The ratio of length to width is 3/2. Let's set up equations of ratios using first the 25 cm length of the paper and then the 20 cm width:
3 25 cm
--- = ----------
2 x
Solving for x, 3x = 50 cm, and x =16 2/3 cm. This is possible, since 16 2/3 is less than the paper width 20 cm.
3 20 cm
--- = -----------
2 x
Solving for x: 3x = 40 cm; x = 40/3 cm, or x= 13 1/3 cm. This is possible, but does not make maximum use of the 20 by 30 cm paper.
Answer: the largest flag Jake can draw on the paper given is 20 cm (length) by 16 2/3 cm.
Answer:
$12,137.39
Step-by-step explanation:
Use the Compound Amount formula:
A = P (1 + r/n)^(nt), where r is the interest rate as a decimal fraction, n is the number of times the interest is compounded each year, and t is the number of years.
Here, A = $9000(1 + 0.075/12)^(12*4), or
= $9000(1.3486) = $12,137.39
Answer:
see explanation
Step-by-step explanation:
The directed lines have the form < x, y >
With the usual notation
right in x- direction → +
left in x- direction → -
up in y- direction → +
down in y- direction → -
21
EF = < 4, 1 >
22
FG = < 4, - 2 >
23
GH = < - 4, - 2 >
24
EH = < 4, - 3 >
Step-by-step explanation:
Applying rules of exponents to solve the given problems;
4^3 x 4^5 =
5^8 ÷ 5^-2 =
(6^3 ) ^ 4 =
For these problems, the applicable rules of exponents are;
aᵇ x aⁿ = aᵇ⁺ⁿ
aᵇ ÷ aⁿ = aᵇ⁻ⁿ
(aᵇ)ˣ = aᵇˣ
For the first problem; 4³ x 4⁵
aᵇ x aⁿ = aᵇ⁺ⁿ
4³ x 4⁵ = 4³⁺⁵ = 4⁸
Second problem: aᵇ ÷ aⁿ = aᵇ⁻ⁿ
5⁸ ÷ 5⁻² = 5⁸⁻⁽⁻²⁾ = 5⁸⁺² = 5¹⁰
Third problem; (aᵇ)ˣ = aᵇˣ
(6³)⁴ = 6³ˣ⁴ = 6¹²
The answer would be 2 I’m pretty sure
