HELP ASAP!!!!!!!!!!!!!!!!!!!!
a. 49
b. 12
c. 7
d. 1

Answer:

Joe Don actually ran 113.2 yards

Step-by-step explanation:

Length of the football field = 100 yd

Width of the football field = 53 yd

The football field is rectangular in shape. So in order to find the diagonal of the field, we need to split it up into two right triangles.

In one of the right triangles, the length and width becomes the two sides of it and diagonal is the third side.

In order to find the third side (diagonal), we need to apply the Pythagoras theorem.

So,

Diagonal² = 100² + 53²

Diagonal² = 10000 + 2809

Diagonal² = 12809

Taking square root on both sides

[tex]\sqrt{Diagonal^{2} } =\sqrt{12809}[/tex]

Diagonal = 113.17 yd

Diagonal = 113.2 yd (approx)

Thus Joe Don actually ran 113.2 yards

Answer:

  b.  –2x^5 + x^3/2 + x/4 + 1

Step-by-step explanation:

In the given expression, the degrees of the terms, in the order given, are ...

  1, 5, 3, 0

When the terms are properly arranged in standard form, they will be ...

  5, 3, 1, 0 . . . . descending order

The standard form of this expression is ...

  -2x^5 +x^3/2 +x/4 +1 . . . . . . matches choice B

Answer: b.–2x^5 + x^3/2 + x/4 + 1

Step-by-step explanation:

The total number of rectangles and triangles used for the net of this solid is 4.

To find the total number of rectangles and triangles used for the net of a solid, we need to examine the shape of the solid. If the given solid has 4 faces, like a regular tetrahedron or a pyramid, then it will have 3 triangles and 1 rectangle. Hence, the total number of rectangles and triangles used for the net of this solid is 4, so the answer is option b.

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Answer:

Option B is correct.

Obtuse triangle.

Step-by-step explanation:

Acute triangle:

In acute triangle, all angles are less than 90 degree.

Obtuse triangle:

In obtuse triangle, the triangle has an angle more than 90 degree.

Right triangle.

In right triangle, it has a right angle 90 degree.

Isosceles triangle.

In isosceles triangle, it can be acute, right or obtuse depends only on the apex angle.

Therefore, the triangle whose angles are all more that 90 degree is, Obtuse triangle.

Answer:

k = 22.5; xy = 22.5

your welcome

XD

Answer:

1. C) Inverse Property of Multiplication

2. None of these. The answer is Associative property of Addition

3. C) Inverse Property of Multiplication

4. None of these. The answer is Associative property of Multiplication

Step-by-step explanation:

1.

Given a number a , the inverse property of multiplication states that

a * 1/a = 1  

and 1/a is called the multiplicative inverse of a.

So 8 * 1/8 = 1 features the inverse property of multiplication where a = 8.

2.

Given three numbers a,b,c , the associative property of addition states that a + (b + c) = (a + b) + c

In this case, a = -5, b = 8 and c = 10

3.

-4 x -1/4 = 1 features the inverse property of multiplication where a = -4.

4.

Given three numbers a,b,c , the associative property of multiplication states that  (a * b) * c = a * (b * c)

In this case, a = a, b = b and c = 3

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

 y = (x-5)^2 - 25

 y+25 = (x-5)^2

 x-5 = +/-sqrt(y+25)

 And you get TWO inverses:

 x = 5 + sqrt(y+25), for x>=5

 x = 5 - sqrt(y+25), for x<=5

Answer:

[tex]y=5\pm \sqrt{(25+x)}[/tex]

Step-by-step explanation:

We are asked to find the inverse for the function [tex]y=x^2-10x[/tex].

We know that to find inverse, we interchange x and y values and then solve for y.

After interchanging x and y values, we will get:

[tex]x=y^2-10y[/tex]

Switch sides:

[tex]y^2-10y=x[/tex]

[tex]y^2-10y-x=x-x[/tex]

[tex]y^2-10y-x=0[/tex]

Now, we will use quadratic formula to solve for y.

[tex]y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

[tex]y=\frac{-(-10)\pm \sqrt{(-10)^2-4(1)(-x)}}{2(1)}[/tex]

[tex]y=\frac{10\pm \sqrt{100+4x}}{2}[/tex]

[tex]y=\frac{10\pm \sqrt{4*25+4x}}{2}[/tex]

[tex]y=\frac{10\pm \sqrt{4(25+x)}}{2}[/tex]

[tex]y=\frac{10\pm 2\sqrt{(25+x)}}{2}[/tex]

[tex]y=5\pm \sqrt{(25+x)}[/tex]

Therefore, the inverse function for our given function would be [tex]y=5\pm \sqrt{(25+x)}[/tex].

