You want to place a towel bar that is 24 1⁄4 centimeters long in the center of a door that is 70 1⁄3 centimeters wide. How far should you place the bar from each edge of the door?

Final answer:

To write each number in scientific notation, express it as a product between 1 and 10 and a power of 10. Each adult spent an average of 570.55 hours per day online at home.

Explanation:

To write each number in scientific notation, we need to express it as a product of a number between 1 and 10 and a power of 10.

163 million can be written as 1.63 x 10⁸

93 billion can be written as 9.3 x 10¹⁰

To convert the units to hours per day, we need to divide the total number of hours by the number of adults.

So, each adult spent an average of (93 x 10¹⁰) / (163 x 10⁸) = 570.55 hours per day online at home. Scientific notation simplifies large numbers, facilitating computations and providing a concise representation of quantities in mathematical contexts.

In this question , it is given that

A right triangle ABC is shown. Leg AC has length 24, leg BC has length 32, and hypotenuse AB has length 40.

And we have to find the values of sin A and cos A .

[tex]sin A  = \frac{opposite}{hypotenuse} = \frac{32}{40}[/tex]

[tex]sin A = \frac{4}{5}[/tex]

And

[tex]cos A = \frac{adjacent}{hypotenuse} = \frac{24}{40}[/tex]

[tex]cos A = \frac{3}{5}[/tex]

And these are the required values of sin A and cos A .

Answer:

The answer above is correct! The answer is option D.

Step-by-step explanation:

Hope this helped clarify :D

Translating vertex A to vertex D, and then rotate △ABC around

point A to align the sides and angles will bring about a rigid

transformation.

This is the transformation which preserves the Euclidean

distance between every pair of points. This could be as a result

of the following:

Option D when done will preserve the distance between the

points when vertex A is translated to vertex D as they contain

the same angle with other sides and angles being made to

align.

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Final answer:

The x- and y-components of the given vector are calculated using trigonometric functions, resulting in -5.16 km along the x-axis and 7.37 km along the y-axis, considering the direction of the vector relative to the axes.

Explanation:

The question asks to find the x- and y-components of the vector d⟷ = (9.0 km, 35° left of +y-axis). To solve this, we use trigonometric functions, specifically sine and cosine, because the vector makes an angle with the axis.

Given the vector makes a 35° angle to the left of the +y-axis, this effectively means it is 35° above the -x-axis (or equivalently, 55° from the +x-axis in the second quadrant). We can calculate the components as follows:

The negative sign in the x-component indicates that the direction is towards the negative x-axis. Therefore, the x- and y-components of the vector are -5.16 km and 7.37 km, respectively.

1. [tex]\(\mathf{d}\): \(d_x = -3.8 \text{ km}, d_y = 8.2 \text{ km}\)[/tex]

2. [tex]\(\mathf{v}\): \(v_x = -4.0 \text{ cm/s}, v_y = 0\)[/tex]

3. [tex]\(\mathf{a}\): \(a_x = 10.3 \text{ m/s}^2, a_y = -14.7 \text{ m/s}^2\)[/tex]

To find the [tex]\( x \)[/tex]- and [tex]\( y \)[/tex]-components of vectors given in terms of magnitude and direction, we need to decompose the vectors using trigonometric functions. Let's go through each problem step by step.

1. Find the [tex]\( x \)[/tex]- and [tex]\( y \)[/tex]-components of the vector [tex]\( \mathf{d} = (9.0 \text{ km}, 25^\circ \text{ left of } +\mathbf{y}\text{-axis}) \)[/tex].

First, understand that "25° left of +[tex]\( y \)[/tex]-axis" means the vector is rotated 25° counterclockwise from the [tex]\( y \)[/tex]-axis.

Decomposition:

- Magnitude, [tex]\( d = 9.0 \text{ km} \)[/tex]

- Angle from the [tex]\( y \)-axis, \( \theta = 25^\circ \)[/tex]

To find the components:

- [tex]\( d_x = d \sin(\theta) \)[/tex]

- [tex]\( d_y = d \cos(\theta) \)[/tex]

However, since the angle is counterclockwise from the [tex]\( y \)-axis[/tex] and to the left, the [tex]\( x \)[/tex]-component is negative.

Therefore:

- [tex]\( d_x = -9.0 \sin(25^\circ) \)[/tex]

- [tex]\( d_y = 9.0 \cos(25^\circ) \)[/tex]

Calculating these:

- [tex]\( d_x \approx -9.0 \times 0.4226 \approx -3.8 \text{ km} \)[/tex]

- [tex]\( d_y \approx 9.0 \times 0.9063 \approx 8.2 \text{ km} \)[/tex]

So, the components are:

- [tex]\( d_x \approx -3.8 \text{ km} \)[/tex]

- [tex]\( d_y \approx 8.2 \text{ km} \)[/tex]

2. Find the [tex]\( x \)[/tex]- and [tex]\( y \)[/tex]-components of the vector [tex]\( \mathf{v} = (4.0 \text{ cm/s}, -x \text{-direction}) \).[/tex]

Since the vector is given in the [tex]\(-x\)[/tex]-direction, it means the entire magnitude is in the [tex]\( x \)[/tex]-direction and negative.

