^2-2xy+6x+5xy-y^2+3y$ (again by distributive property)

⇒ $(2x+y)* (5x-y+3) = 10x^2-y^2+3xy+6x+3y$ ( by operating the like terms.)

If f(x)= -1x2+3, find f(1)
Someone help me, I'm so confused ;-;

Answers

Answer:

-1×(1)×2+3=1 This is the answer I think

10. 3x – x +4 = 4(2x-1) 11. 4(2x +1) = 5x + 3x +9

12. 10+ x = 5 (1/5x + 2)

13. 8 (x + 2) = 2x + 16

14. 3+ 3/2x + 4 = 4x – 5/2x

15. 3/2 (2x +6) = 3x +9

16. ½ (2-4x) +2x = 13

17. 12 + 2x – x = 9x +6

18. 4x +1 = 2 (2x + 3)

19. 4 (x + 3) - 4 = 8(½x + 1)

20. X + 5x + 4 = 3(2x – 1)

21. 5 (x + 2) - 3x = 2 (x +5)
solve all equations and show your work

Answers

Answer:

10. x=4/3 or 1.3

11. No solution

12. All real numbers

13. x=0

14. No solution

15. All real numbers

16. No solution

17. x=3/4 or 0.75

18. No solution

Step-by-step explanation:

5/12+2/9+1/6 adding fractions

Answers

5/12 = 15/36
2/9 = 8/36
1/6 = 6/36
15 + 8 + 6 = 29/36
Happy to Help!

Find the lcd which is 36
15/36+8/36+6/36
=29/36

Which long division problem can be used to prove the formula for factoring the difference of two perfect cubes?

Answers

Some of the possible options of the questions are;

A) $(a - b) | \overline {a^2 + a \cdot b + b^2}$

B) $(a + b) | \overline {a^2 - a \cdot b + b^2}$

C) $(a + b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$

D) $(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$

The difference of two perfect cubes has a binomial factor and a trinomial factor

The option that gives the long division problem that can be used to prove the difference of two perfect cubes is option D

D) $\underline {(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}}$

Reason:

The formula for factoring the difference of twoperfect cubes is presented as follows;

a³ - b³ = (a - b)·(a² + a·b + b²)

Given that a factor of the difference of two cubes is (a - b), and that we

have; (a³ + 0·a·b² + 0·a²·b - b³) = (a³ - b³), both of which are present in

option D, by long division of option D, we have;

${} \hspace {33} a^2 + a \cdot b + b^2\n(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a^2 \cdot b - b^3}\n{} \hspace {33} \underline{a^3 - a^2 \cdot b }\n{} \hspace {55} a^2 \cdot b + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3\n {} \hspace {55} \underline{a^2 \cdot b - a \cdot b^2}\n{} \hspace {89} a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3\n{} \hspace {89} \underline{a \cdot b^2 - b^3}\n{}\hspace {89} 0$

By the above long division, we have;

$(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$ = a² + a·b + b²

Which gives;

$(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$ = (a³ + 0·a·b² + 0·a·b² - b³)/(a - b)

We get;

(a³ + 0·a·b² + 0·a·b² - b³)/(a - b) = a² + a·b + b²

(a - b)·(a² + a·b + b²) = (a³ + 0·a·b² + 0·a·b² - b³) = (a³ - b³)

(a - b)·(a² + a·b + b²) = (a³ - b³)

(a³ - b³) = (a - b)·(a² + a·b + b²)

Therefore;

The long division problem that can be used to prove the formula for

factoring the difference of two perfect cubes is

$(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$ , which is option D

D) $(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$

Learn more here:

brainly.com/question/17022755

Answer:

The correct options, rearranged, are:

Options:

$A)(a^2+ab+b^2)/(a-b)\n\nB)(a^2-ab+b^2)/(a+b)\n\nC)(a^3+0a^2+0ab^2-b^3)/(a+b))\n\n D)(a^3+0a^2+0ab^2-b^3)/(a-b)$

And the asnwer is the last option (D).

Explanation:

You need to find which long division can be used to prove the formula for factoring the difference of two perfect cubes.

The difference of two perfect cubes may be represented by:

$a^3-b^3$

And it is, as a very well known special case:

$a^3-b^3=(a-b)(a^2+ab+b^2)$

Then, to prove, it you must divide the left side, $a^3-b^3$ , by the first factor of the right side, $a-b$

Note that, to preserve the places of each term, you can write:

$(a^3-b^3)=(a^3+0a^2+0ab^2-b^3)$

Then, you have:

$(a^3+0a^2+0ab^2-b^3)=(a-b)(a^2+ab+b^2)$

By the division property of equality, you can divide both sides by the same factor, which in this case will be the binomial, and you get:

$(a^3+0a^2+0ab^2-b^3)/(a-b)=(a^2+ab+b^2)$

That is the last option (D).

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