find the area of one segment formed by a square with sides of 6 inches inscribed in a circle (hint: use the ratio of 1:1:square root of 2 to find the radius of the circle)

Newton's Law of Cooling states that the change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature over time.

Therefore when expressed mathematically, this is equivalent to:

dT = - k (T – Ts) dt

dT / (T – Ts) = - k dt

Integrating:

ln [(T2– Ts) / (T1– Ts)] = - k (t2 – t1)

Before we plug in the values, let us first convert the temperatures into absolute values R (rankine) by adding 460.

R = ˚F + 460

T1 = 200 + 460 = 660 R

Ts = 70 + 460 = 530 R

ln [(T2– 530) / (660 – 530)] = - 0.6 (2 - 0)

T2 = 569.16 R

T2 = 109 ºF

Answer: After 2 hours, it will be 109 ºF

Answer:

109 degrees

Step-by-step explanation:

To tell if there is a horizontal asymptote, for rational functions, compare the degrees of the numerator and denominator; if they are equal, the asymptote is the ratio of their leading coefficients. If the numerator's degree is less, the asymptote is at y=0; if greater, there is no horizontal asymptote. Polynomials do not have horizontal asymptotes.

To determine if a function has a horizontal asymptote, we look at the behavior of the function as x approaches infinity (x →∞). For rational functions, the degrees of the numerator and denominator play a crucial role. If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients of the numerator and denominator. For instance, as x approaches infinity, the term with the highest power dominates and the constant terms become less significant, leading to the horizontal asymptote at y = 1.

In a scenario where the degree of the numerator is less than the degree of the denominator, the function will have a horizontal asymptote at y = 0, since the function approaches zero as x increases without bound. Conversely, if the degree of the numerator is greater than the degree of the denominator, no horizontal asymptote exists because the function grows without limit.

Remember, polynomials do not have horizontal asymptotes, as their values continue to increase or decrease indefinitely with x.

In the right triangle with sides 15, 20, and 25, the exact values of sin A and cos A are 4/5 and 3/5, respectively.

In the right triangle ABC, we can use the given side lengths to find the trigonometric ratios for angle A. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (AB) is equal to the sum of the squares of the lengths of the legs (AC and BC).

Let's denote the angle A as angle A:

Find sin A:

sin A = (opposite / hypotenuse) = BC / AB = 20 / 25 = 4/5.

Find cos A:

cos A = (adjacent / hypotenuse) = AC / AB = 15 / 25 = 3/5.

So, the exact values are sin A = 4/5 and cos A = 3/5.

In summary, in the right triangle ABC with side lengths AC = 15, BC = 20, and AB = 25, the exact values of sin A and cos A are 4/5 and 3/5, respectively.

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Getting two 6's in four rolls can happen in 6 ways. 4C2 = 6

The probability of getting two 6's is 1/36. Therefore,

6 times 1/36 = 1/6 or 0.16666666...

 Converting to a percent to the nearest tenth is 16.7%

digit #1 - 9 numbers (1-9)

digit #2 = 9 numbers (0-9 minus the first digit)

digit #3 = 8 numbers ( 0-9 minus the first 2 digits)

9*9*8 = 648 combinations

arc length = given angle/360 x 2 x PI x radius

120/450 x 2 x 3.14 x 23 = 48.17

 round to 48 cm

Answer:

Option A is correct

the approximate length of the minor arc is, 48 cm

Step-by-step explanation:

Length of an arc is given by:

[tex]l = r \theta[/tex]              .....[1]

here r is the radius of the circle from the center and [tex]\theta[/tex] is the angle in radian.

From the given figure, we have;

r = 23 cm

Use conversion:

1 degree = 0.0174533 radian

then

120 degree = 2.094396

[tex]\theta = 2.094396[/tex] radian.

Substitute these in [1] we have;

[tex]l = 23 \cdot 2.094396 = 48.171108[/tex] cm

Therefore, the approximate length of the minor arc is, 48 cm

The next three terms of the sequence are 6/125, 12/625 and 24/3125

The nth term of a geometric sequence is given as:

Tn = ar^n-1

Given the geometric sequence

3/4,3/10,3/25,...

Find the common ratio

r = 3/10 * 4/3

r = 2/5

Find the 4th, 5th and 6th terms

T4 = (3/4)(2/5)^3
T4 = 3/4 * 8/125
T4 = 6/125

T5 = (3/4)(2/5)^4
T5 = 3/4 * 16/625
T5 = 12/625

T6 = (3/4)(2/5)^5
T6 = 3/4 * 32/3125
T6 = 24/3125

Hence the next three terms of the sequence are 6/125, 12/625 and 24/3125

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the measurement of an inscribed angle is half of the measure of the intercepted arc.

the inscribed angle is 110 degrees, so we multiply 110 by 2.

110*2 = 220 degrees

The measure of the intercepted arc is 220 degrees

220° is the measure of the intercepted arc

The piecewise function is:

[tex]\[ f(x) = \begin{cases} 4 & \text{if } -1 \leq x \leq 1 \\ x - 1 & \text{if } 3 \leq x \leq 5 \end{cases} \][/tex]

To determine the piecewise function represented by the given coordinates, let's examine the points and their corresponding intervals.

The coordinates are (-1, 4), (1, 4), (3, 2), and (5, 4).

