The pairs of polygons below are similar. Give the scale factor of figure A to figure B.

1/8
2
1/4
8
1/2
4

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 : (+,-)

Step-by-step explanation:

[tex]2\pi {r}^{2} + 2\pi rh[/tex]

[tex]2(3.14) {(20)}^{2} + 2(3.14)(20)(11)[/tex]

[tex]6.28(400) + 6.28(220) \\ 2512 + 1381.6 = 3893.6 {mi}^{2} [/tex]

Answer:

4

Step-by-step explanation:

Answer:

C11=-2

C22=8

hope this will help you more

good luck

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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Answer:

about 5 milliliters

Step-by-step explanation:

Answer:

I hope this helps!

Step-by-step explanation:

ACB

Answer:

i) The angular difference in their longitude is approximately 50.65°

ii) The line of longitude where B is located is approximately 30.65°E

Step-by-step explanation:

The given parameters are;

The angle of latitude where the points A and B are located = 60°S

The distance between points A and B = 2,816 km

(i) The angular difference in their longitude, θ, is given as follows;

θ = 2,816/(2 × π × 6,371 × cos(60°)) × 360 ≈ 50.65°

The angular difference in their longitude, θ ≈ 50.65°

(ii) The location of the point A relative to B = Due west

The line of longitude on which A is located = 20°W

∴ The line of longitude where B is located, P = θ - 20°

∴ P = 50.65° - 20°W = 30.65°E

The line of longitude where B is located ≈ 30.65°E.

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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Answer:

3 3/4 or 3 6/8

Step-by-step explanation:

hope this helps <3

Answer:

1. True

2. False

3. True

4. False

Step-by-step explanation:

Therefore, 60 different combinations can be made wearing one of each type

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Answer:

Helloooooooooo theree

Answer:

don't

Step-by-step explanation: it hard

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:

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

No solution is when a system of equations has no solution. This means that the equations are inconsistent, and there is no set of values that can make the equations true.

For example, if the system of equations is 2x + 3y = 4 and 2x + 3y = 5, then there is no solution. This is because no matter what values are chosen for x and y, the equations cannot both be true. This means that the system of equations is inconsistent, and there is no solution.

To symbolically represent no solution, you can use the following notation:

2x + 3y = 4, 2x + 3y = 5   No solution

This notation is used to indicate that the equations are inconsistent and that there is no solution.

Complete question: What is the solution to the equation 2x + 3y = 4 and 2x + 3y = 5?

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Answer:

D

Step-by-step explanation:

find area of first rectangle

18 x 6 = 108

find area of second rectangle

20 x 5 = 100

find area of two triangles

7 x 8 =56

56/2 = 28

each triangle has an area of 28

28+28+100+108 = 264ft

Answer:

59

Step-by-step explanation:

-20w-35

I used the distributive property to distribute terms, and then simplified

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 number of times which the two computers back up at the same time in twenty four hours as required is; 144.

It follows from the task content that the first computer backs up after every five minutes and the second computer backs up every two minutes.

On this note, the least common multiple of both backup times represents when the two computers back up at the same time.

Hence, since the LCM of 5 and 2 is; 10.

It follows that the two computers back up at the same time after every ten minutes.

Ultimately, the number of times for which they backup at the same time in 24 hours; since 60 minutes = 1 hour is;

= ( 24 × 60 ) / 10

= 24 × 6

= 144 times.

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Answer:

y=4x+2

Step-by-step explanation:

-8x+2y=4

y=4x+2

−8x+2(4x+2)=4-8x+2(4x+2)=4

y=4x+2

−8x+8x+4=4-8x+8x+4=4

y=4x+2

0+4=40+4=4

y=4x+2

4=4

y=4x+2

Answer:

140

Step-by-step explanation:

60+80=?......All you need to do is just add 60 and 80 and get the product

I hope that helped.

A linear, negative relationship is the type of relationship which is shown in graph of Time vs. Position of Battery Operated Car .

A linear equation is an algebraic expression such as y=mx+b. The slope is m, and the y-intercept is B. A "linear equation in two variables" is the term used to describe the previous sentence, which has variables named y and x. Two different variables can be found in bivariate linear equations. Examples of linear equations include 2x - 3 = 0, 2y = 8, m + 1 = 0, x/2 = 3, x + y = 2, and 3x - y + z = 3. The term "linear" refers to an equation having the form y=mx+b, where m stands for the slope and b for the y-intercept.

given

The area's 46-16 opposed to 14 is different from the area's 18 and T in a different way. The place is there. It isn't linearly getting farther apart. The lee is actually growing as a result.

