sale 90%Luca buys a swing set during the sale. If the original price was $270, how much does Luca pay?

(x - 10)² is the graph x² by translation of 10 units moved to the right.

Answer:

free energy, in thermodynamics, energy-like property or state function of a system in thermodynamic equilibrium. Free energy has the dimensions of energy, and its value is determined by the state of the system and not by its history.

Answer:

OMG️️

Step-by-step explanation:

What is this❓

what have you wrote✍️

Answer:

hye nice what have written tell then I will answer you.

Answer:

ggggggggggggggggggggg

Answer:

12

Step-by-step explanation:

the angle is defined by equation ((n-2)*180)/n,where n is number of sides of a regular polygon

so here 180n-360=150n

30n=360

n=12

Answer: 32

Step-by-step explanation:

The roman numeral XXXII is 32 and XXIII is 23.

The number for this Roman numeral XXXII is, 32

Given that,

We have to write the number for this Roman numeral  XXXII.

Since, We know that,

X represent in number = 10

I represent in number = 1

Hence, The number for this Roman numeral  XXXII is,

⇒ XXXII

⇒ (10 + 10 + 10 + 1 + 1)

⇒ 32

Therefore, the number for this Roman numeral  XXXII is, 32

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Answer:

6/50

Step-by-step explanation:

0.12 as a fraction is 6/50.

Answer:

3/25

Step-by-step explanation:

12/100=6/50=3/25 .

Answer:

K cool

Step-by-step explanation:

Answer:

Step-by-step explanation:

Answer:

the 20 dollars = the slope

the fee = the y-intercept

if a line has a slope of 20 and passes through the point (7,200), then what is the y-intercept?

y-intercept is 60

Answer:

[tex]\frac{7}{11}[/tex]

Step-by-step explanation:

We require 2 equations with the repeating digits (63) placed after the decimal point.

let x = 0.636363..... (1) multiply both sides by 100

100x = 63.6363... (2)

Subtract (1) from (2) thus eliminating the repeating digits

99x = 63 ( divide both sides by 99 )

x = [tex]\frac{63}{99}[/tex] = [tex]\frac{7}{11}[/tex] ← in simplest form

Answer:

Your answer should be 2.

The sentence representing the equation is 912 is the same as 15 decreased by a number.

An equation is a mathematical statement that shows that two mathematical expressions are equal.

Given an equation, 912 = 15 − x

Here, 912 is equal to a number which is being subtracted from 15.

Hence, The sentence representing the equation is 912 is the same as 15 decreased by a number.

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He needed to make a total of 1980 + 630 = $2610

$2610 / 45 = 58

Answer: 58

Answer:

$22.48 for 4 pairs of jeans is a better deal.

Step-by-step explanation:

To find the price of one pair of jeans, you divide.

39.55 ÷ 7 = price of 1 pair of jeans

39.55 ÷ 7 = $5.65

22.48 ÷ 4 = price of 1 pair of jeans

22.48 ÷ 4 = $5.62

The price difference between the two prices is 3 cents. So, $22.48 for 4 pairs is a better deal that $39.55 for 7 pairs of jeans.

Hope this helps!

Answer:

B

Step-by-step explanation:

Using the probability concept, it is found that since the number of red sweets would be a decimal number, the probability cannot be [tex]\frac{7}{10}[/tex]

A probability is the number of desired outcomes divided by the number of total outcomes.

In this problem:

The probability of red is:

[tex]p = \frac{n}{n + 8}[/tex]

Supposing [tex]p = \frac{7}{10}[/tex], we solve for n:

[tex]\frac{n}{n + 8} = \frac{7}{10}[/tex]

[tex]10n = 7n + 56[/tex]

[tex]3n = 56[/tex]

[tex]n = \frac{56}{3}[/tex]

[tex]n = 18.67[/tex]

Since the number of red sweets would be a decimal number, the probability cannot be [tex]\frac{7}{10}[/tex]

A similar problem is given at https://brainly.com/question/15536019

Answer:

  10%

Step-by-step explanation:

The change is 90-100 = -10. As a percentage of the original amount, that is ...

  -10/100 × 100% = -10%

The change from 100 to 90 is a reduction of 10%.

Answer:

Step-by-step explanation:

the diagonal of a square splits the square into 2 right triangles. So we can use Pythagorean's theorem.

where c is the hypotenuse. So the diagonal is the hypotenuse here, and thus c = 10. Now, since we are dealing with a square, all the sides are the same length, so a = b. So we have:

a² + a² = c²

2a² = 100

a² = 50

a = √50

a = 5 × √2 or 7,071067811865475

--------------------------

Answer: 50

Step-by-step explanation  :<

Answer:

1400

Step-by-step explanation:

Answer:

Below.

