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Law Incorporation [45]
3 years ago
15

Write two fractions that are equivalent to 5

Mathematics
1 answer:
Dima020 [189]3 years ago
8 0

Answer:

5/1, 25/5

Step-by-step explanation:

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What is the annual compound interest rate for an investment account modeled by the function y=12⋅1.18t ?
NikAS [45]

Answer:

The interest rate for the given investment is 18%.

Step-by-step explanation:

We are given that,

The function representing the investment model is, y=12(1.18)^t.

It is required to find the interest rate for the investment.

Since, the function can be re-written as,

y=12(1.18)^t

i.e. y=12(1+0.18)^t

So, on comparing with the formula for compound interest i.e. A=P(1+r)^t, where 'r' is the interest rate.

We have that,

The interest rate for the given investment is 0.18 i.e. 18%.

7 0
3 years ago
DO U KNOW SLOPE??? PLZ HELP
IgorLugansk [536]

To calculate slope; rise/run

Triangle ABC: run-1 rise-5 slope-5

5 rise/ 1 run (5/1)=5

Triangle DEF: run-2 rise 10 slope-5

10 rise/ 2 run (10/2)=5

B. They are equal because the two triangles are similar



7 0
3 years ago
Nathaniel has $2 worth of dimes and quarters. He has 6 more dimes than quarters. By following the steps below, determine the num
Aleksandr-060686 [28]

Answer:

  • 10 dimes and 4 quarters

Step-by-step explanation:

  • Number of dimes = d, each d = 10 c
  • Number of quarters = q, each q = 25 c
  • Total = $2 = 200 c

<u>Equations according the given:</u>

  • 10d + 25q = 200
  • d = q + 6

<u>Solve for q by substitution:</u>

  • 10(q + 6) + 25q = 200
  • 10q + 60 + 25q = 200
  • 35q = 140
  • q = 140/35
  • q = 4

<u>Find the value of d:</u>

  • d = 4 + 6 = 10

Nathaniel has 10 dimes and 4 quarters

8 0
3 years ago
Write an equation of a line that passes through the point (1, 2) and is perpendicular to the line y=-1/4x+2​
Dmitriy789 [7]

Answer:

y = 4x - 2 would be the equation

7 0
3 years ago
What is the smallest integer $n$, greater than $1$, such that $n^{-1}\pmod{130}$ and $n^{-1}\pmod{231}$ are both defined?
olasank [31]

First of all, the modular inverse of n modulo k can only exist if GCD(n, k) = 1.

We have

130 = 2 • 5 • 13

231 = 3 • 7 • 11

so n must be free of 2, 3, 5, 7, 11, and 13, which are the first six primes. It follows that n = 17 must the least integer that satisfies the conditions.

To verify the claim, we try to solve the system of congruences

\begin{cases} 17x \equiv 1 \pmod{130} \\ 17y \equiv 1 \pmod{231} \end{cases}

Use the Euclidean algorithm to express 1 as a linear combination of 130 and 17:

130 = 7 • 17 + 11

17 = 1 • 11 + 6

11 = 1 • 6 + 5

6 = 1 • 5 + 1

⇒   1 = 23 • 17 - 3 • 130

Then

23 • 17 - 3 • 130 ≡ 23 • 17 ≡ 1 (mod 130)

so that x = 23.

Repeat for 231 and 17:

231 = 13 • 17 + 10

17 = 1 • 10 + 7

10 = 1 • 7 + 3

7 = 2 • 3 + 1

⇒   1 = 68 • 17 - 5 • 231

Then

68 • 17 - 5 • 231 ≡ = 68 • 17 ≡ 1 (mod 231)

so that y = 68.

3 0
3 years ago
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