what percentage of 2m is 40cm​

The dimensions of the the plot of land is a length of 19 meters and width of 12 meters.

An equation is an expression that shows the relationship between two or more numbers and variables.

Let x represent the length of the field and y represent the width of the field, hence:

Deion has 62 m of fencing:

2(x + y) = 62

x + y = 31

y = 31 - x     (1)

The area of the land is 228 square meters. hence:

xy = 228

x(31 - x) = 228

x² - 31x + 228 = 0  

x = 12 or x = 19

when x = 12; y = 31 - 12 = 19

when x = 19; y = 31 - 19 = 12

The dimensions of the the plot of land is a length of 19 meters and width of 12 meters.

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It is true that the line segment AB equals 26

The given parameters are:

AB = x +16

BC = 4x + 11

AC = 77

The two-column proof is as follows:

AC = AB + BC                     Line segment formula

77 = x + 16 + 4x + 11            Substitution property of equation

77 = 5x + 27                        Addition property of equation

5x = 50                               Subtraction property of equation  

x = 10                                    Division property of equation

AB = 10 +16                          Substitution property of equation

AB = 26

Hence, the line segment AB has been proved to equal 26

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Answer:

Multiply:

[tex](x+y)by (x+y)[/tex]

[tex] : \implies(x + y)(x + y)[/tex]

[tex] : \implies \: x(x + y) + y(x + y)[/tex]

[tex] : \implies {x}^{2} + xy + xy + {y}^{2} [/tex]

[tex] : \implies{x}^{2} + 2xy + {y}^{2} [/tex]

Multiply:

[tex]a+b \: by \: a^2-b^2[/tex]

[tex]: \implies( {a}^{2} + {b}^{2} ) \times (a + b)[/tex]

[tex]: \implies \: {a}^{2} (a + b) - {b}^{2} (a + b)[/tex]

[tex]: \implies \: {a}^{3} + {a}^{2} b - {ab}^{2} - {b}^{3} [/tex]

Multiply:

[tex](a+5) by (a^2-2a-3)[/tex]

[tex]: \implies{(a + 5) \times ( {a}^{2} - 2a - 3) }[/tex]

[tex]: \implies \: a({a}^{2} - 2a - 3) + 5( {a}^{2} - 2a - 3)[/tex]

[tex]: \implies(a \times {a}^{2} - a \times 2a - a \times 3) + (5 \times {a}^{2} - 5 \times 2a - 5 \times 3)[/tex]

[tex]: \implies{a}^{3} - {2a}^{2} - 3a + 5 {a}^{2} - 10a - 15 [/tex]

[tex]: \implies{ {a}^{3} + {3a}^{2} - 13a - 15}[/tex]

Multiply:

[tex](a^2-ab+b^3) by (a+b)[/tex]

[tex]: \implies{(a + b) \times ( {a}^{2} - ab + {b}^{3} )}[/tex]

[tex]: \implies \: a( {a}^{2} - ab + {b}^{3}) + b( {a}^{2} - ab + {b}^{3} ) [/tex]

[tex]: \implies {a}^{3} - {a}^{2} b + a {b}^{3} + {a^2b} - {ab}^{2} + {b}^{4} [/tex]

[tex]: \implies{ {a}^{3}+ab^3 - ab^2+ {b}^{4} }[/tex]

Step-by-step explanation:

[tex] \blue{ \frak{Seolle_{aph.rodite}}}[/tex]

Answer:

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Answer:

The IQR of the given distribution is

Step-by-step explanation:

The given distribution has the five-number

28, 34, 43, 59, 62

Divide these numbers in two equal parts.

(28, 34), 43,( 59, 62)

Now divide each parenthesis in two equal parts.

(28), (34), 43,( 59), (62)

It means first quartile is the average of 28 and 34. Third quartile is the average of 59 and 62.

The interquartile range (IQR) of this distribution is

Therefore the IQR of the given distribution is 29.5.

