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2 years ago

How many squares will it take to fill the large square

Mathematics

1 answer:

anastassius [24]2 years ago

4 0

Answer:

i believe it would take 16. let me know if thats right?

Step-by-step explanation:

You might be interested in

Indicate the equation of the given line in standard form. The line that is the perpendicular bisector of the segment whose endpo

noname [10]

Well the line that bisects RS, will cut RS in two equal halves, therefore, that line will cut RS perpendicularly at the midpoint of RS.

now, what the dickens is the midpoint of RS anyway?

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
R&({{ -1}}\quad ,&{{ 6}})\quad 
% (c,d)
S&({{ 5}}\quad ,&{{ 5}})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left( \cfrac{5-1}{2}~~,~~\cfrac{5+6}{2} \right)\implies \stackrel{midpoint}{\left(2~~,~~\frac{11}{2} \right)}$

so, we know that perpendicular line, will have to go through (2, 11/2)

now, a perpendicular line to RS, will have a negative reciprocal slope to it. Well, what is the slope of RS anyway?

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ -1}}\quad ,&{{ 6}})\quad 
% (c,d)
&({{ 5}}\quad ,&{{ 5}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{5-6}{5-(-1)}\implies \cfrac{5-6}{5+1}\implies -\cfrac{1}{6}$

and let's check the reciprocal negative of that,

$\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad -\cfrac{1}{6}\\\\
slope=-\cfrac{1}{{{ 6}}}\qquad negative\implies +\cfrac{1}{{{ 6}}}\qquad reciprocal\implies + \cfrac{{{ 6}}}{1}\implies 6$

so, then, what's is the equation of a line whose slope is 6, and goes through 2, 11/2?

$\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ 2}}\quad ,&{{ \frac{11}{2}}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 6
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-\cfrac{11}{2}=6(x-2)
\\\\\\
y-\cfrac{11}{2}=6x-12\implies -6x+y=-12+\cfrac{11}{2}\implies \stackrel{\textit{standard form}}{-6x+y=-\cfrac{13}{2}}$

6 0

2 years ago

what is the value of the expression -2 (3x minus 2) + x + 9 when x equals negative 3 choices are a -16 B -2 C 28 D 34 which one

Nezavi [6.7K]

Answer:

C 28

Step-by-step explanation:

Put the given value where x is in the expression and do the arithmetic.

-2(3(-3) -2) +(-3) +9

= -2(-9 -2) -3 +9

= -2(-11) -3 +9

= 22 -3 +9

= 19 +9

= 28

6 0

2 years ago

In a population of 10,000 geese the growth rate is 4% per year. What is the total increase in population over 4 years? Round you

omeli [17]

Answer:

Step-by-step explanation:

10000*4%= 400 Geese year1

10400*4%=416 Geese year 2

10816*4%= 432.64 Geese year 3

11,248.64*4%= 449.95 Geese. year 4

11248.64+449.95= 11698.59

11698.59-10000=1698.59

Increased by 1699 Geese.

3 0

2 years ago

Read 2 more answers

Find the (a) mean, (b) median, (c) mode, and (d) midrange for the data and then (e) answer the given question. Listed below

mafiozo [28]

Answer:

a) $\bar X = 369.62$

b) $Median=175$

c) $Mode =450$

With a frequency of 4

d) $MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

<u>e)</u> $s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

Step-by-step explanation:

We have the following data set given:

49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000

Part a

The mean can be calculated with this formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

Replacing we got:

$\bar X = 369.62$

Part b

Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:

$Median=175$

Part c

The mode is the most repeated value in the sample and for this case is:

$Mode =450$

With a frequency of 4

Part d

The midrange for this case is defined as:

$MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

Part e

For this case we can calculate the deviation given by:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And replacing we got:

$s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

5 0

2 years ago

For a 2 week period Jan earned $150 less than twice toms earnings together Jen and tom earned 1380$ how much did tom earn?

nika2105 [10]

Given:
Tom's earnings: x
Jan's earnings: 2x - 150
Total earnings: 1380

x + 2x - 150 = 1380
3x = 1380 + 150
3x = 1530
3x/3 = 1530/3
x = 510

Tom's earnings: x = 510
Jan's earnings: 2x - 150 = 2(510) - 150 = 1,020 - 150 = 870
total earnings: 1,380

510 + 870 = 1,380
1,380 = 1,380

6 0

2 years ago