Step-by-step explanation:

1. 10^8 × 10^4 = 10(8+4) = 10^12.

2. (11^5)^4 = 11^(5*4) = 11^20.

3. 8^6 ÷ 8^3 = 8^(6-3) = 8^3.

4. (12^2) × 12^4 = 12^6.

5. (13^4) ÷ 13^5 = 13^(4-5) = 13^-1.

6. (5^2 × 5^3) ÷ 5^4 = 5^5 / 5^4 = 5.

7. 18^4 ÷ 18^6 = 18^-2.

8. (19^2)^4 ÷ 19^8​ = 19^8 / 19^8  = 18^(8-8) = 18^0 = 1.

The true option is: (d) Hugo has 15 songs in his music player. He will add 10 songs every month. Hugo collects songs for r months. For what values of r will Hugo have at least 135 songs?

The inequality is given as:

[tex]\mathbf{135 \le 10r + 15}[/tex]

Rewrite as:

[tex]\mathbf{10r + 15\ge 135 }[/tex]

From the options, we can see that the inequality represents songs in a music player.

Linear inequalities can be represented as:

[tex]\mathbf{mx + b \ge y}[/tex]

Where:

m represents the rate i.e. 10

b represents the y-intercept or base i.e. 15

>= represents at least

So, the inequality can be interpreted as:

Hence, the true option is (d)

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Answer:

u need to use the quadratic formula

Step-by-step explanation:

I think this is about it

Answer:

r = 5q / (p + 10)

Step-by-step explanation:

p = 5(q - 2r)/r

multiply both sides by r

pr = 5(q - 2r)

distribute

pr = 5q  - 10r

add 10r to both sides

pr + 10r = 5q

Factor out r

r(p + 10) = 5q

divide both sides by p + 10

r = 5q / (p + 10)

Answer: 629.51

Step-by-step explanation: 30% of 864.72 is 259.416

864.72 - 259.416 = 605.304      4% of 605.3 = 24.212     605.3 + 24.21 = 629.51

It looks like given matrices are supposed to be

[tex]\begin{array}{ccccccc}\begin{bmatrix}3&2\\2&3\end{bmatrix} & & \begin{bmatrix}1&-1\\2&-1\end{bmatrix} & & \begin{bmatrix}1&2&3\\0&2&3\\0&0&3\end{bmatrix} & & \begin{bmatrix}3&1&1\\1&3&1\\1&1&3\end{bmatrix}\end{array}[/tex]

You can find the eigenvalues of matrix A by solving for λ in the equation det(A - λI) = 0, where I is the identity matrix. We also have the following facts about eigenvalues:

• tr(A) = trace of A = sum of diagonal entries = sum of eigenvalues

• det(A) = determinant of A = product of eigenvalues

(a) The eigenvalues are λ₁ = 1 and λ₂ = 5, since

[tex]\mathrm{tr}\begin{bmatrix}3&2\\2&3\end{bmatrix} = 3 + 3 = 6[/tex]

[tex]\det\begin{bmatrix}3&2\\2&3\end{bmatrix} = 3^2-2^2 = 5[/tex]

and

λ₁ + λ₂ = 6   ⇒   λ₁ λ₂ = λ₁ (6 - λ₁) = 5

⇒   6 λ₁ - λ₁² = 5

⇒   λ₁² - 6 λ₁ + 5 = 0

⇒   (λ₁ - 5) (λ₁ - 1) = 0

⇒   λ₁ = 5 or λ₁ = 1

To find the corresponding eigenvectors, we solve for the vector v in Av = λv, or equivalently (A - λI) v = 0.

• For λ = 1, we have

[tex]\begin{bmatrix}3-1&2\\2&3-1\end{bmatrix}v = \begin{bmatrix}2&2\\2&2\end{bmatrix}v = 0[/tex]

With v = (v₁, v₂)ᵀ, this equation tells us that

2 v₁ + 2 v₂ = 0

so that if we choose v₁ = -1, then v₂ = 1. So Av = v for the eigenvector v = (-1, 1)ᵀ.

• For λ = 5, we would end up with

[tex]\begin{bmatrix}-2&2\\2&-2\end{bmatrix}v = 0[/tex]

and this tells us

-2 v₁ + 2 v₂ = 0

and it follows that v = (1, 1)ᵀ.