Decomposition:

- Magnitude, [tex]\( v = 4.0 \text{ cm/s} \)[/tex]

- Direction: [tex]\(-x\)[/tex]

Therefore:

- [tex]\( v_x = -4.0 \text{ cm/s} \)[/tex]

- [tex]\( v_y = 0 \text{ cm/s} \)[/tex]

So, the components are:

- [tex]\( v_x = -4.0 \text{ cm/s} \)[/tex]

- [tex]\( v_y = 0 \text{ cm/s} \)[/tex]

3. Find the [tex]\( x \)[/tex]- and [tex]\( y \)[/tex]-components of the vector [tex]\( \mathbf{a} = (18 \text{ m/s}^2, 35^\circ \text{ left of } -y \text{-axis}) \)[/tex].

"35° left of -[tex]\( y \)[/tex]-axis" means the vector is rotated 35° counterclockwise from the negative [tex]\( y \)[/tex]-axis.

Decomposition:

- Magnitude, [tex]\( a = 18 \text{ m/s}^2 \)[/tex]

- Angle from the [tex]\(-y \)[/tex]-axis, [tex]\( \theta = 35^\circ \)[/tex]

To find the components:

- [tex]\( a_x = a \sin(\theta) \)[/tex]

- [tex]\( a_y = -a \cos(\theta) \)[/tex]

However, since the angle is counterclockwise from the [tex]\(-y \)[/tex]-axis and to the left, the [tex]\( x \)[/tex]-component is positive.

Therefore:

- [tex]\( a_x = 18 \sin(35^\circ) \)[/tex]

- [tex]\( a_y = -18 \cos(35^\circ) \)[/tex]

Calculating these:

- [tex]\( a_x \approx 18 \times 0.5736 \approx 10.3 \text{ m/s}^2 \)[/tex]

- [tex]\( a_y \approx -18 \times 0.8192 \approx -14.7 \text{ m/s}^2 \)[/tex]

So, the components are:

- [tex]\( a_x \approx 10.3 \text{ m/s}^2 \)[/tex]

- [tex]\( a_y \approx -14.7 \text{ m/s}^2 \)[/tex]

The correct question is:

1. Find the [tex]$x$[/tex] - and [tex]$y$[/tex]-components of the vector [tex]$d \boxtimes=(9.0 \mathrm{~km}, 25 \boxtimes$[/tex] left of [tex]$+\mathrm{y}$[/tex]-axis).

2. Find the [tex]$\mathrm{x}$[/tex] - and [tex]$\mathrm{y}$[/tex]-components of the vector [tex]$\mathrm{v} \boxtimes=(4.0 \mathrm{~cm} / \mathrm{s},-\mathrm{x}$[/tex]-direction [tex]$)$[/tex].

3. Find the [tex]$x$[/tex] - and [tex]$y$[/tex]-components of the vector [tex]$a \boxtimes=(18 \mathrm{~m} / \mathrm{s} 2,35 \boxtimes$[/tex] left of [tex]$-y$[/tex]-axis [tex]$)$[/tex].

a) radius 6 inches, height 3 inches

b) diameter 6 inches, height 3 inches

c) diameter 4 inches, height 6 inches

d) diameter 6 inches, height 6 inches

Now you can probe those options to see which leads to an approximate volume of 28 cubic inches.

a) radius 6 in, height 3 in

=> V = (1/3)*3.14*(6in)^2 * 3in = 113.04 in^3 => not possible

Answer: b. 23 cm

Step-by-step explanation:

We know that area of a triangle is given by :-

[tex]\text{Area}=\dfrac{1}{2}bh[/tex], where b is the base of triangle and h is the height of the triangle.

Given : The area of a triangle is [tex]92\ cm^2[/tex]

The base of the triangle = 8 cm

Let h be the height of the triangle , then we have

[tex]92=\dfrac{1}{2}(8)h\\\\\Rightarrow\ h=\dfrac{92}{4}\\\\\Rightarrow\ h = 23[/tex]

Hence, the height of the triangle = 23 cm

Answer:

the radius is perpendicular to the chord.

Step-by-step explanation:

The geometry is drawn in the image shown below in which AB is the chorh and O is the centre of the circle. Om is the radius which bisects the chord.(Given) So, AN = NB

From the image, considering ΔAON and ΔBON,

AO = BO (radius of circle)

AN = NB (given)

ON = ON (common)

So,

ΔAON ≅ ΔBON

Hence, ∠ANO = ∠BNO

Also, ∠ANO + ∠BNO = 180° (Linear Pair)

So,

∠ANO = ∠BNO = 90°

Hence, it is perpendicular to the chord.

