What is the answer
5 10 25 100 × ÷ ÷

Answer:

if the terms are approaching zero then it is convergent.

Therefore the stated series is convergent

Step-by-step explanation:

Answer:

310/[tex]3^{3}[/tex] = 310/27 =11.48

Step-by-step explanation:

Answer:

310/ = 310/27 =11.48

Step-by-step explanation:

Answer: 1. When you estimate, it is not an exact measurement. 3ft 8 in gets rounded to 4ft and 12 ft 3 in rounds to 12ft. now find the perimeter. P=2l+2w P= 2*12 +2*4 P=32feet

2. 3ft 8in = 3 8/12 or reduced to 3 2/3  12ft 3in = 12 3/12 or reduced to 12 1/4   The fractional part is referring to a fraction of a foot.

3. The perimeter of the room is P=2l+2w or P=2(12 1/4) + 2(3 2/3) p=24 1/2 + 7 1/3  P= 31 5/6 feet

4. The estimate and the actual are very close. They are 1/6 of a foot apart.

5a. Total baseboard 31 5/6ft - 2 1/4 ft = 29 7/12 feet needed.

5b. Take the total and divide it by 8ft  = 29 7/12 divided by 8= 3.7  You are not buying a fraction of a board so you would need 4 boards.

Answer: 23%

Step-by-step explanation: 77 + 23 is 100

Answer:

77 out of 100

Step-by-step explanation:

For a fraction, it would look like 77/100,

and for percentage, 77%.

Answer:

[tex]-\frac{3(\sqrt{6}-\sqrt{2})}{2}\text{ or } \frac{-3\sqrt{6}+3\sqrt{2}}{2}}\text{ or }\frac{3(\sqrt{2}-\sqrt{6})}{2}[/tex]

Step-by-step explanation:

There are multiple ways to achieve and even express the exact answer to this problem. Because the exact value of [tex]6\cos(105^{\circ}})[/tex] is a non-terminating (never-ending) decimal, it does not have a finite number of digits. Therefore, you cannot express it as an exact value as a decimal, as you'd either have to round or truncate.

Solution 1 (Cosine Addition Identity):

Nonetheless, to find the exact value we must use trigonometry identities.

Identity used:

[tex]\cos(\alpha +\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta[/tex]

Notice that [tex]45+60=105[/tex] and therefore we can easily solve this problem if we know values of [tex]\cos(45^{\circ})[/tex], [tex]\cos(60^{\circ})[/tex], [tex]\sin (45^{\circ})[/tex], and [tex]\sin(60^{\circ})[/tex], which is plausible as they are all key angles on the unit circle.

Recall from either memory or the unit circle that:

Therefore, we have:

[tex]\cos(105^{\circ})=\cos(45^{\circ}+60^{\circ}}),\\\cos(45^{\circ}+60^{\circ}})=\cos 45^{\circ}\cos 60^{\circ}-\sin 45^{\circ}\sin 60^{\circ},\\\cos(45^{\circ}+60^{\circ}})=\frac{\sqrt{2}}{2}\cdot \frac{1}{2}-\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2},\\\cos(105^{\circ})=\frac{\sqrt{2}}{4}-\frac{\sqrt{6}}{4},\\\cos(105^{\circ})={\frac{-\sqrt{6}+\sqrt{2}}{4}}[/tex]

Since we want the value of [tex]6\cos 105^{\circ}[/tex], simply multiply this by 6 to get your final answer:

[tex]6\cdot {\frac{-\sqrt{6}+\sqrt{2}}{4}}=\frac{-3\sqrt{6}+3\sqrt{2}}{2}}=\boxed{\frac{3(\sqrt{2}-\sqrt{6})}{2}}[/tex]

Solution 2 (Combination of trig. identities):

Although less plausible, you may have the following memorized:

[tex]\sin 15^{\circ}=\cos75^{\circ}=\frac{\sqrt{6}-\sqrt{2}}{4},\\\sin 75^{\circ}=\cos15^{\circ}=\frac{\sqrt{6}+\sqrt{2}}{4}[/tex]

If so, we can use the following trig. identity:

[tex]\cos(\theta)=\sin(90^{\circ}-\theta)[/tex] (the cosine of angle theta is equal to the sine of the supplement of angle theta - the converse is also true)

Therefore,

[tex]\cos (105^{\circ})=\sin (90^{\circ}-105^{\circ})=\sin(-15^{\circ})[/tex]

Recall another trig. identity:

[tex]\sin(-\theta)=-\sin (\theta)[/tex] and therefore:

[tex]\sin (-15^{\circ})=-\sin (15^{\circ})[/tex]

Multiply by 6 to get:

[tex]6\cos (105^{\circ})=-6\sin (15^{\circ})=-6\cdot \frac{\sqrt{6}-\sqrt{2}}{4}=\boxed{-\frac{3(\sqrt{6}-\sqrt{2})}{2}}[/tex] (alternative final answer).

