Mr. Barth is painting an arrow on the school parking lot. He draws the edges between the following points on the coordinate plane: (–2, 2), (5, 2), (5, 6), (12, 0), (5, –6), (5, –2), (–2, –2).

Answer:

A = 1800m^2

Step-by-step explanation:

(90)(25) - (30)(15)

2250 - 450 = 1800

Answer:

28%

Step-by-step explanation:

Answer:

Step-by-step explanation:

you have studied this in grade 8 grade 9 and grade 10 if you are studying under CBSE, so, better call textbook

Answer:

4

Step-by-step explanation:

Answer:

The equation of the normal is y = 4 - 2·x

Step-by-step explanation:

The given equation of the circle, is presented as follows;

(x - 3)² + (y + 2)² = 20

The point of the normal of the circle = (1, 2)

The equation of the normal to a circle, x² + y² + D·x + E·y + F = 0 at a point P(x₁, y₁) is given as follows;

[tex]\dfrac{y - y_1}{x - x_1} = \dfrac{2 \cdot y_1 + D}{2 \cdot x_1 + E}[/tex]

Expanding the given equation of the circle, gives;

(x - 3)² + (y + 2)² = x² + y² - 6·x + 4·y + 13 = 20

∴ x² + y² - 6·x + 4·y + 13 - 20 = x² + y² - 6·x + 4·y - 7 = 0

x² + y² - 6·x + 4·y - 7 = 0

∴ x₁ = 1, y₁ = 2, D = -6, E = 4, and F = 13

Which gives;

[tex]\dfrac{y - 2}{x - 1} = \dfrac{2 \times 2 + 4}{2 \times 1 + (-6)} = \dfrac{8}{-4} = -2[/tex]

∴ y - 2 = -2 × (x - 1) = 2 - 2·x

y = 2 - 2·x + 2 = 4 - 2·x

The equation of the normal to the circle with equation (x - 3)² + (y + 2)² = 20, at the point (1, 2) is y = 4 - 2·x

Answer:

Answer:

p-value~ 0.075

Step-by-step explanation:

( Choice D on Khan Academy)

The closest answer choice is (Choice A) A p-value ≈ 0.01, is the best approximation among the answer choices given.

In a statistical hypothesis test, the p-value is a probability value that measures the strength of evidence against the null hypothesis.

It is the likelihood of observing a test statistic that is as extreme as or more extreme than the observed test statistic, assuming the null hypothesis is true.

To find the approximate p-value of the test, we need to calculate the proportion of simulated sample proportions that are greater than or equal to the sample proportion of 20%.

From the dot plot, we can see that only a few simulated sample proportions are greater than or equal to 20%, which suggests that the p-value is small.

Counting the dots from the dot plot, we can see that there are a total of 8 + 17 + 5 + 5 + 2 + 2 + 1 = 40 simulated sample proportions that are greater than or equal to 20%.

Therefore, the proportion of simulated sample proportions that are greater than or equal to 20% is 40/40 = 1.

Since the p-value is the probability of observing a sample proportion as extreme or more extreme than the observed sample proportion, given that the null hypothesis is true, we can approximate the p-value as the proportion of simulated sample proportions that are greater than or equal to the observed sample proportion.

Therefore, the approximate p-value is 1, which is very small and suggests strong evidence against the null hypothesis.

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The measurement of angle P in the diagram given in the question is 61.93 °

The trigonometric ratios which has been proven as regarding right angle triangles are given below:

Sine θ = Opposite / Hypothenus

Cos θ = Adjacent / Hypothenus

Tan θ = Opposite / Adjacent

With above information in mind, we can easily determine the value of angle P in the question. This is illustrated below:

From the question given above, the following data were obtained:

Sine θ = Opposite / Hypothenus

Sine P = 15 / 17

Sine P = 0.8824

Take the inverse of sine

P = Sine⁻¹ 0.8824

P = 61.93 °

Thus, the value of angle P in the diagram is 61.93 °

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Answer:

The y intercept is at (0,12) and the line has a positive slope of rising up 1.5 squares and moving horizontly 1 square

Step-by-step explanation:

Answer:

I hope this helps!

Step-by-step explanation:

ACB

Answer:

Estimate of difference is 4.

Step-by-step explanation:

10.04 = 10 to nearest whole number

6.3 = 6 to nearest whole number.

10 - 6 = 4.

The direction of A with respect to B. N B 73° N A is 287

A direction is the way that something is moving, directing, or moving toward something or someone. The directions could be Up, Down, Left, and Right, or North, South, East, and West.

