Given:
The side of square = 12 in.
Scale factor of enlargement = 3 in : 2 m
To find:
The proportion that is use to solve the side length, x, of the enlarged square.
Solution:
Let, the side of length of enlarged square = x m
In case of enlargement the corresponding sides are proportional.



Divide both sides by 3.


Therefore, the required proportion is
and the side length of the square after enlargement is 8 m.
Answer:
(- 2, - 1 )
Step-by-step explanation:
Given the 2 equations
2x - 3y = - 1 → (1)
x + 4y = - 6 → (2)
Rearrange (2) expressing x in terms of y by subtracting 4y from both sides
x = - 6 - 4y → (3)
Substitute x = - 6 - 4y into (1)
2(- 6 - 4y) - 3y = - 1 ← distribute and simplify left side
- 12 - 8y - 3y = - 1
- 12 - 11y = - 1 ( add 12 to both sides )
- 11y = 11 ( divide both sides by - 11 )
y = - 1
Substitute y = - 1 into (3) for corresponding value of x
x = - 6 - 4(- 1) = - 6 + 4 = - 2
Solution is (- 2, - 1 )
Answer:
1 rubber cube dude give since we'll
Answer:
that is 6 and 5/ 12 cups of flour are required
Step-by-step explanation:
total number of cups required to make a loaf of bread
= 3 + ¾
making their denominators equal and that to 4
= 12/ 4 + 3/ 4
adding numerators
= (12 + 3)/ 4
= 15/ 4
total cups of flour required to make pretzels
= 2 + ⅔
making their denominators equal and that to 3
= 6/ 3 + 2/ 3
adding nunerators
= (6 + 2)/ 3
= 8/ 3
Total cups required:-
= 15/ 4 + 8/ 3
taking lcm of denominators to make the fractions equivalent
= {(15 × 3) + (8 × 4)}/ 12
= {45 + 32}/ 12
= 77/ 12

<em>that is 6 and 5/ 12 cups of flour are required</em>.
Answer:
8 days
Step-by-step explanation:
On day 8, Isabella will save 256 nickels, bringing her total to 510.
_____
The number of nickels saved on day n is 2^n. The total is 2^(n+1)-2.
_____
The above can be written down from your knowledge of binary sequences. If you want a more formal development, read on.
__
The number of nickels saved on day n is a geometric sequence with first term 2 and common ratio 2. The n-th term of the sequence is ...
an = a1·r^(n-1) = 2·2^(n-1) = 2^n
The sum of n terms of the sequence is ...
S = a1(r^n -1)/(r -1) = 2(2^n -1)/(2-1)
S = 2^(n+1) -2
__
We want S > 500, so ...
500 < 2^(n+1) -2
502 < 2^(n+1)
251 < 2^n
log(251) < n·log(2)
n > log(251)/log(2)
n > 7.97 . . . . . . . . 8 days or more to save more than 500 nickels