Hi there!  

»»————-　★　————-««

I believe your answer is:  

[tex]V=1000\text{mm}^3[/tex]

»»————-　★　————-««  

Here’s why:  

⸻⸻⸻⸻

I am assuming by the infomation given that the figure is a cube.

⸻⸻⸻⸻

[tex]\boxed{\text{Finding the volume of the cube...}}\\\\S = 10mm; V= s^3\\--------------\\\rightarrow V = 10^3\\\\\rightarrow V = 10 * 10 * 10\\\\\rightarrow \boxed{V=1000\text{mm}^3}[/tex]

⸻⸻⸻⸻

»»————-　★　————-««  

Hope this helps you. I apologize if it’s incorrect.  

500 kwh

$105 - $30 = $75

$75 / $0.15 = 500

Answer:

$105-$30 service fee, this leaves only the electricity used. $75. now to find how many kilowatt hours used you divide $75/.15=500 answer 500 kilowatt hours.

Step-by-step explanation:

see above

Answer:

The length of the rectangle is 10 inches, and the width is 4 inches.

Step-by-step explanation:

Given that the length of a rectangle is 6 inches more than the width, and the perimeter is 28 inches, the following calculation must be performed to find the length and the width:

(X + X + 6) x 2 = 28

2X + 2X + 12 = 28

4X = 28 - 12

X = 16/4

X = 4

Therefore, the length of the rectangle is 10 inches, and the width is 4 inches.

Answer:

Line 3

Step-by-step explanation:

→ Calculate gradient

[tex]\frac{6-3}{4-2} =1.5[/tex]

Answer:

Line 3

Step-by-step explanation:

(0,0) & ( 4 , 6)

[tex]Slope = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\=\frac{6-0}{4-0}\\\\=\frac{6}{4}\\\\=\frac{3}{2}\\\\=1\frac{1}{2}[/tex]

Answer:

[tex]CL=76secs[/tex]

Step-by-step explanation:

From the question we are told that:

Traffic Phases p=3

Phase time Lengths

25 seconds, 30 seconds, and 15 seconds

Red Per Phase [tex]T=2sec[/tex]

Generally the equation for Cycle Length is mathematically given by

 [tex]CL=Phase\ Length+Red\ Per\ Phase[/tex]

 [tex]CL=(25+2)+(30+2)+(15+2)[/tex]

 [tex]CL=76secs[/tex]

Answer:length 30, width 18

Step-by-step explanation:

60% +100%=160%

160% × 2 = 320 %

96/320 = 0.3 ×100 =30 ( length)

30 × 0.6 =18 (width)

Check: (18 + 30) 2 = 96

Answer:

60 ft

Step-by-step explanation:

The answer has to be in feet units

Now that we know the height is 5 cm equivalent to 150 feet, what is the width of the building in feet units

5 cm = 150 ft

Rule: multiply cm by 30 to get the ft

2 cm = ?

2 cm × 30 = 60 ft

2 cm = 60 ft

Answer:

There is no significance evidence that students who completed two years of college had a higher average than students who had only completed high school.

Step-by-step explanation:

The hypothesis :

H0 : μ1 = μ2

H1 : μ1 > μ2

Given :

n1 = 35 ; x1 = 75.1 ; s1 = 12.8

n2 = 50 ; x2 = 72.1 ; s2 = 14.6

Pooled variance = Sp² = (df1*s1² + df2*s2²) ÷ (n1 + n2 - 2)

df1 = n1 - 1 = 35 - 1 = 34

df2 = n2 - 1 = 50 - 1 = 49

(x1 - x2) ÷ Sp(√(1/n1 + 1/n2))

Sp² = (34*12.8^2 + 49*14.6^2) / (35+50-2)

Sp² = (5570.56 + 10444.84) / 83

Sp² = 192.95662

Sp = √192.95662

Sp = 13.89

Test statistic = (75.1 - 72.1) / 13.89 * √(1/35 + 1/50)

Test statistic = 3 / (13.89 * 0.2203892)

Test statistic = 0.980

df = n1 + n2 - 2

df = 35 + 50 - 2 = 83

Using the Pvalue calculator :

Pvalue(0.980, 83) = 0.165

α = 0.1

Pvalue > α ; We fail to reject the H0; and conclude that there is no significance evidence that students who completed two years of college had a higher average than students who had only completed high school.

