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Angelina_Jolie [31]
3 years ago
9

Which problem can be solved using this inequality? x + 5 ≤ 20

Mathematics
2 answers:
kondor19780726 [428]3 years ago
5 0

Bob has marbles. He has a chance to get more.

Make an inequality that shows how many marbles Bob can get if he already has 5 and cannot exceed 21? (meaning he can have 20)

I tried. It should be right though! ^-^

⭐ Please consider brainliest! ⭐

✉️ If any further questions, inbox me! ✉️

Fed [463]3 years ago
3 0

Answer:

Beth is purchasing school supplies. She knows that the price for her pencil case is $5. She only has $20 to spend. What is the most amount of money she can spend on the remaining supplies?

Step-by-step explanation:

b

20 dollars is the whole value of the right side of the inequality that is greater or equal to x+5

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Given the functions, f(x) = x2 + 2 and g(x) = 4x - 1, perform the indicated operation. When applicable, state the domain restric
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Answer:

3rd option

Step-by-step explanation:

substitute x = f(x) into g(x)

g(f(x))

= g(x² + 2)

= 4(x² + 2) - 1

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Y = (x + 4)2 - 1<br> Standard form
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For part A, The answer is that the car gets better gas mileage. We can see it from the graph that the number of gallons used is on the X axis, and the distance traveled using those number of gallons is on the Y axis. The easiest way to compare would be to look at the 1 gallon of gas. You can see that you can travel 25 miles on 1 gallon of gas. The truck on the other hand will get you 18 miles per gallon. Imagine putting 1 in for X, the Y value would be 18 if you did this. The graph just shows us a visual way of saying the same thing. To determine how much farther the car with a girl on 8 gallons of gas, you would just multiply 8 by 25 for the number of miles traveled by the car. You would multiply 8 by 18 to find the number of miles traveled for the truck. The answers are 200 miles for the car and 144 miles for the truck. 200-144=56 miles farther for the car.
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What’s the domain and range of:<br> log(√(2x-1) + 3 )<br> Please explain how you got it too!!
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Two main facts are needed here:

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2. The square root \sqrt x exists for x\ge0.

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By (2) we know that \sqrt{2x-1} exists if 2x-1\ge0, or x\ge\dfrac12.

By (1), we know that \log(\sqrt{2x-1}+3) exists if \sqrt{2x-1}+3>0, or \sqrt{2x-1}>-3. But as long as the square root exists, it will always be positive, so this condition will always be met.

Ultimately, then, we only require x\ge\dfrac12, so the function has domain \left[\dfrac12,\infty).

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