A wire of length 300m is used to fence a rectangular plot of land.The length of the land is twice its width. Find the length and the width of the land. ​

Answer: 1.89 minutes

Step-by-step explanation: If we take 1 and 1/3 hours and turn that into minutes we get 80 minutes we then divide it by the number of miles 42 1/3. We would get about 1.89, so it took about 1.89 minutes

Answer:

60

Step-by-step explanation:

100 is the greatest value, 40 is the lowest 100-40=60

Answer: 60 i think

Step-by-step explanation:

a. The limit as x approaches 3 from the positive side of √(x² - 9) is 0. b. The limit as x approaches 0 of |x| is 0. c. The limit as x approaches 4 of (x² - 16)/(x³ - 64) is indeterminate (0/0). d. The limit as x approaches a of (√x)/(x - a²) is (√a)/(a - a²).

Here are the step-by-step calculations for each limit

a. lim x→3+ √(x² - 9):

Substitute x = 3 into the expression:

lim x→3+ √(x² - 9) = √(3² - 9) = √(9 - 9) = √0 = 0

b. lim x→0 |x|:

Substitute x = 0 into the expression:

lim x→0 |x| = |0| = 0

c. lim x→4 (x² - 16)/(x³ - 64):

Substitute x = 4 into the expression:

lim x→4 (x² - 16)/(x³ - 64) = (4² - 16)/(4³ - 64) = (16 - 16)/(64 - 64) = 0/0

The result is an indeterminate form, meaning we need further manipulation or information to evaluate the limit.

d. lim x→a (√x)/(x - a²):

Substitute x = a into the expression:

lim x→a (√x)/(x - a²) = (√a)/(a - a²)

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Answer:

See below.

Step-by-step explanation:

So first, we can separate the entire figure into a semi-circle and an isosceles triangle.

AREA:

The area for a semi-circle is \(\frac{1}{2}\pi r^2\).

The diameter is 8cm, so the radius is 4cm.

Area of the semi-circle is:

\(\frac{1}{2}(4)^2\pi=\frac{1}{2}(16\pi)=8\pi cm^2\)

The area for a triangle is \(\frac{1}{2}bh\).

The base is the same as the diameter (8), and we are given the height as 10. Thus:

\(\frac{1}{2} (8)(10)=8(5)=40cm^2\)

The total area is \((8\pi +40 )cm^2\)

PERIMETER:

The perimeter of a semicircle is: \(\pi r + 2r\) (this is derived from dividing the circumference by 2 and then adding on the diameter).

Thus, the perimeter is:

\(4\pi +8\)

However, we ignore the 8 since the 8 is not part of the perimeter.

The perimeter of the triangle is the two slant lengths. We know the base and the height, so we can use the Pythagorean Theorem:

\(a^2+b^2=c^2\)

\(4^2+10^2=c^2\)

\(c^2=116\)

\(c=\sqrt{116}=2\sqrt{29\)

Two of them will be \(4\sqrt{29}\)

Thus, the total perimeter is \(4\pi + 4\sqrt{29}\)

Answer:

Im not sure but hope you find it!!

Step-by-step explanation:

Answer:

105

Step-by-step explanation:

douwdek0 The Titanic could stay afloat with four of its 16 watertight compartments flooded, more than anyone could imagine on a ship of its size. ... As the ice bumped along its starboard side, it punched holes in the ship's steel plates, flooding six compartments. In a little over two hours, the Titanic filled with water and sank. the answer is 128 if u read all this but brainlest????

Total distance of the triathlon race is 23.34 kilometers.

Reg enters a triathlon race. He swims 1.25 kilometers, bikes 20.5 kilometers, and runs 1.59 kilometers.

The total distance of the race is 23.34 kilometers.

Reg enters a triathlon race.

He swims 1.25 kilometers, bikes 20.5 kilometers, and runs 1.59 kilometers.

