Which temperature conversion is correct? °c = k - 273 k = 273 x °c k = °c - 273 °c = k + 273?

Answer: The answer is the number of hours of daylight each day.

Step-by-step explanation:  We are to give the best  example of a natural phenomenon of periodic behavior.

A periodic behavior means the behavior which repeats itself in equal intervals of time or periods. Since we are talking about a natural phenomenon, so the rotation of earth around the sun can be an example.

In other words, we can say the number of hours of daylight each day will be the proper example, because the sun gives light for almost same number of hours each day.

Thus, the natural phenomenon is the number of hours of daylight each day.

A = 180-110 = 70

 angle A is 70 degrees

There are 11.3 feet far above the ground does the ladder touch the wall.

The Pythagoras theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the square of the other two sides.

Given that;

The bottom of a ladder must be placed 4 feet from a wall. the ladder is 12 feet long.

Now, Let the length of the ladder above the ground does the ladder touch the wall = x

Hence, We get;

x² = 12² - 4²

x² = 144 - 16

x² = 128

x = √128

x = 11.3 feet

Thus, the length of the ladder above the ground does the ladder touch the wall = 1.3 feet

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Answer:

[tex]d=22t+10...(1)[/tex]

[tex]d=28t...(2)[/tex]

Step-by-step explanation:

Let t represent the number of hours and d represent the total distance after t hours.

We have been given that John bikes 22 km per hour and starts at mile 10. This means that slope of line will be 22 and y-intercept will be 10. So total distance covered by John in t hours will be:

[tex]d=22t+10...(1)[/tex]

We are also told that Gwynn bikes 28 km per hour and starts at mile 0. This means that slope of line will be 28 and y-intercept will be 0. So total distance covered by Gwynn in t hours will be:

[tex]d=28t+0[/tex]

[tex]d=28t...(2)[/tex]

Therefore, our required system of linear equations will be:

[tex]d=22t+10...(1)[/tex]

[tex]d=28t...(2)[/tex]

Answer:

2.86 quarts ( approx )

Step-by-step explanation:

Given,

The initial quantity of the Mr. Gittleboro's radiator that contains 30% antifreeze = 10 quarts,

Let x quarts of pure antifreeze replaced x quarts of Mr. Gittleboro's radiator to bring it up to a required 50% antifreeze,

So, the quantity of 30% antifreeze radiator after drained off x quarts = (10-x) quarts

Also, the quantity of final antifreeze radiator 50% antifreeze = 10 quarts

Thus, we can write,

30% of (10-x) + 100% of x = 50% of 10

30(10-x) + 100x = 500

300 - 30x + 100x = 500

300 + 70x = 500

70x = 200

x = 2.85714285714 quarts ≈ 2.86 quarts

[tex]\boxed{{\mathbf{2}}{\mathbf{.86 quartz}}}[/tex]  of [tex]30\%[/tex]  antifreeze replaced by pure antifreeze to maintain [tex]50\%[/tex]  antifreeze in the radiator.

Further explanation:

Given:

It is given that radiator contains [tex]30\%[/tex]  antifreeze and the radiator holds 10 quartz.

Step by step explanation:

Step 1:

Consider [tex]x[/tex]  as the number of quartz of pure antifreeze that is replaced by [tex]x[/tex]  number of quartz of Mr. Gittleboro’s radiator to maintain the level of [tex]50\%[/tex] antifreeze in the radiator.

It is given that radiator contains [tex]30\%[/tex]  antifreeze and the radiator holds 10 quartz.

The amount of [tex]30\%[/tex]  antifreeze radiator after drained off [tex]x{\text{ quartz}}[/tex]  is [tex]\left({10-x}\right){\text{quartz}}[/tex] .

The radiator only holds [tex]{\text{10 quartz}}[/tex] .

The amount of final antifreeze radiator [tex]50\%[/tex]  antifreeze is [tex]{\text{10 quartz}}[/tex] .

Step 2:

The equation for maintain the level of [tex]50\%[/tex]  antifreeze in the radiator by replacing the [tex]x[/tex]  number of quartz of [tex]30\%[/tex]  antifreeze by pure antifreeze as,

[tex]\begin{aligned}30\%{\text{ of}}\left({10-x}\right)+100\% {\text{ of }}x&=50\% {\text{ of 10}}\\\frac{{30}}{{100}}\times\left({10-x}\right)+\frac{{100}}{{100}}{\text{}}\times{\text{}}x&=\frac{{50{\text{ }}}}{{100}}\times{\text{10}}\\{\text{30}}\left({10-x}\right)+100x&=500\\\end{aligned}[/tex]

Now simplify the further equation using the distributive property.

