All categories

3 years ago

X-Y=5, 6X-6Y =?

Mathematics

2 answers:

denis-greek [22]3 years ago

3 0

Answer:

Yes, it is 30

Step-by-step explanation:

And yes, you thought it through correctly. 6(X-Y) = 6*5

6(X-Y) = 30

tatiyna3 years ago

3 0

Answer:

$\displaystyle 6x - 6y = 30$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Terms/Coefficient

Factoring

Step-by-step explanation:

Step 1: Define

Identify.

x - y = 5

Step 2: Solve

Factor: $\displaystyle 6x - 6y = 6(x - y)$
Substitute in expression: $\displaystyle 6x - 6y = 6(5)$
[Order of Operations] Evaluate: $\displaystyle 6x - 6y = 30$

You might be interested in

Place Temperature (° Fahrenheit) Death Valley, CA 134 Barrow, AK negative 53 Question What is the difference of the temperatures

Mandarinka [93]

Answer:

D: 187 °F

Step-by-step explanation:

You need to subtract -53 from 134. But since that 53 degrees is negative and you are subtracting it, you change the negative sign to a positive sign and add positive 53 to positive 134 to get 187.

8 0

3 years ago

Can someone please help me find the domain, range, intercepts, and asymptotes pf this function?

marishachu [46]

1) given function

y = - 2 ^ ( -x + 2) + 1

2) domain: domain is the set of the x-values for which the function is defined.

The exponential function is defined for all the real numbers, so the domain of the given function is all the real numbers.

3) x-intercept => y = 0

=> y = - 2 ^ ( -x + 2) + 1 = 0 => 2^ ( -x + 2) = 1

=> - x + 2 = 0 => x = 2

The x-intercept is x = 0

4) y-intercept => x = 0

=> y = - 2 ^ ( -x + 2) + 1= - 2 ^ ( 0 + 2) 1 = - (2)^(2) + 1 =- 4 + 1 = - 3

=> The y-intercept is - 3

5) limit when x -> negative infinite

Lim f(x) when x -> ∞ = - ∞

6) limit when x -> infinite

Lim f(x) when x - > infinite = 1

=> asymptote = y = 1

7) range is the set of values of the fucntion: y

Given that the function is strictly decreasing from -∞ to ∞, the range is from - ∞ to less than 1

Range (-∞,1)

3 0

3 years ago

By graphing the system of constraints find the values of x and y that maximize the objective function, find the maximum value.

sveticcg [70]

P = 4 IS THE ANSWER KKKKKKKKKKKKKKK

5 0

3 years ago

Read 2 more answers

In Tijuana, the morning temperature was 20°C and by the afternoon, the temperature fell by 11°C. What is the temperature during

Vaselesa [24]

The answer is 21 ok????

3 0

3 years ago

Read 2 more answers

About A red and a blue die are thrown. Both dice are fair. The events A, B, and C are defined as follows: A: The sum on the two

Flauer [41]

$p(A) = \frac{1}{2}$

$p(B) = \frac{1}{6}$

$p(C) = \frac{1}{6}$

$P(A | C)=\frac{1}{2}$

Solution:

The probability of an event is given as:

$\text { probability of an event }=\frac{\text { number of favorable outcomes }}{\text { total number of outcomes }}$

In throwing one die, the total number of outcomes = 6 { 1, 2, 3, 4, 5 , 6}

First let us calculate p(A):

The event is defined as: The sum on the two dice is even

Sum on two dice is even if and only if either both dice turn up odd or both even.

The odd outcomes in thowing a single die = 3 {1, 3, 5}

The even outcomes in throwing a single die = 3 {2, 4, 6}

The probability that both turn up odd is:

$\text { probability of both die turing up odd }=\frac{3}{6} \times \frac{3}{6}=\frac{1}{4}$

Similarly, the probability that both turn up even is:

$\text { probability of both die turing up even }=\frac{3}{6} \times \frac{3}{6}=\frac{1}{4}$

probability that the sum on two dice is even = probability that both turn up odd + probability that both turn up even

$\text { probability of sum on two dice is even }=\mathrm{p}(\mathrm{A})=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

Thus $p(A) = \frac{1}{2}$

Let us calculate p(B):

The event B is defined as: The sum on the two dice is at least 10

The total possible outcomes of two die is given as:

{(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6) }

Since each individual die can turn up any of the numbers 1, 2, 3, 4, 5, 6 the event "sum of the two dice will be at least 10" is :

atleast 10 means that sum can be 10 or greater than 10

{(4,6), (6,4), (5,5), (5,6), (6,5), (6,6)}

Here favourable outcomes = 6

Total number of outcomes = 36

Hence, the probability that the sum of the two dice will be at least 10 is:

$\text { probability that the sum of the two dice will be at least } 10=\frac{6}{36}=\frac{1}{6}$

Thus $p(B) = \frac{1}{6}$

Let us calculate p(C):

The event C is defined as: The red die comes up 5

Favourable outcomes = {(1,5),(2,5),(3,5),(4,5),(5,5),(6,5)}

$\text { probability of red die comes up } 5 \text { is the event }=\frac{6}{36}=\frac{1}{6}$

Thus $p(C) = \frac{1}{6}$

B) What is p(A l C)

$P(A | C)=\frac{p(A \cap C)}{P(C)}$

$\mathrm{A} \cap \mathrm{C}=\{(1,5),(3,5),(5,5)\}$

$p(A \cap C)=\frac{3}{36}=\frac{1}{12}$

$P(A | C)=\frac{p(A \cap C)}{P(C)}=\frac{\frac{1}{12}}{\frac{1}{6}}=\frac{1}{2}$

Thus $P(A | C)=\frac{1}{2}$

8 0

4 years ago