In order to calculate the total distance of the race, we need to find the sum of the distance Reg swam, biked and ran. Therefore, Total distance = Distance swam + Distance biked + Distance ranTotal distance = 1.25 + 20.5 + 1.59 km

Total distance = 23.34 km.

The summary of this problem is that the total distance of the triathlon race is 23.34 kilometers.

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Answer:

1.42 minutes

Step-by-step explanation:

There are 4 total swimmers, who all swam an equal part of the distance.

The overal time was 5.69 minutes, so the average time is

5.69 min ÷ 4 = 1.42 minutes, on average.

Answer:

Step-by-step explanation:

x<−4
hope this helps

Answer:

54.87

69.08

Step-by-step explanation:

Answer:

9.4.6=216

216.1,3=280

Step-by-step explanation:

\( \huge\hookrightarrow \mathbb{ \color{gold}Answer}\)

We know,

\( \boxed{volume = l \times b \times h}\)

so,

\(\hookrightarrow{volume = 6 \times 4 \times 9}\)

\(\hookrightarrow volume = 216 \: m {}^{3} \)

now,

\( \boxed{density = \frac{mass}{volume} }\)

So,

\( \hookrightarrow1.3 = \dfrac{m}{216} \)

\(\hookrightarrow \: m = 216 \times 1.3\)

\( \hookrightarrow \: m = 280.8\)

hence, mass of air in the room is 280.8 kg

Here is the basic salary-to-hourly formula for this method: First, divide the employee’s annual salary by 52 weeks (the number of weeks in a year). Divide the weekly wages by 40 hours.

The parachutist started the free fall at a height of 68.12 meters above the ground.(b) The height at which the fall begins is 68.12 meters. The units of height are meters.

Given parameters: Initial velocity (u) = 0 m/s

           Final velocity (v) = 2.8 m/s

           Initial height (h) = ?

           Final height (s) = 65 m

Acceleration (a) = 2.4 m/s²Time taken (t) = ?

The formula for calculating the time taken in a free fall with an initial velocity is given by;

                                   v = u + at

Wherev = final velocityu = initial velocity

                       a = accelerationt = time taken

Rearranging the formula to get the time taken;

                       => t = (v - u) / a=> t = (2.8 - 0) / 2.4=> t = 1.17 seconds

(a) The parachutist is in the air for 1.17 seconds.

Since the parachutist starts from rest at a height (h) above the ground, her final velocity can be calculated using the formula;

                                 v² = u² + 2asWherev = final velocityu = initial velocitya = acceleration due to gravitys = height fallen

Rearranging the formula to get the initial height (h)

                                          ;=> h = s - u² / 2a

                                            => h = 65 - 0² / 2(2.4)

                                            => h = 68.12 meters (approx)

Therefore, the parachutist started the free fall at a height of 68.12 meters above the ground.(b) The height at which the fall begins is 68.12 meters. The units of height are meters.

Therefore, the answer in (a) is 1.17 seconds (s), while the answer in (b) is 68.12 meters (m).

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Answer:

Step-by-step explanation:

The probability of a continuous random variable taking any specific value is always zero, so the statement p(x = 15) = 0.05 is false.

To make the addition problem 997 + 543 easy to calculate mentally, we can use the property of arithmetic called regrouping or carrying.

By regrouping the numbers based on their place values, we can simplify the addition process.

To add 7 + 5, we can use the "make-a-ten" strategy. Since 7 + 5 equals 12, we can think of it as taking 2 from the units place and carrying it over to the tens place. This results in 1 being added to the tens place, and the units place becomes 2. Therefore, 7 + 5 = 12.

Equations:

7 + 5 = 12

7 = 7 + 0 (regrouping)

12 = 10 + 2 (regrouping)

To compute 234 - 47 using the standard algorithm, we can use base-ten drawings alongside.

   2 3 4

-    4 7

__________

   1 8 7

We start by subtracting 7 from 4 in the ones place, which requires regrouping. We take 10 from the tens place and add it to the ones place, resulting in 14 - 7 = 7. Then, we subtract 4 from 3 in the tens place, resulting in 3. Finally, we subtract 0 from 2 in the hundreds place, resulting in 2. Therefore, 234 - 47 = 187.

