Which equation would best help solve the following problem? Rennie kicks a field goal with an initial vertical velocity of 31 m/s. How long will it take the football to hit the ground?

–4.9t2 + 31t = 0


–4.9t2 – 31t = 0


16t2 + 31t = 0

–16t2+ 31t = 0

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: it’s D $325

Step-by-step explanation:

the fee the school should charge to maximize their revenue is $325

Answer:

B.

Step-by-step explanation:

False, The measure of Arc AC is not 65 degree if angle ABC is equal to 78 degree.

What is Circle?

The circle is a closed two dimensional figure , in which the set of all points is equidistance from the center.

Given that;

The statement is,

The measure of Arc AC is 65 degree if angle ABC is equal to 78 degree.

Now,

Since, Sum of Measure of arc AC and angle ABC is equal to 360 degree.

So, In angle ABC is 78 degree.

Then, Measure of arc AC = 360 - 78

                                       = 282 degree.

Thus, The measure of Arc AC is not 65 degree if angle ABC is equal to 78 degree.

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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.

The solution to the equation 8(x-1) - 2x = -(x + 50) is x = -6.

To solve the equation 8(x-1) - 2x = -(x + 50).

First, distribute the 8 to the terms inside the parentheses:

8x - 8 - 2x = -(x + 50)

Simplify the left side of the equation:

6x - 8 = -(x + 50)

Distribute the negative sign to the terms inside the parentheses:

6x - 8 = -x - 50

Isolate the x terms on one side and the constant terms on the other side:

6x + x = -50 + 8

Combine like terms:

7x = -42

Divide both sides of the equation by 7:

7x / 7 = -42 / 7

x = -6

Therefore, the solution to the equation 8(x-1) - 2x = -(x + 50) is x = -6.

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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.

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:

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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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 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]

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: 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.

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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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].