Ask question

Ask question

All categories

2 years ago

9

PLEASE HELP!!!Can sometime help me solve this?

2 answers:

castortr0y [4]2 years ago

4 0

The final answer is (65, 60).

x + y = 125
- y - y
-------------------
x = 125 - y

12(125 - y) + 6y = 1140

1,500 - 12y + 6y = 1,140
1,500 - 6y = 1,140
- 1,500 - 1,500
------------------------------
-6y = -360
------- --------
-6 -6

y = 60

x + 60 = 125
- 60 - 60
-------------------
x = 65

damaskus [11]2 years ago

3 0

Hope you understand. tell me if u need help

You might be interested in

Matthew's first 7 tests in geometry class this year were 82, 78, 81, 89, 79, 92, and 87, respectively. What is the

jok3333 [9.3K]

Answer:c

Step-by-step explanation:82+78+81+89+79+92+87=588

And 588/7 =84

4 0

2 years ago

Read 2 more answers

Irma needs to rent a car for a business trip from Phoenix to San Jose (720 miles one way). She will drive round trip and will be

cricket20 [7]

The cost per day is $65 per day, the total cost for 8days will be:
8×65
=520
suppose that E is the number of miles she can drive without exceeding $600.
Thus
0.24×E+520=600
hence solving for E we get
0.24E=600-520
0.24E=80
hence
E=80/0.24
E=333.3333
E~333

8 0

3 years ago

jonny [76]

Answer:

A. f(n) = (n + 1)^2 – 1

Step-by-step explanation:

Figure 1

3 = 4 - 1 = 2² - 1 = (1 + 1)² - 1

Figure 2

8 = 9 - 1 = 3² - 1 = (2 + 1)² - 1

Figure 3

15 = 16 - 1 = 4² - 1 = (3 + 1)² - 1

<u>The function is:</u>

f(n) = (n + 1)² - 1

7 0

2 years ago

((X + 2) + (y - 5) =1

Viktor [21]

Answer:

on like term

Step-by-step explanation:

x+2)=(y-5)=2+5=7

x-y{x-y]=7x-y

by Emmanuel okorie

3 0

2 years ago

For the problem 1/5g- 1/10- g + 1 3/10g -1/10, Tyson created an equivalent expression using the following steps. 1/5g+-1g+1 3/10

professor190 [17]

The true statements are:

Tyson's expression is not equivalent to the original expression
The equivalent expression is: $\frac{1}{2}g- \frac 1{5}$

<h3>What are equivalent expressions?</h3>

Equivalent expressions are expressions that have equal values

The original expression is given as:

$\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10}$

Collect like terms

$\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10} = \frac 15g - g + 1 \frac{3}{10}g- \frac 1{10} -\frac{1}{10}$

Evaluate the like terms

$\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10} = \frac{1}{2}g- \frac 1{5}$

Tyler's equivalent expression is given as:

$\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10}$

Collect like terms

$\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10} = \frac15g-g+ 1 \frac3{10}g-\frac 45g-\frac 1{10}+1 \frac{1}{10}$

Evaluate the like terms

$\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10} = -\frac 3{10}g$

The simplified expressions of the original expression, and Tyson's equivalent expressions are not equal.

Hence, Tyson's expression is not equivalent to the original expression

Read more about equivalent expressions at:

brainly.com/question/9603710

3 0

1 year ago