The perimeter of the whole garden would be 66m.

Hope this helps! :)

Answer:

Gary is now 24years

Step-by-step explanation:

let the age of Jan be x and that of Gary be x+15

in six years time they will be as follows

Jan =x+6

Gary=x+15+6=x+21

2(x+6)=x+21

2x+12=x+21

collect the like terms

2x-x=21-12

x=9

Gary =9+15=24years

Hello,

I have only found 113 solutions (i have num 15 given)

nb= 1         ::: 27*n +  98= 30*n + 56===> n= 1  4

nb= 2         ::: 28*n +  95= 30*n + 67===> n= 1  4

nb= 3         ::: 29*n +  65= 30*n + 17===> n= 4  8

nb= 4         ::: 29*n +  65= 30*n + 18===> n= 4  7

nb= 5         ::: 29*n +  65= 30*n + 47===> n= 1  8

nb= 6         ::: 29*n +  65= 30*n + 48===> n= 1  7

nb= 7         ::: 29*n +  74= 30*n + 16===> n= 5  8

nb= 8         ::: 29*n +  74= 30*n + 18===> n= 5  6

nb= 9         ::: 29*n +  74= 30*n + 56===> n= 1  8

nb= 10        ::: 29*n +  74= 30*n + 58===> n= 1  6

nb= 11        ::: 30*n +  16= 29*n + 74===> n= 5  8

nb= 12        ::: 30*n +  17= 29*n + 65===> n= 4  8

nb= 13        ::: 30*n +  18= 29*n + 65===> n= 4  7

nb= 14        ::: 30*n +  18= 29*n + 74===> n= 5  6

nb= 15        ::: 30*n +  47= 29*n + 65===> n= 1  8

nb= 16        ::: 30*n +  48= 29*n + 65===> n= 1  7

nb= 17        ::: 30*n +  56= 27*n + 98===> n= 1  4

nb= 18        ::: 30*n +  56= 29*n + 74===> n= 1  8

nb= 19        ::: 30*n +  58= 29*n + 74===> n= 1  6

nb= 20        ::: 30*n +  67= 28*n + 95===> n= 1  4

nb= 21        ::: 36*n +  97= 40*n + 25===> n= 1  8

nb= 22        ::: 38*n +  59= 40*n + 27===> n= 1  6

nb= 23        ::: 38*n +  65= 40*n + 27===> n= 1  9

nb= 24        ::: 38*n +  69= 40*n + 15===> n= 2  7

nb= 25        ::: 39*n +  78= 45*n + 6===> n= 1  2

nb= 26        ::: 39*n +  82= 40*n + 15===> n= 6  7

nb= 27        ::: 39*n +  82= 40*n + 17===> n= 6  5

nb= 28        ::: 39*n +  82= 40*n + 65===> n= 1  7

nb= 29        ::: 39*n +  82= 40*n + 67===> n= 1  5

nb= 30        ::: 40*n +  15= 38*n + 69===> n= 2  7

nb= 31        ::: 40*n +  15= 39*n + 82===> n= 6  7

nb= 32        ::: 40*n +  17= 39*n + 82===> n= 6  5

nb= 33        ::: 40*n +  25= 36*n + 97===> n= 1  8

nb= 34        ::: 40*n +  27= 38*n + 59===> n= 1  6

nb= 35        ::: 40*n +  27= 38*n + 65===> n= 1  9

nb= 36        ::: 40*n +  65= 39*n + 82===> n= 1  7

nb= 37        ::: 40*n +  67= 39*n + 82===> n= 1  5

nb= 38        ::: 46*n +  87= 50*n + 39===> n= 1  2

nb= 39        ::: 46*n +  87= 52*n + 9===> n= 1  3

nb= 40        ::: 47*n +  68= 50*n + 29===> n= 1  3

nb= 41        ::: 47*n +  83= 50*n + 26===> n= 1  9

nb= 42        ::: 47*n +  98= 51*n + 6===> n= 2  3

nb= 43        ::: 47*n +  98= 53*n + 2===> n= 1  6

nb= 44        ::: 48*n +  63= 50*n + 29===> n= 1  7

nb= 45        ::: 48*n +  73= 52*n + 9===> n= 1  6

nb= 46        ::: 49*n +  63= 51*n + 7===> n= 2  8