The ability to specify where one thing is in relation to another is known as position in mathematics. Defining direction entails explaining the direction in which something moves, such as forward, backward, or in a full or half turn.

It is possible to define direction using relative terms like up, down, in, out, left, right, forward, backward, or sideways. A direction can also be denoted by one of the four cardinal directions: north, south, east, or west.

Angles at a point add up to 360 degrees.

⇒ Bearing from A to B is = 360° - 73° = 287°.

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Answer:

x = 2

Step-by-step explanation:

4x + 3 = 11

(4x + 3) - 3 = 11 - 3

4x = 8

4x/4 = 8/4

x = 2

As you can see, to isolate x, we must do the opposite operations to both sides.

Answer:

here is ur answer

Step-by-step explanation:

4x +2y = 4

4x = 4 -2y

x = 4/4 -2y

x= -2 y

The answer to this question will  t = 8. The first step to solving this equation is to combine the like terms by adding the two numbers together.

In this case, 5t + 35 = 40t. The next step is to divide both sides of the equation by the coefficient of the variable, which in this case is 5. When you do this, you get 40t/5 = t. Therefore, the answer to 5t + 35 is t = 8.

A coefficient is an integer that is either multiplied by the variable it is associated with or written alongside the variable. A coefficient is, in other words, the numerical factor of a term that contains both constants and variables. For instance, the coefficient in the expression 2x is 2.

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Answer:

about 5 milliliters

Step-by-step explanation:

x2 - 3x - 5 = 0 has a unique solution of type -11, which is what the equation entails.

The "discriminant" is the value that aids in identifying the many types of solutions when using the quadratic formula to solve a problem. It is a unique function of a co-efficient in a mathematical equation (a quadratic equation), and the value it yields provides details about the roots of the equation. Calculating the discriminant involves square rooting the "b" term and subtracting four times the "a" term from the "c" term. Alongside "D," there is a (delta) that represents discrimination. A quadratic equation with two distinct real number solutions is said to have a positive discriminant. A quadratic equation with a discriminant of zero has repeated real number solutions. Both of the solutions are not real numbers, as indicated by a negative discriminant.

given

x2 - 3x - 5 = 0

D=b2−4ac

D= −32−4×1×5

D=92−20

D= −11

x2 - 3x - 5 = 0 has a unique solution of type -11, which is what the equation entails.

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It will take a time of n/60m minutes for the jetski to cover n metres at the same speed.

Speed = distance/ time

Speed = m/s

For the distance of n meters.

distance = speed * time

n = m/s*time

n = m /s* time

Time = n(m/sec)

Time = n/m * sec

Time = n/60m min

Thus, it will take  n/60m minutes for the jetski to cover n metres at the same speed.

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The basis for the eigenspace corresponding to lambda=5,1,4 are None,[tex]\left[\begin{array}{c}-1 \\\frac{1}{2} \\0\end{array}\right][/tex] and [tex]$\left[\begin{array}{l}2 \\ 1 \\ 1\end{array}\right]$[/tex]

[tex]$$A=\left[\begin{array}{ccc}5 & -12 & 10 \\0 & 7 & -3 \\0 & 6 & -2\end{array}\right]$$[/tex]

Eigenspace corresponding to lambda=5,1,4

The eigenspace E_lambda corresponding to the eigenvalue lambda is the null space of the matrix a [tex]\mathrm{A}-(\lambda) \mathrm{I}"[/tex]

for lambda=5

[tex]$$\mathrm{E}_5=\mathrm{N}(\mathrm{A}-5 \mathrm{I})$$[/tex]

Reducing the matrix A-5I by elementary row operations

[tex]$$\begin{aligned}A-5 I & =\left[\begin{array}{ccc}5-5 & -12 & 10 \\0 & 7-5 & -3 \\0 & 6 & -2-5\end{array}\right] \\& =\left[\begin{array}{ccc}0 & -12 & 10 \\0 & 2 & -3 \\0 & 6 & -7\end{array}\right] \\& \sim\left[\begin{array}{ccc}0 & -12 & 10 \\0 & 1 & -\frac{3}{2} \\0 & 6 & -7\end{array}\right] R_2 \rightarrow \frac{R_2}{2} \\& \sim\left[\begin{array}{ccc}1 & 0 & -8 \\0 & 1 & -\frac{3}{2} \\0 & 6 & -7\end{array}\right] R_1 \rightarrow R_1+2 R_2\end{aligned}$$[/tex]