To compute 23 + 19 using the standard algorithm, we can use base-ten drawings alongside.

  2 3

+   1 9

________

   4 2

We start by adding 9 to 3 in the ones place, resulting in 12. Then, we add 1 to 2 in the tens place, resulting in 3. Therefore, 23 + 19 = 42.
Therefore, to simplify the mental calculation of 997 + 543, we can utilize the arithmetic property of regrouping or carrying.

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Complete question here:
Show how to use a property of arithmetic to make the addition problem 997+543 easy to calculate mentally. Write equations to show your use of a property of arithmetic. State the property you use and show where you use it. Draw a diagram and use it to explain how to use the "make-a-ten" strategy to add 7 + 5. Write a series of equations that go with the steps in your diagram. Write equations for this diagram. Use base-ten drawings along side the standard algorithm to diagram the regroupings necessary to compute 234 - 47 using the standard algorithm. Clearly indicate your regroupings and results. You may use several diagrams or one - as long as you are clear. Use base-ten drawings along side the standard algorithm to diagram the regroupings necessary to compute 23 + 19 using the standard algorithm. Clearly indicate your regroupings and results. You may use several diagrams or one - as long as you are clear.

Using quantifiers; a) ∀ student ∈ this class, ∃ exactly 2 mathematics classes ∈ this school that the student has taken and b) ∃ person, ∀ country ∈ the world (country ≠ Libya), the person has visited that country.

a) "Every student in this class has taken exactly two mathematics classes at this school."

In this statement, we have two main quantifiers:

Universal quantifier (∀): This quantifier denotes that we are making a statement about every individual student in the class. It indicates that the following condition applies to each and every student.

Existential quantifier (∃): This quantifier indicates the existence of something. In this case, it asserts that there exists exactly two mathematics classes at this school that each student has taken.

So, when we combine these quantifiers and their respective conditions, we get the statement: "For every student in this class, there exists exactly two mathematics classes at this school that the student has taken."

b) "Someone has visited every country in the world except Libya."

In this statement, we also have two main quantifiers:

Existential quantifier (∃): This quantifier signifies the existence of a person who satisfies a particular condition. It asserts that there is at least one person.

Universal quantifier (∀): This quantifier denotes that we are making a statement about every individual country in the world (excluding Libya). It indicates that the following condition applies to each and every country.

So, when we combine these quantifiers and their respective conditions, we get the statement: "There exists at least one person who has visited every country in the world (excluding Libya)."

In summary, quantifiers are used to express the scope of a statement and to indicate whether it applies to every element or if there is at least one element that satisfies the given condition.

Therefore, Using quantifiers; a) ∀ student ∈ this class, ∃ exactly 2 mathematics classes ∈ this school that the student has taken and b) ∃ person, ∀ country ∈ the world (country ≠ Libya), the person has visited that country.

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Answer:

8a^3

Step-by-step explanation:

The area of a square can be found by using the following expression:

where "A" is the area and "l" is the length of each side of the square. The square on this problem has an area of 9 cm², so we can find the sides by replacing A for 9 on the expression above.

To solve it we need to take the square root on both sides of the equation.

Each side of the square has a length of 3 cm, so the sides are 3 cm x 3 cm.

