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azamat
2 years ago
12

Matt can buy an 8-ounce jar of salsa for 2.79 or a 15 ounce jar of salsa for 5.54. which jar has the best unit rate

Mathematics
1 answer:
harkovskaia [24]2 years ago
4 0
8 oz of salsa. has a unit rate of around 35 cents per ounce.
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Find x, round to the nearest tenth.
Paladinen [302]

Answer:

Step-by-step explanation:

Law of Cosines

x = √(40²+12² - 2·40·12·cos(37°)) ≅ 31.3 units

7 0
3 years ago
-40-2(3m+1/2)=7m-2<br> Solve for m please<br> Thank you.
Anvisha [2.4K]

Answer:

{-3}

Step-by-step explanation:

Раскроем скобки:

- 40 - 6m - 1 = 7m - 2

Перенесем все, что с m, в левую часть:

- 13m = 39

m = -3

4 0
2 years ago
Read 2 more answers
Out of the 25 students in Mrs. Coulter's class, 15 are girls. In the school there are 390 boys. If the ratio of boys to students
GREYUIT [131]

Answer:

975 students

Step-by-step explanation:

This is a dense question. Let's look at it piece by piece:

1. The ratio of boys to students in the class is 10/25. If there are 25 students and 15 girls, that leaves 10 boys.

2. The question says that the ratio is in proportion of boys to total students in the school. "In proportion" in this case means that if we convert 10/25 to a fraction with numerator 390, the denominator is the amount of students in school.

3. 10x39=390. What you do the numerator you must do to the denominator: 25x39=975.

4. This means 975 students go to the school.

6 0
2 years ago
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Charactersitics of an algebraic expression
artcher [175]

It would have a variable, and/or coefficients and sometimes constants

6 0
3 years ago
Linear Algebra: Permutation Matrices Let M be the matrix {{0,0,0,1,0}, {0,0,1,0,0}, {0,1,0,0,0}, {0,0,0,0,1}{1,0,0,0,0}}. What i
FromTheMoon [43]

Multiplying M by any matrix A would return new matrix, B, in which

• the 1st row of B is equal to the 4th row of A,

• the 2nd row of B is equal to the 3rd row of A,

• the 3rd row of B is equal to the 2nd row of A,

• the 4th row of B is equal to the 5th row of A, and

• the 5th row of B is equal to the 1st row of A.

The pattern here is

1 => 4 => 5 => 1

2 => 3 => 2

Let {4, 3, 2, 5, 1} denote the matrix M, where each number refers to the row of the identity matrix, I.

Using this notation, the pattern above gives

M² = {5, 2, 3, 1, 4}

M³ = {1, 3, 2, 4, 5}

M⁴ = {4, 2, 3, 5, 1}

M⁵ = {5, 3, 2, 1, 4}

M⁶ = {1, 2, 3, 4, 5}

so that <em>n</em> = 6.

(Notice that the first cycle has length 3 and the second one has length 2; the minimum <em>n</em> needed here is then LCM(2, 3) = 6.)

6 0
3 years ago
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