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love history [14]

3 years ago

15

Rosita earns money by walking dogs after school and on the weekends . She walks Madeline every other day ,Buddy every forth day

, and Ernie every third day . Today she walked all thee dogs . What day did Rosita walk all three dogs?

1 answer:

sweet-ann [11.9K]3 years ago

5 0

Answer:

<u><em>Rosita will walk all three dogs in 12 days.</em></u>

Explanation:

Madeline is walked every other day means every two days: 2.

Buddy is walked every fourth day: 4.

Ernie is walked every third day: 3.

You need to find the least common multiple (LCM) of 2, 4, 3.

Decompose each number into its prime factors:

2 = 2
4 = 2²
3 = 3

The LCM is the product of the common factors and non common factors each raised to its the greatest power:

LCM = 2² × 3 = 12

Then, Rosita will walk all three dogs in 12 days.

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Step-by-step explanation:

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= 205/14

Approx. 14.64 cents

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1.) The equation for the pH of a substance is pH = –log[H+], where H+ is the concentration of hydrogen ions. What is the approxi

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Step-by-step explanation:

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2 years ago

HELP ASAP PLZZZZ

Tcecarenko [31]

QUESTION 1

The given system of equations is

$3d - e = 7...eqn(1)$
$d + e = 5...eqn(2)$

To solve by linear combination, we add equation (1) to equation (2) to get,

$3d + d= 7 + 5$

$4d = 12$

We divide through by 4 to obtain,

$d = \frac{12}{4}$

$d = 3$

We put d=3 into equation (2) to get,

$3+ e = 5$

$e = 5 - 3$

$e = 2$

$\boxed {The \: solution \: is \: (3, 2)}$

QUESTION 2

The given system is

4x + y = 5 ...eqn(1)

3x + y = 3 ...eqn(2)

To solve by linear combination, we subtract equation (2) from equation (1) to eliminate y from the equation.

This will give us,

$4x - 3x = 5 - 3$

This implies that,

$x = 2$

Put x=3 into equation (1) to get,

$4(2) + y = 5$

$8+ y = 5$

$y = 5 - 8$

$y = - 3$

The solution is

$(2,-3)$

QUESTION 3

We want to solve the system;

a – 2b = –2 ....eqn(1)

2a + 2b = 14...eqn(2)

by linear combination.

We need to add equation (1) to equation (2) to eliminate b.

This implies that,

$2a + a = 14 + - 2$

Simplify,

$3a = 12$

Divide both sides by 3 to get,

$a = 4$
Put a=4 into equation (2) to obtain,

$2(4) + 2b = 14$

$8 + 2b = 14$
$2b = 14 - 8$

$2b = 6$

$b = 3$

The ordered pair in the form (a, b) is

$(4,3)$

QUESTION 4

The given system of equations is

11x + 4y = 18 ...eqn(1)

3x + 4y = 2 ...eqn(2)

We subtract equation (2) from equation (1) to get,

$11x - 3x = 18 - 2$

$8x = 16$

$x = 2$

Put x=2 into equation (2) to obtain,

$3(2) + 4y = 2$

This implies that,

$6 + 4y = 2$

$4y = 2 - 6$

$4y = - 4$

$y=-1$

The correct answer is (2,-1).

QUESTION 5

The given system is ;

2d + e = 8...eqn1

d – e = 4...eqn2

We add the two equations to eliminate e.

This implies that,

$2d + d = 8 + 4$

$3d = 12$

We divide both sides by 3 to get,

$d = 4$

We put d=4 into equation (2) to get,

$4 - e = 4$

$- e = 4 - 4$

$- e = 0$

$e = 0$

The solution is

$(4,0)$

7 0

3 years ago

Given A(-2, 5) and B(13, -7), find the midpoint of AB.

Dmitrij [34]

Answer:

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Step-by-step explanation:

You would use the mid-point formula for this

$(\frac{x_{2} +x_{1} }{2} , \frac{y_{2}+y_{1} }{2} )$

if you plug that in it is ( $\frac{13-2}{2}, \frac{-7+5}{2}$ )

resulting in (11/2,-2/2) = (11/2,-1)

4 0

1 year ago