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2 years ago

What’s the area of the rectangle in terms of x if the length is (x+8) and the width is (2x-1)?

Mathematics

1 answer:

WARRIOR [948]2 years ago

5 0

Answer:

Step-by-step explanation:

$A=LW\\ \\ A=(x+8)(2x-1)\\ \\ A=2x^2-x+16x-8\\ \\ A=2x^2+15x-8$

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3 years ago

A number, y, is equal to twice the sum of a smaller number and 3. The larger number is also equal to 5 more than 3 times the sma

Nata [24]

Answer:

The equations represent the situation is $y=2x+6$ and $y=5+3x$ .

Step-by-step explanation:

Consider the provided information.

First convert the statement into mathematical representation.

Consider the smaller number is x and the larger number is y.

y, is equal to twice the sum of a smaller number and 3.

This information can be written as: $y=2(x+3)$

Which can be further solve as:

$y=2x+6$

The larger number is also equal to 5 more than 3 times the smaller number.

This information can be written as: $y=5+3x$

Thus, the equations represent the situation is $y=2x+6$ and $y=5+3x$ .

Which can be further solve as:

$2x+6=5+3x$

$x=1$

Now, substitute the value of x in $y=5+3x$ .

$y=5+3$

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3 0

3 years ago

Read 2 more answers

HELPP!!

SSSSS [86.1K]

Answer:

There are 15 integers between 2020 and 2400 which have four distinct digits arranged in increasing order.

Step-by-step explanation:

This can be obtained by after a simple counting of number from 2020 and 2400 as follows:

The first set of integers are:

2345, 2346, 2347, 2348, and 2349.

Therefore, there are 6 integers in first set.

The second set of integers are:

2356, 2357, 2358, and 2359.

Therefore, there are 4 integers in second set.

The third set of integers are:

2367, 2368, and 2369.

Therefore, there are 3 integers in third set.

The fourth set of integers are:

2378, and 2379.

Therefore, there are 2 integers in fourth set.

The fifth and the last set of integer is:

2389

Therefore, there is only 1 integers in fifth set.

Adding all the integers from each of the set above, we have:

Total number of integers = 6 + 4 + 3 + 2 + 1 = 15

Therefore, there are 15 integers between 2020 and 2400 which have four distinct digits arranged in increasing order.

7 0

3 years ago