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wariber [46]
3 years ago
14

Write each of the rations belowIn their simplest form9:15​

Mathematics
1 answer:
creativ13 [48]3 years ago
8 0

jhe6b3g35va6bwuwbb6wb6n2ybw6bwbeybeyebywbywbywbwbywbywbywbywb6wb6wn6wn6wbwb

6wb

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Brenda and her best friend went on a vacation together. They both paid for a flight, and Brenda had a fancy room for $150 per ni
Mila [183]

Answer:

A. 2x + 150y = 1,500

   2x + 120y = 1,320

The flights were $300 each, and they stayed 6 nights.

B. $1,380

Step-by-step explanation:

Let x be the cost of the flight (in one side) and y be the number of nights on their vacation.

A. Brenda had a fancy room for $150 per night, then she paid $150y for y nights.  Brenda's total is $1,500, then

150y+2x=1,500

Brenda's friend had a smaller room for $120 per night, then she paid $120y for y nights.  Brenda's friend total is $1,320, then

120y+2x=1,320

Subtract two equations:

150y+2x-120y-2x=1,500-1,320\\ \\30y=180\\ \\y=6

Substitute into the first equation:

150\cdot 6+2x=1,500\\ \\900+2x=1,500\\ \\2x=600\\ \\x=300

Hence

2x + 150y = 1,500

2x + 120y = 1,320

The flights were $300 each, and they stayed 6 nights.

B. If a third friend joined them and stayed in a mid-range room for $130 per night, then he paid \$130\cdot 6=\$780

The total cost is

\$780+\$600=\$1,380

7 0
4 years ago
Please find the slope, using formula
koban [17]

Answer:

-.2857

Step-by-step explanation:

3 0
3 years ago
Find each of the following products of conjugate pairs.See if you can work out a pattern
yKpoI14uk [10]

Answer: is 0

Step-by-step explanation:

4 0
4 years ago
How do you solve 5(3+k)=45
Helen [10]
We simplify the equation to the form, which is simple to understand
5(3+k)=45

Reorder the terms in parentheses
+(+15+5k)=45

Remove unnecessary parentheses
+15+5k=+45

We move all terms containing k to the left and all other terms to the right.
+5k=+45-15

We simplify left and right side of the equation.
+5k=+30

We divide both sides of the equation by 5 to get k.
k=6
7 0
3 years ago
Read 2 more answers
Which monomials are factors of 14xy^4
wlad13 [49]
The following are the factors of 14xy⁴:

2, 7, 14, x, y, y², y³, y⁴, xy, xy², xy³, xy⁴, 2x, 2y, 2y², 2y³, 2y⁴, 2xy, 2xy², 2xy³, 2xy⁴, 7x, 7y, 7y², 7y³, 7y⁴, 7xy, 7xy², 7xy³, 7xy⁴, 14x, 14y, 14y², 14y³, 14y⁴, 14xy, 14xy², 14xy³

I did not list the original monomial, as we know it is a factor of itself.
4 0
4 years ago
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