Answer:

Hi,

Step-by-step explanation:

sin(a-b)=sin(a) cos(b)+ cos(a) sin(b)

a=45° and b=30°

[tex]sin(45^o-30^o)=sin(45^o)*cos(30^o)+cos(45^o)*sin(30^o)\\\\=\dfrac{\sqrt{2}}{2}* \dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{2}*\dfrac{1}{2}\\\\\\\boxed{sin(15^o)=\dfrac{\sqrt{2}}{4}*(1+\sqrt{3} )}\\[/tex]

Answer:

2

Step-by-step explanation:

the same as...

(2) is the answer

If the population is highly skewed, the sample size needed for the central limit theorem to apply usually has to be the same as that when the population is not highly skewed.

The central limit theorem states in probability theory that, in many instances, when independent random variables are added together, their correctly normalized sum tends toward a normal distribution, even if the original variables are not normally distributed.

If the population is highly skewed, the sample size needed for the central limit theorem to apply usually has to be the same as that when the population is not highly skewed.

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The total area with land based on the information given is 3x² + 60x.

Let the width = x

Let the length = (1.5 × x) = 1.5x

Area = 1.5x × x = 1.5x²

The area along with land:

Width = x + 20

Length = 3 × x = 3x

The total area with land:

= Length × Width

= 3x(x + 20)

= 3x² + 60x.

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The length of the intervals of the given frequency table is

500 ft.

The given list shows the lengths in feet of the 25 longest bridges in the United States.

we have to determine the median and mode of the bridge data.

The median is the middle number of the data set arranged in ascending order.

1010, 1053, 1200, 1207, 1380, 1470, 1495, 1596, 1600, 1600, 1600, 1632, 1750, 1800, 1850, 2000, 2150, 2150, 2300, 2310, 2800, 3500, 3800, 4200, 4260.

The median is 1750.

Mode is the number that appears most often in the data set.

1010, 1053, 1200, 1207, 1380, 1470, 1495, 1596, 1600, 1600, 1600, 1632, 1750, 1800, 1850, 2000, 2150, 2150, 2300, 2310, 2800, 3500, 3800, 4200, 4260

The mode is 1600.

Therefore the correct option is Median: 1750, Mode: 1600

So the length of the intervals of the given frequency table is

500 ft.

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[tex]\text{Nth term of an arithmetic series} = a +(n-1)d \\\\\text{Nth term of an geometric series}= ar^{n-1}\\\\\text{where,}\\\\\text{a = first term.}\\\\\text{d = common difference.}\\\\\text{r = common ratio.}[/tex]

Answer:

[tex]31.8\%[/tex]

Step-by-step explanation:

The area of the circle is [tex]A=\pi r^2=\pi(4)^2=16\pi[/tex]

The area of the triangle is [tex]A=\frac{bh}{2}=\frac{8*4}{2}=\frac{32}{2}=16[/tex]

Hence, the probability of a randomly selected point within the circle falls in the red shaded area is [tex]\frac{16}{16\pi}=\frac{1}{\pi}\approx0.318\approx31.8\%[/tex]

Answer: Yes it is possible

Example: Yeast is a single-celled fungus.

There are probably other types of fungus that are single-celled. However, some other fungi are multi-celled. You will likely need more information about the organism under the microscrope before you can classify it properly.

An x-intercept of the continuous function in the table is (-1, 0)

The x-intercept of a line is the point where the line crossed the x-axis or the  point where the value of y is zero.

From the table, the x-intercept are all the point where the value of f(x) is zero. Hence the Which is an x-intercept of the continuous function in the table is (-1, 0)

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The probability that a student is a female or major in civil engineering is 62%

At a certain college, 49% of the students are female, and 21% of the students major in civil engineering. Furthermore, 8% of the students both are female and major in civil engineering. What is the probability that a randomly selected female student majors in civil engineering?

Let A represent Female and B represents civil engineering.

The above representation means that the given parameters are:

The required probability is calculated as:

P(A or B) = P(A) + P(B) - P(A and B)

This gives

P(A or B) = 49% + 21% - 8%

Evaluate

P(A or B) = 62%

Hence, the probability that a student is a female or major in civil engineering is 62%

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Answer:

67m²

Step-by-step explanation:

12m + 6m + 4m + 36 m 8m = 67m²

Considering the given function, the ordered pair (5,8) means that the signed up for 5 weeks of yoga classes and 8 weeks of kickboxing classes.