[tex]\sim\left[\begin{array}{ccc}1 & 0 & -8 \\ 0 & 1 & -\frac{3}{2} \\ 0 & 0 & 2\end{array}\right] R_3 \rightarrow R_3-6 R_2$$\\\sim\left[\begin{array}{ccc}1 & 0 & -8 \\ 0 & 1 & -\frac{3}{2} \\ 0 & 0 & 1\end{array}\right] R_3 \rightarrow \frac{\mathrm{R}_3}{2}$$\\\sim\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & -\frac{3}{2} \\ 0 & 0 & 1\end{array}\right] \mathrm{R}_1 \rightarrow \mathrm{R}_1+8 \mathrm{R}_3$[/tex]

[tex]$\sim\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] R_2 \rightarrow R_2+\frac{2 R_3}{2}$[/tex]

The solutions x of A-5I=0 satisfy x_1=x_2=x_3=0 that is, the null space solves the matrix

[tex]$$\left[\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right]\left[\begin{array}{l}x_1 \\x_2 \\x_3\end{array}\right]=\left[\begin{array}{l}0 \\0 \\0\end{array}\right]$$[/tex]

Hence The null space is [tex]\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right] E_5[/tex] has no basis

[tex]$$\begin{aligned}& \text { case: } 2 \\& \text { for } \lambda=1 \\& \mathrm{E}_5=\mathrm{N}(\mathrm{A}-(1) \mathrm{I})\end{aligned}$$[/tex]

we reduce the matrix A-I by elementary row operations as follows.

[tex]$$\begin{aligned}A-1 & =\left[\begin{array}{ccc}5-1 & -12 & 10 \\0 & 7-1 & -3 \\0 & 6 & -2-1\end{array}\right] \\& =\left[\begin{array}{ccc}1 & -3 & \frac{5}{2} \\0 & 6 & -3 \\0 & 6 & -3\end{array}\right] R_1 \rightarrow \frac{R_1}{4} \\& \sim\left[\begin{array}{ccc}1 & -3 & \frac{5}{2} \\0 & 1 & -\frac{1}{2} \\0 & 6 & -3\end{array}\right] R_2 \rightarrow \frac{R_2}{6}\end{aligned}[/tex]

[tex]$$$\sim\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 6 & -3\end{array}\right] R_1 \rightarrow R_1+3 R_2$\\$\sim\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 0 & 0\end{array}\right] R_3 \rightarrow R_3-6 R_2$[/tex]

Thus, the solutions x of (A-I) X=0 satisfy

[tex]$\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 0 & 0\end{array}\right]\left[\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$[/tex]

x_3=t

[tex]$\Rightarrow \mathrm{x}_1=-\mathrm{t}, \mathrm{x}_2=\frac{\mathrm{t}}{2}$[/tex]

[tex]$\vec{x}=\left[\begin{array}{c}-t \\ \frac{t}{2} \\ t\end{array}\right]=\left[\begin{array}{c}-1 \\ \frac{1}{2} \\ 1\end{array}\right] t$[/tex]

The Basis for the nullspace A-I will be: [tex]$\left.\left(\begin{array}{c}-1 \\ \frac{1}{2} \\ 1\end{array}\right]\right)$[/tex]

case:3

lambda=4

[tex]$$\mathrm{E}_5=\mathrm{N}(\mathrm{A}-(4) \mathrm{I})$$[/tex]

we reduce the matrix A-4I by elementary row operations as follows.

[tex]$\begin{aligned} A-4 \mid & =\left[\begin{array}{ccc}5-4 & -12 & 10 \\ 0 & 7-4 & -3 \\ 0 & 6 & -2-4\end{array}\right] \\ & =\left[\begin{array}{ccc}1 & -12 & 10 \\ 0 & 3 & -3 \\ 0 & 6 & -6\end{array}\right] \\ & \sim\left[\begin{array}{ccc}1 & -12 & 10 \\ 0 & 1 & -1 \\ 0 & 6 & -6\end{array}\right] R_2 \rightarrow \frac{R_2}{3}\end{aligned}$[/tex]

[tex]$\begin{aligned} & \sim\left[\begin{array}{ccc}1 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 6 & -6\end{array}\right] \mathrm{R}_1 \rightarrow \mathrm{R}_1+12 \mathrm{R}_2 \\ & \sim\left[\begin{array}{ccc}1 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 0 & 0\end{array}\right] \mathrm{R}_3 \rightarrow \mathrm{R}_3-6 \mathrm{R}_2\end{aligned}$[/tex]