Answer:

25, 2, & 3 are rational numbers

π is a irrational number.

Explanation:

I think we need to find out the irrational number.

Here we learn main difference between rational and irrational number.

Rational Number:

All numbers whose written in fraction form where numerator and denominator are whole number.

All the whole numbers are rational number.

For example: 2 (because we can write 2 as [tex] \frac{2}{1} [/tex]), [tex] \frac{5}{6} [/tex] etc

0.666666.... (endless repeating digits)

In given question: 25, 2 and 3 are rational numbers

Irrational Numbers:

All numbers that are not rational are considered irrational. An irrational number can be written as a decimal form, but not as a fraction form.

An irrational number has endless non-repeating digits to the right of the decimal point.

π = 3.141592...... (endless non repeating digits)

√2 =1.414213... (endless non repeating digits)

Answer:

25, 2, 3

Step-by-step explanation:

A rational number is a terminating decimal or a non-terminating, repeating decimal.

25, 2, and 3 are terminating decimals, so they are rational.

There are 10 different ways to select a committee of 2 people from a group of 5 candidates.

To calculate the number of ways to select a committee of 2 people from a group of 5 candidates, we can use the combination formula. The number of combinations of n items taken r at a time is given by the formula:

nCr = n! / (r!(n-r)!)

Here, n is the total number of candidates (5) and r is the number of people to be selected for the committee (2). Plugging in the values, we get:

5C2 = 5! / (2!(5-2)!)

Simplifying this expression:

5C2 = (5 * 4) / (2 * 1)

Therefore, there are 10 different ways to select the committee.

Area of circle is 345.4 cm²

Given that;

Angle of sector = 45°

Area of sector = 13.75π cm²

Find:

Area of the circle

Computation:

Area of sector = [tex]\frac{\theta}{360}[/tex][Area of circle]

13.75π = [tex]\frac{\theta}{360}[/tex][Area of circle]

13.75(3.14) = [tex]\frac{45}{360}[/tex][Area of circle]

13.75(3.14) = [tex]\frac{45}{360}[/tex][Area of circle]

43.175 = 0.125[Area of circle]

Area of circle = 345.4 cm²

Find out more information about 'Area of circle'

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Answer:

Option A is correct

Step-by-step explanation:

Given is a table which shows data from a survey about the amount of time students spend doing homework each week.

          College    HS

High    50            16

Low     5                0

Q1       7.5             9.5

Q3      15              14.5

 IQR    7.5             5

Median  11             13

Mean     13.8         10.7

Std dev 6.4            5.3

1.5IQR         11.25         7.5

Q1-1.5IQR     -9.38      -12

Q3+1.5IQR    37.5       29.5

WE find that there is only one outlier in college with 50 as high.

Except that all others are well within IQR range.  Hence outliers are minimum

So option A is right.

Final answer:

To determine the spread of data on student homework hours, IQR and standard deviation are used. The college data's spread is best described by IQR, while the high school data's spread is best described by standard deviation, making Option C the most suitable choice.

Explanation:

To measure the spread of the data for the amount of time students spend doing homework each week, we have two primary measures: the Interquartile Range (IQR) and the standard deviation (σ). IQR represents the difference between the third quartile (Q3) and the first quartile (Q1) and describes the spread of the middle 50% of the data. The standard deviation indicates how much the values in the dataset deviate from the mean, which can help in understanding the overall spread including potential outliers.

When dealing with outliers, it's important to consider that potential outliers are typically identified as values that are more than 1.5 × IQR above Q3 or below Q1. We would use the minimum (Low) and maximum (High) values provided alongside the IQR to check for outliers. For the college data, with a High of 50 and a Low of 5, the range is 45. For high school, the range is 16. Outliers could potentially affect the standard deviation significantly, making IQR a better measure for the middle spread of data.

Based on this, the college data with a wider range may be more affected by outliers, whereas the high school data with a smaller range may not. Therefore, the spread for the college data can be better described by the IQR, while the standard deviation can be a good measure for the high school data. This suggests that Option C: "The college spread is best described by the IQR. The high school spread is best described by the standard deviation." is the most suitable choice.