Then the decomposition of A into PDP⁻¹ is obtained with

[tex]P = \begin{bmatrix}-1 & 1 \\ 1 & 1\end{bmatrix}[/tex]

[tex]D = \begin{bmatrix}1 & 0 \\ 0 & 5\end{bmatrix}[/tex]

where the n-th column of P is the eigenvector associated with the eigenvalue in the n-th row/column of D.

(b) Consult part (a) for specific details. You would find that the eigenvalues are i and -i, as in i = √(-1). The corresponding eigenvectors are (1 + i, 2)ᵀ and (1 - i, 2)ᵀ, so that A = PDP⁻¹ if

[tex]P = \begin{bmatrix}1+i & 1-i\\2&2\end{bmatrix}[/tex]

[tex]D = \begin{bmatrix}i&0\\0&i\end{bmatrix}[/tex]

(c) For a 3×3 matrix, I'm not aware of any shortcuts like above, so we proceed as usual:

[tex]\det(A-\lambda I) = \det\begin{bmatrix}1-\lambda & 2 & 3 \\ 0 & 2-\lambda & 3 \\ 0 & 0 & 3-\lambda\end{bmatrix} = 0[/tex]

Since A - λI is upper-triangular, the determinant is exactly the product the entries on the diagonal:

det(A - λI) = (1 - λ) (2 - λ) (3 - λ) = 0

and it follows that the eigenvalues are λ₁ = 1, λ₂ = 2, and λ₃ = 3. Now solve for v = (v₁, v₂, v₃)ᵀ such that (A - λI) v = 0.

• For λ = 1,

[tex]\begin{bmatrix}0&2&3\\0&1&3\\0&0&2\end{bmatrix}v = 0[/tex]

tells us we can freely choose v₁ = 1, while the other components must be v₂ = v₃ = 0. Then v = (1, 0, 0)ᵀ.

• For λ = 2,

[tex]\begin{bmatrix}-1&2&3\\0&0&3\\0&0&1\end{bmatrix}v = 0[/tex]

tells us we need to fix v₃ = 0. Then -v₁ + 2 v₂ = 0, so we can choose, say, v₂ = 1 and v₁ = 2. Then v = (2, 1, 0)ᵀ.

• For λ = 3,

[tex]\begin{bmatrix}-2&2&3\\0&-1&3\\0&0&0\end{bmatrix}v = 0[/tex]

tells us if we choose v₃ = 1, then it follows that v₂ = 3 and v₁ = 9/2. To make things neater, let's scale these components by a factor of 2, so that v = (9, 6, 2)ᵀ.

Then we have A = PDP⁻¹ for

[tex]P = \begin{bmatrix}1&2&9\\0&1&6\\0&0&2\end{bmatrix}[/tex]

[tex]D = \begin{bmatrix}1&0&0\\0&2&0\\0&0&3\end{bmatrix}[/tex]

(d) Consult part (c) for all the details. Or, we can observe that λ₁ = 2 is an eigenvalue, since subtracting 2I from A gives a matrix of only 1s and det(A - 2I) = 0. Then using the eigen-facts,

• tr(A) = 3 + 3 + 3 = 9 = 2 + λ₂ + λ₃   ⇒   λ₂ + λ₃ = 7

• det(A) = 20 = 2 λ₂ λ₃   ⇒   λ₂ λ₃ = 10

and we find λ₂ = 2 and λ₃ = 5.

I'll omit the details for finding the eigenvector associated with λ = 5; I ended up with v = (1, 1, 1)ᵀ.

• For λ = 2,

[tex]\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}v = 0[/tex]

tells us that if we fix v₃ = 0, then v₁ + v₂ = 0, so that we can pick v₁ = 1 and v₂ = -1. So v = (1, -1, 0)ᵀ.

• For the repeated eigenvalue λ = 2, we find the generalized eigenvector such that (A - 2I)² v = 0.

[tex]\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}^2 v = \begin{bmatrix}3&3&3\\3&3&3\\3&3&3\end{bmatrix}v = 0[/tex]

This time we fix v₂ = 0, so that 3 v₁ + 3 v₃ = 0, and we can pick v₁ = 1 and v₃ = -1. So v = (1, 0, -1)ᵀ.

Then A = PDP⁻¹ if

[tex]P = \begin{bmatrix}1 & 1 & 1 \\ 1 & -1 & 0 \\ 1 & 0 & -1\end{bmatrix}[/tex]

[tex]D = \begin{bmatrix}5&0&0\\0&2&0\\0&2&2\end{bmatrix}[/tex]