Thus, the solutions x of (A-4IX)=0 satisfy

[tex]$$\left[\begin{array}{ccc}1 & 0 & -2 \\0 & 1 & -1 \\0 & 0 & 0\end{array}\right]\left[\begin{array}{l}x_1 \\x_2 \\x_3\end{array}\right]=\left[\begin{array}{l}0 \\0 \\0\end{array}\right]$$[/tex]

x_3=t

[tex]$\Rightarrow \mathrm{x}_1=2 \mathrm{t}, \mathrm{x}_2=\mathrm{t}$[/tex]

[tex]$$\vec{x}=\left[\begin{array}{c}2 t \\t \\t\end{array}\right]=\left[\begin{array}{l}2 \\1 \\1\end{array}\right] t$$[/tex]

The Basis for the nullspace A-4 I will be [tex]\left(\left[\begin{array}{l}2 \\ 1 \\ 1\end{array}\right]\right)[/tex]

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Answer: difference means subtract

x-7

HOPE THIS HELPS

Answer:

The coordinates (8,9) lie in quadrant 1

Step-by-step explanation:

The coordinates on quadrant ___ go like this:

Quadrant l : (+,+)

Quadrant ll : (-,+)

Quadrant lll : (-,-)

Quadrant llll : (+,-)

Answer:The cost of each can is $ 0.31

.

Step-by-step explanation: You need to take the amount of money each pack costs and divide it by the number of cans to find the unit rate. Hope this helps!!

brianlist please ?

To find the unit cost, simply divide the amount of money by the amount.

4.80÷12= 0.4 dollars

So each soda is only 0.4 dollars.

How can you check this?

0.4 dollars times 12 should equal the total cost of soda.

0.4x12=4.80, so it's correct.

Answer:

fault lines.

Step-by-step explanation:

Answer:

135

Step-by-step explanation:

m^2 = 3

Cube m^2

(m^2)^3 = 3^3

m^(2*3) = 27

m^6 = 27

Now multiply by 5

5 * m^6 = 5*27

5 m^6 =135

Linear equations are equations that can be written in the form ax + b = c, where a, b, and c are real numbers and a ≠ 0.

The equation can be solved for the variable x by subtracting b from both sides and then dividing both sides by a. This will give the solution for x, which is x = (c - b) / a.

For example, consider the equation 2x + 4 = 10. This is a linear equation since it can be written in the form ax + b = c, where a = 2, b = 4, and c = 10. To solve for x, subtract b from both sides to get 2x = 6, and then divide both sides by a to get x = 3. Therefore, the solution to the equation is x = 3.

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Answer:

59

Step-by-step explanation:

The Given quadratic equation has two solutions: 1 and 3/2. A quadratic problem was solved by utilizing the factoring method.

A quadratic equation is defined.

At least one squared term must be present because a quadratic is a second-degree polynomial equation. It is also known as quadratic equations. ax^2 + b*x + c = 0 is the quadratic equation's general form.

where x is the unknown variable and a, b, and c are constant terms.

In this case, I'm solving a quadratic problem by factoring.

The following quadratic equation is

2x^2 - 5x + 3 = 0.

2x^2 - (3+2)x + 3 = 0

2x^2 -3x -2x + 3 = 0.

x(2x-3) -1(2x -3) = 0.

(2x-3)(x-1) = 0.

2x - 3 =0 or x -1 =0

x = 3/2 or x =1

So the solutions of a quadratic equation are 3/2 and 1.

Given Question is incomplete complete question here:

What are the solutions of the quadratic equation 2x^2-5x+3 =  0?

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The volume of the cube, given the lengths of the small cubes used to make the larger cube, is 8 cm ³

The surface area of the cube is 24 cm ²

The total length of the edges of the cube is  24 cm

The volume of a cube can be found by the formula :

= Side x Side x Side

This is because all the sides of a cube at the same and so the volume would measure them three - dimensionally.

The side length of this cube is the length of two of the smaller cubes as these make up a side of the cube :

= 2 x 1

= 2 cm

The volume is therefore :

= 2 x 2 x 2

= 8 cm ³

The surface area of a cube is:

= 6 x side ²

= 6 x 2 ²

= 24 cm ²

The total length of the edges would be:

= 12 edges on a cube x length of each edge

= 12 x 2

= 24 cm

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Answer:

yes

Step-by-step explanation: