Answer:
5000 grams
( it says i have to type 20 characters so i did this)
It would be triangle because you're cutting it
Answer:
The equation that can be used to find the width is 2(13 + w) = 42
The width of the rectangle is 8 inches
Step-by-step explanation:
The formula of the perimeter of a rectangle is P = 2(l + w), where l is its length and w is its width
∵ One rectangle has a length of 13 inches
∴ l = 13 inches
∵ Its perimeter is 42 inches
∴ P = 42 inches
- Substitute the values of l and P in the formula of the perimeter
∵ 42 = 2(13 + w)
The equation that can be used to find the width is 2(13 + w) = 42
To find W divide both sides by 2
∴ 21 = 13 + w
- Subtract 13 from both sides
∴ 8 = w
The width of the rectangle is 8 inches
<u>Question 8</u>
a^2 + 7a + 12
= (a+3)(a+4)
When factorising a quadratic, the product of the two factors should equal the constant term (12), and the sum of the two factors should equal the linear term (7). To find the two factors, list out the factors of 12 (1x12, 2x6, 3x4) and identify the pair that adds up to 7 (3+4).
An alternative method if you get stuck during your exam would be to solve it algebraically using the quadratic formula and then write it in the factorised form.
a = (-7 +or- sqrt(7^2 - 4(1)(12)) / 2(1)
= (-7 +or- sqrt(1))/2
= -3 or -4
These factors are the negative of the values that would go in the brackets when written in factorised form, as when a = -3 the factor (a+3) would equal 0. (If it were positive 3 instead, then in the factorised form it would be a-3).
<u>Question 10</u>
-3(x - y)/9 + (4x - 7y)/2 - (x + y)/18
Rewrite each fraction with a common denominator so you can combine the fractions into one.
= -6(x - y)/18 + 9(4x - 7y)/18 - (x + y)/18
= (-6(x - y) + 9(4x - 7y) - (x + y)) /18
Expand the brackets and collect like terms.
= (-6x + 6y + 36x - 63y - x - y)/18
= (29x - 58y)/18
= 29/18 x - 29/9 y
Answer:
Quelles caractéristiques ces mots ont-ils en commun ? Cliquez sur tout ce qui s'applique.
- <u><em></em></u>
![\left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%263%5C%5C4%265%266%5C%5C7%268%269%5Cend%7Barray%7D%5Cright%5D)
<u><em></em></u>
Step-by-step explanation:
