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Lyrx [107]
3 years ago
15

You can use an arithmetic operator to return the modulus of a calculation, which is the ____ when you divide one number by anoth

er number.
Mathematics
1 answer:
balandron [24]3 years ago
7 0

Answer:

<u>Residue</u>

Step-by-step explanation:

Let a and b be integers. We define a mod b to be the residue of dividing a by b. For example, if a evenly divides b, then a mod b=0, 20 mod 6= 2. The modulus operator is widely used in programming, and it is convenient when a and b are large numbers.

a mod b is always a nonnegative integer. In fact, 0≤ a mod b≤ |b-1| by the division algorithm. a and b can also be negative integers. Since 8=-(-5)+3 then 8 mod -5= 3.

As a final example, some known properties can be rewritten in terms of mod. a mod 2=0 if and only if a is even. a mod 2=1 if and only if a is odd.

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If y=e5t is a solution to the differential equation
a_sh-v [17]

Answer:

k = 30, y(t) = C_1e^{5t}+C_2e^{6t}

Step-by-step explanation:

Since y=e^{5t} is a solution, then it must satisfy the differential equation. So, we calculate the derivatives and replace the value in the equation. We have that

\frac{d^2y}{dt^2} = 25 e^{5t},\frac{dy}{dt} = 5e^{5t}

Then, replacing the derivatives in the equation we have:

25e^{5t}-11(5)e^{5t}+ke^{5t}=0 e^{5t}(25-55+k) =0

Since e^{5t} is a positive function, we have that

25-55+k = 0 \rightarrow k = 30.

Now, consider a general solution y(t) = Ae^{rt}, A \in \mathbb{R}, then, by calculating the derivatives and replacing them in the equation, we get

Ae^{rt}(r^2-11r+30)=0

We already know that r=5 is a solution of the equation, then we can divide the polynomial by the factor (r-5) to the get the other solution. If we do so, we get that (r-6)=0. So the other solution is r=6.

Therefore, the general solution is

y(t) = C_1e^{5t}+C_2e^{6t}

8 0
3 years ago
What is the answer to 3k-1=7k+2
Phoenix [80]
Answer is given above

5 0
3 years ago
Read 2 more answers
Find an equation for the inverse for each of the following.
IgorLugansk [536]

Answer:

  • <em>To solve these first swap x and y, solve for y and name it inverse function</em>

3. <u>y = -8x + 2</u>

  • x = -8y + 2
  • 8y = -x + 2
  • y = -x/8 + 2/8
  • y = -(18)x + 1/4

f⁻¹(x) = -(18)x + 1/4

-----------------------------------------

4.<u> y = (2/3)x - 5</u>

  • x = (2/3)y - 5
  • (2/3)y = x + 5
  • y = (3/2)x + (3/2)5
  • y = 1.5x + 7.5

f⁻¹(x) = 1.5x + 7.5

-----------------------------------------

5. <u>f(x) = 2x² - 6</u>

  • x = 2y² - 6
  • 2y² = x + 6
  • y² = 1/2x + 3
  • y = \sqrt{1/2x + 3}

f⁻¹(x) = \sqrt{1/2x + 3}

-----------------------------------------

6. <u>y = (x - 3)²</u>

  • x = (y - 3)²
  • \sqrt{x} = y - 3
  • y = 3 + \sqrt{x}

f⁻¹(x) = 3 +  \sqrt{x}

4 0
3 years ago
Find the first, fourth, and tenth terms of the arithmetic sequence described by the given rule
Sergeeva-Olga [200]

Answer:

First term: 5

Fourth term: 5 1/2

Tenth term: 6 1/2

Step-by-step explanation:

Let's find the first, fourth and tenth terms of the arithmetic sequence described by the given rule:

A(n) = 5 + (n-1) (1/6)

First term:

A(1) = 5 + (1-1) (1/6)

A(1) = 5 + (0) (1/6)

A(1) = 5

Fourth term:

A(4) = 5 + (4-1) (1/6)

A(4) = 5 + (3) (1/6)

A(4) = 5 + 3/6 = 5 3/6 = 5 1/2 (simplifying)

Tenth term:

A(10) = 5 + (10-1) (1/6)

A(10) = 5 + (9) (1/6)

A (10) = 5 + 9/6 = 6 3/6 = 6 1/2 (simplifying)

5 0
3 years ago
√−100 = ___ +_____i<br> ...
Natalija [7]

Answer:

Step-by-step explanation:

Rewrite this as

\sqrt{(-1)(100)}

Knowing that i-squared = -1:

\sqrt{(i^2)(100)}

Both i-squared and 100 are perfect squares, so this simplifies to

±10i

7 0
3 years ago
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