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1 year ago

Consider the following "generic rectangle" where the area of each region is given. Write equivalent

Mathematics

1 answer:

marta [7]1 year ago

8 0

Answer:

sum: 3x² -4x -4
product: (x -2)(3x +2)

Step-by-step explanation:

The areas of four regions are given. We can simply add them to find the sum. To express them as a product, we need to look at common factors.

The total of the given area expressions is ...

3x² +2x -6x -4 = 3x² -4x -4 . . . . sum

<h3>Product</h3>

Extending the table to show common factors of each row and column, we have ...

$\begin{array}{c|c|c|}&3x&2\\\cline{1-3}x&3x^2&2x\\\cline{1-3}-2&-6x&-4\\\cline{1-3}\end{array}$

Since each cell of the table is the product of the corresponding common factors, we can write the area as the product ...

(x -2)(3x +2) . . . . product

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A female cockroach can produce 80 offspring at a time. If half are female, and each female produces 80 offspring, how many cockr

IRINA_888 [86]

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3 years ago

Read 2 more answers

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Illusion [34]

Answer:

The answer is "717.25 minutes".

Step-by-step explanation:

In this question first, we arrange the value in ascending order that are:

307,311,321,322,354,363,377,406,425,435,461,464,474,499,505,513,517,534,548

The median is always in an orderly position that is $= \frac{(n+1)}{2}$ .

$\to \frac{(n+1)}{2} = \frac{(20+1)}{2} = 10.5$ position orders

$10^{th} \ and \ 11^{th}$ place an average of observations so, the

$Average = \frac{(425 +435)}{2} = \frac{(860)}{2} =430$

Because as medium stands at 10.5 that median is as below 10 are greater than the level.

$Q_1$ often falls throughout the ordered BELOW average is $= \frac{(n+1)}{2}$ place.

$\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}$ ordered position

Average $5^{th} \ and \ 6^{th}$ location findings

$Q_1 = \frac{(354+363)}{2} = \frac{(717)}{2}= 358.5$

In $= \frac{(n+1)}{2}$ the ordered place, Q3 always falls ABOVE the median.

$\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}$ ordered At median ABOVE

Consequently $Q_3$ falls between $5^{th} \ and \ 6^{th}$ ABOVE the median position

$15^{th}\ and \ 16^{th}$ place average of observations

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$\to IQR = Q_3 - Q_1= 502-358.5= 143.5$

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3 years ago

A spinner has 3 sections colored red, yellow, and blue. The probabilities of the spinner landing on red and yellow or 0.25 and 0

JulijaS [17]

It would be 37.5 times, which rounded up would be about 38 because blue would be .25 of the spinner, so you would expect it to land on blue 25% of the time. Hope this helps!!

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3 years ago

Two brothers want to share a large field equally.

Elan Coil [88]

Yes, the partition gives the two brothers equal shares.

Step-by-step explanation:

Step 1:

Assume the entire field has an area of B. So one brother takes $\frac{1}{12} B$ , $\frac{1}{6}B$ , and $\frac{1}{4} B$

So we need to calculate how much this brother takes in terms of B.

To do this we calculate how much $\frac{1}{12} B + \frac{1}{6} B + \frac{1}{4} B$ is.

Step 2:

To add $\frac{1}{12} B + \frac{1}{6} B + \frac{1}{4} B$ ,

First take the LCM of the denominators 12, 6, and 4

The LCM is 12, we multiply the denominator to get the LCM value, this same value is multiplied to the numerator too.

$\frac{1}{12} B + \frac{1}{6} B + \frac{1}{4} B = \frac{1(1)}{12(1)} B + \frac{1(2)}{6(2)} B + \frac{1(3)}{4(3)} B \\\\\frac{1}{12} B + \frac{1}{6} B + \frac{1}{4} B= \frac{1B +2B+3B}{12} = \frac{B}{2}$