The points (-1, 4) and (1, 4) have the same y-coordinate, indicating that on the interval [tex]\(-1 \leq x \leq 1\)[/tex], the function has a constant value of 4. Therefore, the piecewise function for this interval is f(x) = 4 for [tex]\(-1 \leq x \leq 1\).[/tex]

The points (3, 2) and (5, 4) indicate that on the interval [tex]\(3 \leq x \leq 5\)[/tex], the function is a line passing through these two points. We can find the slope (m) and y-intercept (b) for this line.

[tex]\[ m = \frac{\text{change in } y}{\text{change in } x} = \frac{4 - 2}{5 - 3} = \frac{2}{2} = 1 \][/tex]

Using the point (3, 2), we can find the y-intercept:

2 = 1(3) + b

b = -1

Therefore, the equation for the line on the interval [tex]\(3 \leq x \leq 5\) is \(f(x) = x - 1\) for \(3 \leq x \leq 5\).[/tex]

Putting it all together, the piecewise function is:

[tex]\[ f(x) = \begin{cases} 4 & \text{if } -1 \leq x \leq 1 \\ x - 1 & \text{if } 3 \leq x \leq 5 \end{cases} \][/tex]

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Answer:

b. 10

Step-by-step explanation:

I did the test

Answer:

inverse tangent 4/1

75.96 + 180 bc 3rd quadrant which equals 255.96 degrees.

Step-by-step explanation:

round 256 degrees

Heyyy guys just thought I’d let y’all know the answer is *117649* if you (7^2)^3 that’s the answer and 7^0 is 0 soooo yeah!

Answer:

ASA only

Step-by-step explanation:

Given is a picture of two triangles with one side and one angle congruent.

Comparison of these two triangles given

side = side

one angle = one angle (given)

Second angle = second angle (Vertically opposite angles)

Thus we find here that two angles and one corresponding side are congruent.

HEnce we say that these two triangles are congruent by ASA theorem

ASA theorem can be applied here because the equal side is between the two congruent angles.  

Instead if the side is not between the congruent angles but corresponding side then we can only use AAS

SO here ASA is correct.

Answer:

ASA only

Step-by-step explanation:

We are given that  two triangles in which

An angle of triangle is equal to its corresponding angle of second triangle.

One side of a triangle is equal to one side of other triangle.

ASA  postulate: It states that two angles and included side of one triangle are congruent to its corresponding angles and corresponding side of other triangle , then the two triangles are congruent.

AAS postulate: It states that two angles and non- included side of one triangle are congruent to its corresponding two angles and its corresponding side of another triangle, then the triangles are congruent by AAS postulate.

In triangle AOB and COD

[tex]\angle AOB= \angle COD[/tex]  (Vertical angles are equal )

[tex]\angle ABO=\angle CDO[/tex] ( Given )

[tex]OB=OD[/tex] (Given )

[tex]\triangle AOB\cong \triangle COD[/tex] ( ASA Postulate )

Answer: ASA only

To find the coordinates of point C which divides line AB into a 2:5 ratio, use the section formula. Therefore, the coordinates are (-4/7, -11/7). This provides point C's location on the line segment AB.

The coordinates of C are determined through the section formula, which finds the coordinates of a point dividing a line segment into a specific ratio. In this case, the ratio is 2:5. To solve for the x-coordinate of C, the formula [(m*x2) + (n*x1)] / (m+n) is used, where m and n are the ratio values, x1 and x2 are the x-coordinates of points A(-2, -3) and B(3, 2). Using these, the x-coordinate of C is found to be (2*3 + 5*(-2)) / (2+5) = -4/7.

Similar calculations apply to solving for the y-coordinate of C: [(m*y2) + (n*y1)] / (m+n) = (2*2 + 5*(-3)) / (2+5) = -11/7.

Therefore, the coordinates of point C are (-4/7, -11/7). This is the point C on the line segment AB that is 2/5 of the way from point A to point B.

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To solve the quadratic equation x^2 - 10x + 15 = 0 by completing the square, we manipulate it to the form (x - 5)^2 = 10, then solve for x to get two solutions: x = 8.16 and x = 1.84.

To solve the quadratic equation by completing the square, we need to manipulate the equation x^2 - 10x + 15 = 0 so it takes the form of (x - a)^2 = b, where a is half of the coefficient of x, and b is a constant.

First, we move the constant to the other side of the equation:

x^2 - 10x = -15

Next, we find the value that completes the square. This value is (10/2)^2 = 25.

Adding 25 to both sides, we get:

x^2 - 10x + 25 = 25 - 15

Now the equation is a perfect square:

(x - 5)^2 = 10

To find the solution for x, we take the square root of both sides:

x - 5 = ±10√10

x = 5 ± √10

Thus, the solutions rounded to the nearest hundredth are:

x = 5 + √10 ≈ 8.16

x = 5 - √10 ≈ 1.84

The complete answer in the template is:

Form:
( x - 5 )^2 = 10
Solution:
x = 8.16, 1.84

6. add the averages then take that average

50+74 = 124, 124/2 = 62

answer: 62

8. first put all the numbers in order

4, 5 ,6 ,8 11, 12,14, 15, 23 ,23

 Find the middle number, since there are 10 numbers find the average of the 2 middle numbers

11+12 =23/2 =11.5

 Lower half data = 4, 5, 6, 8, 11

 The middle number of that set is Q1,

Q1=6

9

9, find q3

1, 2, 3, 5, 5, 6, 9

 Find middle number (5)

upper half data = 5, 6, 9

Q3=  5 , 6, 9 = 5+10 +15 = 30

so answer is $30

Answer:m=ca i remember my teacher telling me how to do this.

Step-by-step explanation: