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enyata [817]
3 years ago
13

(x+1)(x+2) = K(3x+7)​

Mathematics
2 answers:
Sati [7]3 years ago
4 0
Hope it helps you friend

ehidna [41]3 years ago
3 0
K = (x + 1) (x + 2) /3x + 7
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Aight ok so if i got two ducks and and i take one how munch is left
vichka [17]

Answer:

1

Step-by-step explanation:

5 0
3 years ago
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Some kids are selling lemonade for $1.50 per cup at a high
BaLLatris [955]

A function that would represent profit based on the number of cups of lemonade is Profit = 1.5n - 14

<u>Solution:</u>

Given, Some kids are selling lemonade for $1.50 per cup at a high school baseball game.  

They spent $14 on all of the items needed for the lemonade stand (cups, lemonade, table oth, sign, etc)

We have to create a function that would represent profit based on the number of cups of lemonade

Now, let the number of cups sold be "n"

Then , we know that,<em> profit = selling price – cost price  </em>

Profit = number of cups sold x price per cup – cost price

Profit = n x $ 1.5 – $ 14  

Profit = 1.5n – 14

Hence, the function is Profit = 1.5n - 14

6 0
3 years ago
Find the volume of this object.
Tems11 [23]

Answer:

408 cm^3

Step-by-step explanation:

You have to find the volume of each object individually. You first find the volume of the sphere, so you input the radius, which is 3, into the equation which comes out to be 108 cm^3. You then have to find the volume of the cylinder, and since you have the diameter of the cylinder, and you need the radius, and the radius is half of the diameter, you would input 5 into the equation as the radius, since half of 10 is 5. That comes out to be 300 cm^3, so then you add the two volumes together, which is 408 cm^3

7 0
1 year ago
Solve -12&gt;3x then graph on a number line that you create and please please show all of your work step by step.
V125BC [204]
The work has one step: divide the expression by 3.
.. -4 > x

Because the symbol is "greater than", the circle at x=-4 is an open circle.

5 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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