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Novosadov [1.4K]
3 years ago
9

WIL GIVE BRAINLIEST!

Mathematics
1 answer:
Grace [21]3 years ago
5 0

Answer:  4x - 13

<h2>Step-by-step explanation: </h2>

<em>First, we need to get triple of x:</em>

Triple of x is: 3x

<em>Next is to get the difference between 3x and 13:</em>

x + 3x -13

4x - 13

Hence, the variable expression is 4x - 13

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This problem is correct??
ale4655 [162]

yes you are correct  evreything is correct

6 0
3 years ago
5270÷312 with remainder​
aleksley [76]

This is the answer for the calculation

6 0
3 years ago
Read 2 more answers
I WILL GIVE 100 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT.What is the solution to the inequality 3+5/7x&gt;1+x?
harina [27]

Answer:

C: X <= 7

Step-by-step explanation:

Reimagine the inequality as

3 - 1 >= X - 5/7*X

2 >= 2/7*X

1 >= 1/7 * X

7 >= X

8 0
2 years ago
2x^2+5x+3/x^2-3x-4 divided by 4x^2+2x-6/x^2-8x+16
Masja [62]

Answer: \frac{x-4}{2x-2}

Step-by-step explanation:

\frac{2x^2+5x+3}{x^2-3x-4} divided by \frac{4x^2+2x-6}{x^2-8x+16} is the same thing as multiplying \frac{2x^2+5x+3}{x^2-3x-4} by \frac{x^2-8x+16}{4x^2+2x-6}.

The equation we get is:

\frac{2x^2+5x+3}{x^2-3x-4}*\frac{x^2-8x+16}{4x^2+2x-6}

With this equation, we can factor each equation:

\frac{(x+1)(2x+3)}{(x+1)(x-4)} *\frac{(x-4)(x-4)}{2(x-1)(2x+3)}

We can cancel out like terms since they would be dividing each other:

\frac{x-4}{2x-2}

8 0
3 years ago
Imagine a pond. In it sits one lilypad, which reproduces once a day. Each of its offspring also reproduces once a day, doubling
sergeinik [125]

Answer: On the 29th day

Step-by-step explanation:

According to this problem, no lilypad dies and the lilypads always reproduce, so we can apply the following reasoning.

On the first day there is only 1 lilypad in the pond. On the second day, the lilypad from the first reproduces, so there are 2 lilypads. On day 3, the 2 lilypads from the second day reproduce, so there are 2×2=4 lilypads. Similarly, on day 4 there are 8 lilypads. Following this pattern, on day 30 there are 2×N lilypads, where N is the number of lilypads on day 29.

The pond is full on the 30th day, when there are 2×N lilypads, so it is half-full when it has N lilypads, that is, on the 29th day. Actually, there are 2^{30} lilypads on the 30th, and 2^{29} lilypads on the 29th. This can be deduced multiplying succesively by 2.  

4 0
3 years ago
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