\(\qquad\qquad\huge\underline{{\sf Answer}}♨\)

Here's the solution ~

As we know, we can calculate the circumference of a circle in terms of its diameter as :

\(\qquad \sf  \dashrightarrow \:c = \pi d\)

where, c = circumference and d = diameter

And also, circumference of circle is terms of radius (r) is :

\(\qquad \sf  \dashrightarrow \:c = 2\pi r\)

Now, let's move on to questions ~

\(\qquad \sf  \dashrightarrow \:3.14 \times 5.9 \)

\(\qquad \sf  \dashrightarrow \: \approx18.53 \: ft\)

・ ．━━━━━━━†━━━━━━━━━．・

\( \qquad \sf  \dashrightarrow \: \approx10.048 \: ft\)

・ ．━━━━━━━†━━━━━━━━━．・

\(\qquad \sf  \dashrightarrow \:3.14 \times 6.1\)

\(\qquad \sf  \dashrightarrow \: \approx19.15 \: ft\)

・ ．━━━━━━━†━━━━━━━━━．・

\(\qquad \sf  \dashrightarrow \:2×3.14 \times 3.7\)

\(\qquad \sf  \dashrightarrow \: \approx2×11.62 \: m\)

\(\qquad \sf  \dashrightarrow \: \approx23.24 \: m\)

・ ．━━━━━━━†━━━━━━━━━．・

\(\qquad \sf  \dashrightarrow \:2×3.14 \times 6.2\)

\(\qquad \sf  \dashrightarrow \: \approx2× 19.47 \: units\)

\(\qquad \sf  \dashrightarrow \: \approx38.94 \: m\)

・ ．━━━━━━━†━━━━━━━━━．・

\(\qquad \sf  \dashrightarrow \:2×3.14 \times 5.1\)

\(\qquad \sf  \dashrightarrow \: \approx2×16.01 \: ft\)

\(\qquad \sf  \dashrightarrow \: \approx \: 32.02m\)

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

Given,

7p-11p+3>3-4p

-4p+3>3-4p

-4p>-4p

which is absured.

So there is no solution of the inequality.

===============================================

Explanation:

Comparing y = x^2+bx+c to y = ax^2+bx+c, we see that a = 1.

The vertex given is (-2,3). In general, the vertex is (h,k). So h = -2 and k = 3.

Plug those three values into the vertex form below

y = a(x-h)^2 + k

y = 1(x-(-2))^2 + 3

y = (x+2)^2 + 3

Then expand everything out and simplify

y = x^2+4x+4 + 3

y = x^2+4x+7

We see that b = 4 and c = 7.

The boat's distance and bearing from the port regions are as follows;

The boat is located around 157.15 kilometers from the harbor on a bearing of 150°.

In mathematics, a bearing is an angle measured clockwise from the north or y-axis direction.

The Coast Guard cutter travels at a speed of 30 kno

Angelica, a Coast Guard cutter, has a bearing of 95 degrees.

The coast guard goes for 2 hours in the first direction.

The shift in direction of the Coast Guard cutter = 185°

Travel time in the new direction = 2 hours

30 knots equals 55.56 km/hr

The distance traveled in the first direction = 55.56 km/hr = 111.12 km in 2 hours

Distance traveled in the other direction = 55.56 km/hr 2 hours = 111.12 km

According to the accompanying graphic, the figure produced by the Coast Guard cutter's course is an isosceles triangle.

The Pythagorean theorem gives the distance from the port to the position of the boat as l = √(111.122^2 + 111.122^2) ≈ 157.15.

The boat is roughly 157.15 kilometers away from the port.

An isosceles right triangle has 45° base angles.

As a result, the bearing is 95° + 45° = 150°.

The angle between the port and the boat is 150°.

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Answer:

x > -25.

Step-by-step explanation:

-39 < 5x + 36

Subtract 36 from both sides:

-39 - 36 < 5x + 36 - 36

-75 < 5x

5x > -75

Divide both sides by 5:

x > -25.

Answer:

The second one is 33

The first one is 17

Step-by-step explanation:

For the first one your just adding three the second one your just adding 8

Let,

Now,

The probability is:

→ \(P(xq) = 0.35\times 0.15\)

             \(= 0.0525\)

→ \(P(yp) = (1-0.35)\times 0.85\)

             \(= 0.65\times 0.85\)

             \(= 0.5525\)

hence,

The likelihood that perhaps the factory's carbon emissions were well within the allowable range:

= \(\frac{0.5525}{(0.5525+0.0525)}\)

= \(\frac{0.5525}{0.605}\)

= \(0.9132\)

Thus the above response is right.