Discriminant 0 in quadratic equation means 1 real solution—a repeated root where parabola touches x-axis once.

When the discriminant of a quadratic equation is 0, it means that the quadratic equation has exactly one real solution. This solution is considered a "double root" or "repeated root," meaning that the parabola defined by the quadratic equation touches the x-axis at exactly one point. Mathematically, this occurs when the quadratic equation has two identical roots.

The general form of a quadratic equation is [tex]\(ax^2 + bx + c = 0\)[/tex], and the discriminant, denoted by [tex]\(b^2 - 4ac\),[/tex] helps determine the nature of the roots.

When the discriminant is zero [tex](\(b^2 - 4ac = 0\))[/tex], the quadratic equation has one real root. This happens when the parabola defined by the equation just touches the x-axis at one point. The solution is given by:

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

The following conditions are:

D < 0      ; there are two non-real or imaginary roots which are complex conjugates

D = 0      ; there is one real root and one imaginary (non-real)

D > 0      ; there are two real distinct roots

Therefore the answer to this question is:

The solution has one real root and one imaginary root.

Answer:

[tex]\frac{60 - 10\sqrt10-6\sqrt3+\sqrt30}{97}[/tex]

Step-by-step explanation:

Hello!

To rationalize the denominator, we have to remove any root operations from the denominator.

We can do that by multiplying the numerator and denominator by the conjugate of the denominator. The conjugate simply means the same terms with different operations.

The answer is [tex]\frac{60 - 10\sqrt10-6\sqrt3+\sqrt30}{97}[/tex].

The solution for the system of equation x - y = 4 and x + y = 2 is (3, -1).

An equation is a combination of different variables, in which two mathematical expressions are equal to each other.

The given pair of equations,

x - y = 4       (1)

And x + y = 2         (2)

To find the solution of the equations,

Add both the equations,

x - y + x + y = 4 + 2

2x = 6

x = 3,

Substitute the value of x = 3 in equation (1),

3 - y = 4

y = -1

The values of x and y are 3 and -1 respectively.

Hence, option (A) is correct.

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The complex solutions to the equation[tex]3x^2+3x+4=0 are x = (-3 + i\sqrt{39})/6 and x = (-3 - i\sqrt{39})/6.[/tex]

To find all complex solutions of the quadratic equation [tex]3x^2+3x+4=0[/tex]e the quadratic formula:

[tex]x = \((-b \pm \sqrt{b^2-4ac})/(2a)\).[/tex]

Here, a = 3, b = 3, and c = 4. Plugging these values into the formula, we get:

[tex]x = \((-3 \pm \sqrt{3^2-4 \cdot 3 \cdot 4})/(2 \cdot 3)\).[/tex]

This simplifies to:

[tex]x = \((-3 \pm \sqrt{-39})/6\).[/tex]

Since the discriminant (under the square root sign) is negative, we know that the solutions will be complex. Using i to represent the square root of -1, we can write the solutions as:

[tex]x = \((-3 \pm i\sqrt{39})/6\).[/tex]

So, the complex solutions are [tex]x = (-3 + i\sqrt{39})/6 and x = (-3 - i\sqrt{39})/6.[/tex]

The value of x makes the denominator of the function equal zero is 10

An equation is a mathematical expression that contains an equals symbol. Equations often contain algebra.

Given that:

y = [tex]\frac{6}{4x-40}[/tex]

Now the denominator is: 4x - 40

So, 4x =40

      x =40/4

      x =10

So, make the denominator 0, put x= 10.

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Answer:

B.

Step-by-step explanation:

There are 84 three digit numbers can be made from the digits 1, ..., 9

What is Combination?

A combination is a technique to determines the number of possible arrangements in a collection of items where the order of the selection does not matter.

Given that;

The numbers are,

⇒ 1, 2, ..., 9

Now,

All the three digit numbers can be made from the digits 1, ..., 9 are;

⇒ [tex]^{9} C_{3}[/tex]

⇒ 9! / 3! 6!

⇒ 9 × 8 × 7 / 6

⇒ 84

Thus, There are 84 three digit numbers can be made from the digits

1, ..., 9.

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Answer: 13 units

Step-by-step explanation:

The distance formula to calculate distance between two points (a,b) and (c,d) is given by :-

[tex]d=\sqrt{(d-b)^2+(c-a)^2}[/tex]

Given points : (13, 20) and (18, 8)

Now, the distance between the points (13, 20) and (18, 8)is given by ;-

[tex]d=\sqrt{(8-20)^2+(18-13)^2}\\\\\Rightarrow\ d=\sqrt{(-12)^2+(5)^2}\\\\\Rightarrow\ d=\sqrt{144+25}\\\\\Rightarrow\ d=\sqrt{169}\\\\\Rightarrow\ d=13\text{units}[/tex]

Hence, the distance between the points (13, 20) and (18, 8) is 13 units.