nb= 47        ::: 49*n +  72= 53*n + 8===> n= 1  6

nb= 48        ::: 49*n +  78= 52*n + 30===> n= 1  6

nb= 49        ::: 49*n +  87= 56*n + 3===> n= 1  2

nb= 50        ::: 50*n +  26= 47*n + 83===> n= 1  9

nb= 51        ::: 50*n +  29= 47*n + 68===> n= 1  3

nb= 52        ::: 50*n +  29= 48*n + 63===> n= 1  7

nb= 53        ::: 50*n +  39= 46*n + 87===> n= 1  2

nb= 54        ::: 52*n +  30= 49*n + 78===> n= 1  6

nb= 55        ::: 57*n +  92= 63*n + 8===> n= 1  4

nb= 56        ::: 58*n +  72= 60*n + 34===> n= 1  9

nb= 57        ::: 58*n +  73= 60*n + 49===> n= 1  2

nb= 58        ::: 58*n +  79= 60*n + 31===> n= 2  4

nb= 59        ::: 58*n +  97= 60*n + 13===> n= 4  2

nb= 60        ::: 59*n +  47= 62*n + 8===> n= 1  3

nb= 61        ::: 59*n +  71= 60*n + 23===> n= 4  8

nb= 62        ::: 59*n +  71= 60*n + 28===> n= 4  3

nb= 63        ::: 59*n +  71= 60*n + 43===> n= 2  8

nb= 64        ::: 59*n +  71= 60*n + 48===> n= 2  3

nb= 65        ::: 59*n +  74= 63*n + 2===> n= 1  8

nb= 66        ::: 59*n +  78= 61*n + 30===> n= 2  4

nb= 67        ::: 59*n +  84= 61*n + 30===> n= 2  7

nb= 68        ::: 59*n +  87= 61*n + 3===> n= 4  2

nb= 69        ::: 60*n +  13= 58*n + 97===> n= 4  2

nb= 70        ::: 60*n +  23= 59*n + 71===> n= 4  8

nb= 71        ::: 60*n +  28= 59*n + 71===> n= 4  3

nb= 72        ::: 60*n +  31= 58*n + 79===> n= 2  4

nb= 73        ::: 60*n +  34= 58*n + 72===> n= 1  9

nb= 74        ::: 60*n +  43= 59*n + 71===> n= 2  8

nb= 75        ::: 60*n +  48= 59*n + 71===> n= 2  3

nb= 76        ::: 60*n +  49= 58*n + 73===> n= 1  2

nb= 77        ::: 61*n +  30= 59*n + 78===> n= 2  4

nb= 78        ::: 61*n +  30= 59*n + 84===> n= 2  7

nb= 79        ::: 65*n +  89= 70*n + 24===> n= 1  3

nb= 80        ::: 68*n +  59= 72*n + 3===> n= 1  4

nb= 81        ::: 68*n +  91= 70*n + 45===> n= 2  3

nb= 82        ::: 69*n +  43= 70*n + 15===> n= 2  8

nb= 83        ::: 69*n +  43= 70*n + 18===> n= 2  5

nb= 84        ::: 69*n +  43= 70*n + 25===> n= 1  8

nb= 85        ::: 69*n +  43= 70*n + 28===> n= 1  5

nb= 86        ::: 69*n +  48= 72*n + 3===> n= 1  5

nb= 87        ::: 69*n +  52= 70*n + 14===> n= 3  8

nb= 88        ::: 69*n +  52= 70*n + 18===> n= 3  4

nb= 89        ::: 69*n +  52= 70*n + 34===> n= 1  8

nb= 90        ::: 69*n +  52= 70*n + 38===> n= 1  4

nb= 91        ::: 69*n +  54= 71*n + 8===> n= 2  3

nb= 92        ::: 69*n +  58= 73*n + 2===> n= 1  4

nb= 93        ::: 69*n +  82= 75*n + 4===> n= 1  3

nb= 94        ::: 69*n +  85= 74*n + 20===> n= 1  3

nb= 95        ::: 70*n +  14= 69*n + 52===> n= 3  8

nb= 96        ::: 70*n +  15= 69*n + 43===> n= 2  8

nb= 97        ::: 70*n +  18= 69*n + 43===> n= 2  5

nb= 98        ::: 70*n +  18= 69*n + 52===> n= 3  4

nb= 99        ::: 70*n +  24= 65*n + 89===> n= 1  3

nb= 100       ::: 70*n +  25= 69*n + 43===> n= 1  8