The function that represents the relationship between the number x of yoga classes that Donovan signs up for and the number y of kickboxing classes is given by:

8x + 12y = 136.

Hence the ordered pair (5,8) means that the signed up for 5 weeks of yoga classes and 8 weeks of kickboxing classes.

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Answer: 0.5 or - 0.834

Step-by-step explanation: Here is the explanation!

Answer:

Yes, x = 0 is a solution to the given equation

Step-by-step explanation:

Hi Student!

The goal of this question is to determine x = 0 is a solution of the expression that was provided.  The first step that we must take is input 0 into all of the x's that we have in the expression.  Then we just simplify both sides and determine if the end expression is true and if it is then x = 0 is a solution.

Plug in the values

Simplify both sides

Looking at the final expression, we can see that 1 is indeed equal to 1 and since the expression is true, we can say that x = 0 is a solution of the expression that was provided in the problem statement.

Answer:

[tex]x=\frac{3}{4},\:x=-1[/tex]

Keys:

For this problem, you need the quadratic formula(listed below).

When you see ± in a quadratic equation, you must know there is going to be at least 2 solutions.

Step-by-step explanation:

solving for x₁ and x₂

[tex]4x^2+x=3\\4x^2+x-3=3-3\\4x^2+x-3=0\\x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot 4\left(-3\right)}}{2\cdot 4}\\[/tex]

[tex]1^2=1\\=\sqrt{1-4\cdot \:4\left(-3\right)}\\=\sqrt{1+4\cdot \:4\cdot \:3}\\=\sqrt{1+48}\\=\sqrt{49}\\=\sqrt{7^2}\\\sqrt{7^2}=7\\=7[/tex]

[tex]x_{1,\:2}=\frac{-1\pm \:7}{2\cdot \:4}\\x_1=\frac{-1+7}{2\cdot \:4},\:x_2=\frac{-1-7}{2\cdot \:4}\\[/tex]

solve for x₁

[tex]\frac{-1+7}{2\cdot \:4}[/tex]

[tex]=\frac{6}{2\cdot \:4}[/tex]

[tex]=\frac{6}{8}[/tex]

[tex]= \frac{6\div2}{8\div2}[/tex]

[tex]=\frac{3}{4}[/tex]

solve for x₂

[tex]\frac{-1-7}{2\cdot \:4}[/tex]

[tex]=\frac{-8}{2\cdot \:4}[/tex]

[tex]=\frac{-8}{8}[/tex]

[tex]=-\frac{8}{8}[/tex]

[tex]=-1[/tex]

Hope this helps!

(1) The missing term in the sequence, a₁₂  = 0.8.

(2) The missing term in the sequence, a₈ = 102.5.

(3) The missing term in the sequence, a₈ = 111.

(4) The missing term in the sequence, a₁₂ = -19.

(5) The missing term in the sequence, a₁₂ = 94.

(6)  The missing term in the sequence, a₆ = 40.

(7)  The missing term in the sequence, a₃₆ = -52.

(8)  The missing term in the sequence, a₂₁ = -58.

The missing term in the sequence is determined as follows;

Tₙ = a + (n - 1)d

T₄ = a + 3d

18.4 = a + 3d  ---(1)

T₅ = a + 4d

16.2 = a + 4d  ---(2)

subtract (1) from (2)

-2.2 = d

18.4 = a + 3(-2.2)

a = 25

a₁₂  = a + 11d

a₁₂  = 25 + 11(-2.2)

a₁₂  = 0.8

a₂ = a + d

57.5 = a + d -- (1)

a₅ = a + 4d

80 = a + 4d  --- (2)

solve (1) and (2)

d = 7.5

a = 50

a₈ =  a + 7d

a₈ = 50 + 7(7.5)

a₈ = 102.5

a₁₀ = a + 9d

141 = a + 9d --- (1)

a₁₃ = a + 12d

186 = a + 12d --- (2)