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Answer: 0.098%

Step-by-step explanation:

The triangle is obtuse, since the square of the longest side is greater than the sum of the squares of the other two sides.

The sum of the areas of the squares formed on the legs of the triangle equals the area of the square formed on the hypotenuse:

a²+b²=c²

a, b and c are side lengths

a=12,b=37 and c=40

12²+37²=40²

144+1369=1600

1513 is not equal to 1600

Since 1513 < 1600, we know that:

This means that the triangle is obtuse, since the square of the longest side is greater than the sum of the squares of the other two sides.

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By side side side congruence theorem if any two triangles have three pairs of congruent sides, then the triangles are congruent.

We know that side side side (S-S-S) congruence theorem is a theorem for triangle congruence.

It states that if any two triangles have three pairs of congruent sides, then the triangles are congruent.

We need to prove:  the SSS congruence theorem

Consider ΔABC and ΔPQR

here, AB = PQ, BC = QR and AC = PR

Construct a congruent copy of triangle ABC that shares a side with PQR.  

∠SQR ≅ ∠ABC and AB = QS

In ΔABC and ΔSQR,

AB = QS                     ..............(by construction)

∠SQR ≅ ∠ABC                 ..............(by construction)

BC = QR                 ..............(given)

ΔSQR ≅ ΔABC         ...............(SAS congruence theorem)

So, by corresponding angles of congruent theorem,

∠A ≅ ∠S and ∠ACB ≅ SRQ      

But the quadrilateral PQRS is a kite, since we have QP = QS and RP = RS

We know that the diagonal divides the kite into two congruent triangles. Here,  the diagonal QR  divides the kite into ΔSQR and ΔPQR.

But ΔSQR ≅ ΔABC

so we have ΔPQR ≅ ΔABC

Hence the side side side congruence theorem proved.

Therefore, if any two triangles have three pairs of congruent sides, then the triangles are congruent.

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False. The best big-o bound that any algorithm that multiplies two n-bit numbers can achieve is O(n^2), not O(n+n) or O(2n).

An algorithm is a set of instructions or a procedure for solving a problem or performing a task. In computer science, an algorithm is typically a step-by-step process for solving a problem or performing a computation. Algorithms can be expressed in various forms, such as natural language, pseudocode, flowcharts, or programming languages. They are used in a wide range of applications, from simple arithmetic operations to complex data processing and artificial intelligence. The efficiency and correctness of an algorithm depend on various factors such as the input size, the data structure, and the computational complexity of individual operations.

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Because each of n bits of a number x must be multiplied by each of n bits of a number y, the best big-o bound that any algorithm that multiplies two n-bit numbers can achieve is O(n+n)-O(n^2

While it is true that each of n bits of a number x must be multiplied by each of n bits of a number y, it is possible to achieve a better big-o bound than O(n+n) or O(2n). The best-known algorithm for multiplying two n-bit numbers, the Karatsuba algorithm, achieves a big-o bound of O(n^log2(3)), which is faster than O(n^2).

Therefore, the statement that the best big-o bound that any algorithm that multiplies two n-bit numbers can achieve is O(n+n) or O(n^2) is false.
Main answer: b. false

The given statement implies that the best big-O bound for multiplying two n-bit numbers is O(n+n)-O(n^2), which simplifies to O(2n)-O(n^2) or simply O(n^2). However, this is not the best bound achievable for multiplying two n-bit numbers. The best known algorithm, the Karatsuba algorithm, has a time complexity of O(n^log2(3)), which is approximately O(n^1.585). This is faster than O(n^2) and thus, the statement is false.

The best big-O bound that any algorithm that multiplies two n-bit numbers can achieve is not O(n^2), but rather O(n^log2(3)), as demonstrated by the Karatsuba algorithm. Therefore, the given statement is false.

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