nb= 101       ::: 70*n +  28= 69*n + 43===> n= 1  5

nb= 102       ::: 70*n +  34= 69*n + 52===> n= 1  8

nb= 103       ::: 70*n +  38= 69*n + 52===> n= 1  4

nb= 104       ::: 70*n +  45= 68*n + 91===> n= 2  3

nb= 105       ::: 74*n +  20= 69*n + 85===> n= 1  3

nb= 106       ::: 76*n +  93= 80*n + 45===> n= 1  2

nb= 107       ::: 79*n +  45= 82*n + 6===> n= 1  3

nb= 108       ::: 79*n +  54= 83*n + 6===> n= 1  2

nb= 109       ::: 80*n +  45= 76*n + 93===> n= 1  2

nb= 110       ::: 87*n +  64= 90*n + 25===> n= 1  3

nb= 111       ::: 87*n +  65= 90*n + 23===> n= 1  4

nb= 112       ::: 90*n +  23= 87*n + 65===> n= 1  4

nb= 113       ::: 90*n +  25= 87*n + 64===> n= 1  3

[tex] {\bold{\red{\huge{\mathbb{QUESTION}}}}} [/tex]

The lengths of the sides of a triangle are 3, 3, 3√2. Can the triangle be a right triangle?

[tex]\bold{ \red{\star{\blue{TO \: \: PROVE }}}}[/tex]

IF ITS A RIGHT ANGLED OR NOT

[tex]\bold{\blue{\star{\red{FORMULA}}}}[/tex]

IF IT WILL FOLLOW PYTHAGORAS THEOREM THEN IT WILL BE A RIGHT ANGLE TRIANGLE.

[tex]{HYPOTENUSE}^{2} \\ ={ PERPENDICULAR}^{2}+{BASE}^{2} [/tex]

[tex]\bold{ \red{\star{\orange{GIVEN }}}}[/tex]

1ST SIDE -> 3

2ND SIDE -> 3

3RD SIDE ->3√2

[tex] \huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}[/tex]

[tex]{HYPOTENUSE}^{2}\\ ={ PERPENDICULAR}^{2}+{BASE}^{2} \\{h}^{2}={p}^{2}+{b}^{2} [/tex]

AS HYPOTENUSE is always greater than other 2 sides so 3√2 can only be hypotenuse if it's a right angle triangle

[tex]{(3 \sqrt{2})}^{2} = {3}^{2} + {3}^{2} \\ 9 \times2 = 9 + 9 \\ 18 = 18[/tex]

[tex] {\red{\star}}{ \blue{HENCE \: PROVED}} { \red{ \star}}[/tex]

[tex] \red \star{Thanks \: And \: Brainlist} \blue\star \\ \green\star If \: U \: Liked \: My \: Answer \purple \star[/tex]

Answer:

3,199 approximately

Step-by-step explanation:

to find how much 20% of 15,998 does we multiply 15,998 with 20 and then divide it by 100

15,998 x 20 / 100 = 3,199

Step-by-step explanation:

so  the equation in standard form that models the given scenario is

2x + 3y = 24

Answer:

5√2

Step-by-step explanation:

Given √10*√5

Using surd, the expression can be evaluated further as :

√10*√5 = √50

√50 can be expressed as :

√50 = √25*2 = √25 * √2

√25 = 5

Hence,

√50 = √25 * √2 = 5√2

Hence, √10*√5 = 5√2

Answer:

D

Step-by-step explanation:

Number 8 is not related to the x, but is related to the the function. so g(x) is 8 units above f (x)

Answer:

D. 8 units above the graph

Step-by-step explanation:

y = mx + b

this formula is basically y = x^2 + 8

+8 part means where it is on the y axis

if it were y = (x+6)^2 + 8

it would also be on 6 places to the left on the x axis

Answer:

B. Similar - AA

Step-by-step explanation:

Two angles in ∆ABC are congruent to two corresponding angles in ∆DEF. Thus, it follows that the third pair of angles of both triangles would also be congruent.