Option B, the slanting upwards, is what we're going to go with. Since it will be linear, it isn't option C. It is untrue that the speaking has not been constant over the distance. Because it is a D-option, this increase-decrease is untrue. Option B is the best choice. Absolutely.

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Answer:

here is ur answer

Step-by-step explanation:

4x +2y = 4

4x = 4 -2y

x = 4/4 -2y

x= -2 y

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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Answer:

D)  (6.6, 7.4)

Step-by-step explanation:

Begin by finding the margin of error based on the given information.

[tex]\boxed{\begin{minipage}{7.3cm}\underline{Margin of Error formula}\\\\$\textsf{Margin of Error}= Z \times \dfrac{S}{\sqrt{n}}$\\\\where:\\\phantom{ww} $\bullet$ $Z =$ $Z$ score\\\phantom{ww} $\bullet$ $S =$ Standard Deviation of a population\\\phantom{ww} $\bullet$ $n =$ Sample Size\\\end{minipage}}[/tex]

Given:

The z-score for 95% confidence level is 1.96.

Therefore, to find the margin of error, substitute the values into the formula:

[tex]\implies \textsf{Margin of Error}= 1.96 \times \dfrac{1.3}{\sqrt{50}}[/tex]

[tex]\implies \textsf{Margin of Error}=0.360341...[/tex]

To find a reasonable range for the true mean number of hours a teenage spend on their phone, subtract and add the margin of error to the given mean of 7 hours:

[tex]\begin{aligned}\implies \textsf{Lower bound}&=7-0.360341...\\&=6.63965...\\&=6.6\;\; \sf (1\;d.p.)\end{aligned}[/tex]

[tex]\begin{aligned}\implies \textsf{Upper bound}&=7+0.360341...\\&=7.360341...\\&=7.4\;\; \sf (1\;d.p.)\end{aligned}[/tex]

Therefore, the reasonable range for the true mean is:

Given:

Quadrilateral STUV is rotated 270° counterclockwise about the origin.

The coordinates of V were (3, -4).

To find:

The coordinates of V’.

Solution:

If a figure rotated 270° counterclockwise about the origin, then the rule of rotation is:

[tex](x,y)\to (y,-x)[/tex]

The coordinates of V were (3, -4). Using the above rotation rule, we get

[tex]V(3,-4)\to V'(-4,-3)[/tex]

Therefore, the coordinates of V’ are (-4,-3).

The measures of the angles of the parallelogram are

The measure of ∠A = 72°

The measure of ∠B = 108°

A parallelogram is a simple quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure

The four types are parallelograms, squares, rectangles, and rhombuses

Properties of Parallelogram

Opposite sides are parallel

Opposite sides are congruent

Opposite angles are congruent

Same-Side interior angles (consecutive angles) are supplementary

Each diagonal of a parallelogram separates it into two congruent triangles

The diagonals of a parallelogram bisect each other

Given data ,

Let the parallelogram be represented as ABCD

Now , the measure of ∠A = ( 5y - 3 )°

The measure of ∠C = ( 3y + 27 )°

For a parallelogram , opposite angles are congruent

So , measure of ∠A = measure of ∠C

Substituting the values in the equation , we get

( 5y - 3 )° =  ( 3y + 27 )°

On simplifying the equation , we get

( 5y - 3 ) =  ( 3y + 27 )

Subtracting 3y on both sides of the equation , we get

2y - 3 = 27

Adding 3 on both sides of the equation , we get

2y = 30

Divide by 2 on both sides of the equation , we get

y = 15

So , the measure of ∠A = ( 5y - 3 )°

The measure of ∠A = ( 5 ( 15 ) - 3 ) )°

The measure of ∠A = ( 75 - 3 )°

The measure of ∠A = 72°

Now , for a parallelogram , same-side interior angles (consecutive angles) are supplementary

So , the measure of ∠A + the measure of ∠B = 180°

Substituting the values in the equation , we get

72° + the measure of ∠B = 180°

Subtracting 72° on both sides of the equation , we get

The measure of ∠B = 108°

Hence , the measures of angles of the parallelogram are 72° and 108° respectively

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Answer:

Image attached

Step-by-step explanation:

Origin looks to be (0, -1), so move each point out by 2