Subtract (1) from (2)

d = 15

a = 6

a₈ = a + 7d

a₈ = 6 + 7(15)

a₈ = 111

a₂₂ = a + 21d

-49 = a + 21d ---- (1)

a₂₅ = a + 24d

-58 = a + 24d --- (2)

subtract (1) from (2)

d = -3

a = 14

a₁₂ = a + 11d

a₁₂ = 14 + 11(-3)

a₁₂ = -19

a₄ = a + 3d

-2 = a + 3d --- (1)

a₈ = a + 7d

46 = a + 7d ---- (2)

Subtract (1) from (2)

d = 12

a = -38

a₁₂ = a + 11d

a₁₂ = -38 + 11(12)

a₁₂ = 94

a₉ = a + 8d

64 = a + 8d --- (1)

a₁₂ = a + 11d

88 = a + 11d --- (2)

Subtract (1) from (2)

d = 8

a = 0

a₆ = a + 5d

a₆ = 0 + 5(8)

a₆ = 40

a₂₀ = a + 19d

-4 = a + 19d ---- (1)

a₂₃ = a + 22d

-13 = a + 22d --- (2)

Subtract (1) from (2)

d = -3

a = 53

a₃₆ = a + 35d

a₃₆ = 53 + 35(-3)

a₃₆ = -52

a₂₈ = a + 27d

5 = a + 27d ---- (1)

a₃₃ = a + 32d

50 = a + 32d --- (2)

Subtract (1) from (2)

d = 9

a = -238

a₂₁ = a + 20d

a₂₁ = -238 + 20(9)

a₂₁ = -58

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Given the diameter of the surface of the clock, the area of the surface of the alarm clock is 3846.5cm².

Note that: Area of a circle is expressed as;

A = πr²

Where r is radius and π is constant pi ( π = 3.14 )

Given that;

A = πr²

A = 3.14 × ( 35cm )²

A = 3.14 × 1225cm²

A = 3846.5cm²

Therefore, given the diameter of the surface of the clock, the area of the surface of the alarm clock is 3846.5cm².

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Answer: x=15, y=3

Step-by-step explanation:

Subtracting the two equations, we get -3y=-9, meaning y=3.

Substituting this into the first equation, we get that x-3=12, and thus, x=15.

Answer:

y = 9

Step-by-step explanation:

given that y varies inversely as x then the equation relating them is

y = [tex]\frac{k}{x}[/tex] ← k is the constant of variation

to find k use the condition y = 12 when x = 6

12 = [tex]\frac{k}{6}[/tex] ← multiply both sides by 6 to clear the fraction

72 = k

y = [tex]\frac{72}{x}[/tex] ← equation of variation

when x = 8 , then

y = [tex]\frac{72}{8}[/tex] = 9

Answer:

B

Step-by-step explanation:

If the definitions of type I and type II regions is the same as in the link provided, then as a type I region the integration domain is the set

[tex]R_{\rm I} = \left\{(x,y) \mid 0 \le x \le 3 \text{ and } \sqrt{x+1} \le y \le 2\right\}[/tex]

and as a type II region,

[tex]R_{\rm II} = \left\{(x,y) \mid 0 \le x \le y^2-1 \text{ and } 1 \le y \le 2\right\}[/tex]

where we solve y = √(x + 1) for x to get x as a function of y.

A. The area of the type I region is

[tex]\displaystyle \iint_{R_{\rm I}} dA = \int_0^3 \int_{\sqrt{x+1}}^2 dy \, dx = \int_0^3 (2 - \sqrt{x+1}) \, dx = \boxed{\frac43}[/tex]

B. The area of the type II region is of course also

[tex]\displaystyle \iint_{R_{\rm II}} dA = \int_1^2 \int_0^{y^2-1} dx \, dy = \int_1^2 (y^2-1) \, dy = \boxed{\frac43}[/tex]

I've attached a plot of the type II region to give an idea of how it was determined. The black arrows indicate the domain of x as it varies from the line x = 0 (y-axis) to the curve y = √(x + 1).

Answer:

175 ft^2

Step-by-step explanation:

split the pentagon into 5 triangles with base length 10ft and height 7ft. each triangle then has an area of 10 * 7 * 1/2 = 35 ft^2

then the pentagon has an area 35*5 = 175 ft^2