Therefore, the three sides of ∆ABC and corresponding sides of ∆DEF will be proportional to each other.

This satisfies the AA Similarity Criterion. Therefore, ∆ABC ~ ∆DEF by AA.

Answer:

(2) 2,640 ft

Step-by-step explanation:

I'm going to assume that in this question, you are walking 1 full circle around the reservoir. That would mean you need to calculate the circumference of the circular reservoir.

The circumference formula is:

C = ⫪d

C stands for Circumference

d stands for diameter

I will use 22/7 instead of pi, so the formula looks more like this:

C = (22/7)(d)

The diameter is 840 feet, so we will substitute the variable d with 840:

C = (22/7)(840)

You can plug this part into the calculator, but by hand, it'll look something like this:

(22/7)*840 = (22*840)/7

18,480/7 = 2.640

Hope it helps (●'◡'●)

Answer:

Z = 50

Step-by-step explanation:

Given the following data;

Z = 187

x = 64

y = 6

Translating the word problem into an algebraic expression, we have;

Z = k√x/y

First of all, we would find the constant of proportionality, k;

187 = k√64/6

187 * 6 = k√64

1122 = 8k

k = 1122/8

k = 140.25

To find z, when x and y = 9

Z = 140.25√9/9

Z = (140.25 * 3)/9

Z = 420.75/9

Z = 46.75 ≈ 50

Note: The values in the latter part of the question isn't explicitly stated, so I assumed a value of 9 for both x and y.

remember the pythagorean theorem:

a² + b² = c²

where c is the hypotenuse.

so:

[tex] {a}^{2} + {b}^{2} = { ( \sqrt{26)}}^{2} [/tex]

the square and the square root cancel each other out, so...

a² + b² = 26

we know that a and b are of equal length given the angles.

so it's

[tex] { \sqrt{13} }^{2} + { \sqrt{13} }^{2} = 26[/tex]

here the squares and square roots also cancel, but to keep the equation from the formula true we need to write them. that makes the difference between optional and B

Option A is correct,

[tex] \sqrt{13} inches[/tex]

Answer: Approximately 72.69 meters

Step-by-step explanation:

[tex]sin(42.3)=\frac{opposite}{hypotenuse} =\frac{h}{108} \\\\108*sin(42.3)=h\\\\h=72.685[/tex]

The height of the antenna by using the Pythagoras theorem is 72.68 meters.

"Trigonometry is one of the branches of mathematics that deals with the relationship between the sides of a triangle (right triangle) with its angles".

For the given situation,

Length of guidewire = 108 meters

Angle of elevation = 42.3 degrees

Height of the antenna be 'h'.

By Pythagoras theorem,

[tex]Sine[/tex] θ = [tex]\frac{Perpendicular}{hypotenuse}[/tex]

On substituting the above values,

⇒ [tex]Sine 42.3 = \frac{h}{108}[/tex]

⇒ [tex]0.6730 =\frac{h}{108}[/tex]

⇒ [tex]h=0.6730[/tex] × [tex]108[/tex]

⇒ [tex]h= 72.68[/tex]

Hence we can conclude that the height of the antenna is 72.68 meters.

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Answer:

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Step-by-step explanation:

Answer:

By these examples you are able to see that the square of an odd number is always 1 more than a multiple of 4.

Step-by-step explanation:

For examples,

Let's consider squares of 3, 11, 25, 37 and 131.

[tex] {3}^{2} = 9[/tex]

8 is a multiple of 4, and 9 is more than 8.

[tex] {11}^{2} = 121[/tex]

120 is a multiple of 4 and 121 is one more than it.

[tex] {25}^{2} = 625[/tex]

624 is a multiple of 4 and 625 is one more than it.

[tex] {37}^{2} = 1369[/tex]

1368 is a multiple of 4 and 1369 is one more than 1368.

[tex] {131}^{2} = 17161[/tex]

17160 is a multiple of 4.