[tex]\begin{aligned}{\text{30}}\left({10-x}\right)+100x&=500\\300-30x+100x&=500\\300+70x&=500\\70x&=500-300\\\end{aligned}[/tex]

Now simplify the further equation to obtain the value of [tex]x[/tex] .

[tex]\begin{aligned}70x&=500-300\hfill\\70x&=200\hfill\\x&=\frac{{200}}{{70}}\hfill\\x&=2.85714\approx 2.86{\text{ quartz}}\hfill\\\end{aligned}[/tex]

Therefore, [tex]2.86{\text{ quartz}}[/tex]  of [tex]30\%[/tex]  antifreeze replaced by pure antifreeze to maintain [tex]50\%[/tex]  antifreeze in the radiator.

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Answer details:

Grade: High school

Subject: Mathematics

Chapter: Linear equation

Keywords: radiator, amount, replaced, Gittlebro, antifreeze, drained, maintain, quartz, number, linear equation, percentage, fraction, final antifreeze.

Answer: 2x-3=x+2

. -x -x

1x -3=2

+3 +3

1x=5

1x divide by 1x, 5 divide by 1x

Which will be:

X=5

Let

x--------> the number

we know that

1) the phrase, "the sum of 14 and a number"

represent the expression --------> [tex](14+x)[/tex]

and

2) the complete phrase, "one fifth the sum of 14 and a number"

represent the expression --------> [tex]\frac{1}{5}(14+x)[/tex]

therefore

the answer is

[tex]\frac{1}{5}(14+x)[/tex]

Given that the rise and run of a ladder are 19 feet and 11 feet respectively:

a. Slope of the ladder = 19/11 = 1.7

b. Slope = inclination of the ladder

c. The slope will be steeper and difficult to climb if the ladder was set closer to the wall of the building.

Recall:

a. run = 11 feet

rise = 19 feet

Slope of the ladder = [tex]\frac{rise}{run} = \frac{19}{11}[/tex] = 1.7

b. The slope of the ladder refers to the inclination of the ladder to the wall.

c. If the ladder was set close to the wall of the building, the more steeper the slope would be. This would make it difficult to climb compared to a gentle slope.

In summary, given that the rise and run of a ladder are 19 feet and 11 feet respectively:

a. Slope of the ladder = 19/11 = 1.7

b. Slope = inclination of the ladder

c. The slope will be steeper and difficult to climb if the ladder was set closer to the wall of the building.

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Using π ≈ 3.14, the radius of a circle with circumference 28.3 m is approximately 4.5 meters.

Step 1:

Write down the formula for the circumference of a circle.

The formula for the circumference (C) of a circle is given by:

[tex]\[ C = 2 \pi r \][/tex]

Step 2:

Substitute the given value for the circumference into the formula.

Given [tex]\(C = 28.3\)[/tex] m, we use the approximation [tex]\(\pi \approx 3.14\)[/tex]:

[tex]\[ 28.3 = 2 \times 3.14 \times r \][/tex]

Step 3:

Solve for the radius (r).

Divide both sides of the equation by [tex]\(2 \times 3.14\)[/tex]:

[tex]\[ r = \frac{28.3}{2 \times 3.14} \][/tex]

Step 4:

Calculate the value of (r).

[tex]\[ r = \frac{28.3}{6.28} \][/tex]

[tex]\[ r \approx 4.5 \text{ m} \][/tex]

So, the radius of the circle with a circumference of 28.3 m, using the approximation [tex]\(\pi \approx 3.14\)[/tex], is approximately 4.5 meters.

To find the rate of change of the x-coordinate at the point (1, 5) with a given y-coordinate rate of change, we use the chain rule to differentiate the equation y = 24 + [tex]x^3[/tex] with respect to time, and solve for dx/dt.

The student's question involves the relationship between the rates of change of the x-coordinate and y-coordinate of a particle moving along a curve. Given the equation y = 24 +[tex]x^3[/tex], we need to find how fast the x-coordinate is changing at the point (1, 5), given that the y-coordinate is increasing at a rate of 4 cm/s. To find this, we can use the chain rule from calculus to relate the rates of change. Differentiating both sides with respect to time t, we get dy/dt = 3[tex]x^2[/tex] dx/dt. Solving for dx/dt, we find the rate of change of the x-coordinate when x = 1 and dy/dt = 4 cm/s.

At the point (1, 5), [tex]\( \frac{{dx}}{{dt}} = \frac{4}{3} \)[/tex] cm/s when [tex]\( \frac{{dy}}{{dt}} = 4 \)[/tex] cm/s.

To find how fast the x-coordinate is changing [tex](\( \frac{{dx}}{{dt}} \))[/tex] when the y-coordinate is increasing [tex](\( \frac{{dy}}{{dt}} \)),[/tex] we'll use implicit differentiation.

Given [tex]\( y = 24 + x^3 \),[/tex] differentiate both sides with respect to time [tex](\( t \)):[/tex]

[tex]\[ \frac{{dy}}{{dt}} = \frac{{d}}{{dt}}(24 + x^3) \][/tex]

Given that [tex]\( \frac{{dy}}{{dt}} = 4 \)[/tex], and when [tex]\( x = 1 \) and \( y = 5 \),[/tex] we can find [tex]\( \frac{{dx}}{{dt}} \):[/tex]

[tex]\[ 4 = 0 + 3x^2 \cdot \frac{{dx}}{{dt}} \][/tex]

[tex]At \( x = 1 \):[/tex]

[tex]\[ 4 = 3(1)^2 \cdot \frac{{dx}}{{dt}} \][/tex]

[tex]\[ 4 = 3 \cdot \frac{{dx}}{{dt}} \][/tex]

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

So, the x-coordinate of the point is changing at a rate of [tex]\( \frac{4}{3} \)[/tex] cm/s at that instant.

Plug in -0.5 for c

a = 3 - 5(-0.5)

Multiply -5 with (-0.5)

a = 3 + 2.5

Simplify

a = 5.5

a = 5.5 is your answer

hope this helps

In the formula 'a = 3-5c', 'a' becomes 5.5 when c=-0.5. This is derived by substituting -0.5 as 'c' in the given formula yielding the result of 5.5 after the calculation.

The subject of your question is mathematics and it's about finding the value of 'a' in the formula: a = 3-5c when c=-0.5.

To answer your question, we'll substitute the value of 'c' into the formula:a = 3 - 5*(-0.5). When you multiply -5 with -0.5, the answer is 2.5.

Therefore, when you add 3 and 2.5, the answer is 5.5. So, when c = -0.5, the value of 'a' will be 5.5.

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Answer:

Step-by-step explanation:

Yes because their are no outlier

4 total quizzes,

 for a 40 average on 4 quizzes she needs 40*4 = 160 total points

for 3 quizzes she has 45 +32 +37 = 114 points

160-114 = 46, she needs at least a 46 on the 4th quiz

Answer:

The answer is the second one for sure y=5/6x−3.

Step-by-step explanation:

The equation of the line is y = (5/6)x - 3 which is the correct answer would be option (B).

The term "equation" refers to mathematical statements that have at least two terms with variables or integers that are equal.

We have been given a line that has a slope of 5/6 and a y-intercept is -3.

Let the required line of the equation would be as

y = mx + c

Where m is the slope of the line

We have given the y-intercept is -3 which means the x-coordinates will be zero.

Substitute the value of x = 0 and y = -3 in the above equation to get c.

⇒ -3 = m(0) + c

⇒ -3 = c

The slope of the line m = 5/6 which is given in the question,

Substitute the value of m = 5/6 and c = -3 in the equation y = mx + c

⇒ y = (5/6)x - 3

Therefore, the equation of the line is y = (5/6)x - 3.

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The rational number that equals to 2.8 (a decimal number that has repeating 8), from the options given, is twenty six over nine (26/9) because when converted to decimal it gives 2.89, which is closest to 2.8

The question is asking which rational number is equal to 2.8 where the bar over the number indicates that it is repeating. The dot over the 8 implies that the number is actually 2.8888..., where the 8 is repeated indefinitely. So we need to find which fraction out of your given options: twenty two over nine, twenty four over nine, twenty six over nine, twenty eight over nine, equals to this decimal number.

To convert these fractions to decimal form:

22/9 = 2.44,
24/9 = 2.67,
26/9 = 2.89,
28/9 = 3.11

By comparing these values, we can see that twenty six over nine (26/9) is the decimal equivalent to 2.8 as it gives us 2.89 which is closest to it.

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Using Pythagoras theorem, we can find that the library is approximately 7.746 miles from the football field, which forms a right-angled triangle with Kristen's home and the park.

This problem is a classic example of use of Pythagoras theorem in right-angle triangle. Here, the distances of Kristen's home, the park, and the football field form a right-angle triangle, with the distance from the park to the library acting as the altitude of the triangle. To find the distance from the library to the football field, we can use the relation:

According to the problem, The library is 2 miles from her home (Base), The football field is 8 miles from the library (Hypotenuse). So the distance from the library to the football field can be calculated as:

So, the library is approximately 7.746 miles from the football field.

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