1. You convert the fractions so that they will have the same denominator because you can't multiply if the denominators aren't the same.
- 2/3 times 5/5 which equals 10/15
- 1/5 times 3/3 which equals 3/15
2. Now that the denominators are the same we can multiply the fractions.
- 10/15 times 3/15 equals 30/15
3. Since we have the answer we need to simplify and 30/15 simplified is 2.
So the answer is 2 or 30/15
Answer:
$55+$9x≥$199
You must work for at least 16 hours to be able to buy the bicycle.
Step-by-step explanation:
Let x represent the number of hours you need to work to buy the bicycle.
You already have $55.
⇒$55+ −−−−−≥ −−−−−
You also earn $9 per hour.
Algebraically, this can be written as 9x.
⇒$55+$9x≥ −−−−−
You need to earn at least $199 to buy the bicycle.
⇒$55+$9x≥$199
The ≥ sign is used because the left-hand side of the inequality must be "greater than or equal to" $199.
Let's find out how many hours you need to work to buy the bicycle.
Subtracting $55 from both sides of the inequality:
⇒$55−$55+9x≥$199−$55
⇒$9x≥$144
Dividing both sides by $9:
⇒$9x$9$=$144$9
∴x≥16
Therefore, you need to work at least 16 hours to afford the bicycle.
I see you do K12 huh? XD Okay so, the answer should be below or above or something. lol Hope this helped!
-Twix
<span>The expressions that cannot be factored using integral coefficients and integral constants is the below:
</span>b. 2x2 + x + 4
d. x2 - 2x - 4
For a. 25x2 - 40x + 16 the factors are (5x – 4) (5x – 4)
For c. 6x2 - 2x - 20 the factors are (2x - 4) (3x + 5)
<u>2x + 3y = 1</u>
<u>y = 3x + 15</u>
There's not much you can do with the first equation, because it has
two variables in it ... 'x' and 'y' . No matter how much you move them
around, you'll never be able to get either one equal to just a number.
Is there any way you could get rid of one of the variables in the first
equation, and have just 1 letter in it to solve for ?
Absolutely ! The second equation tells you something that 'y' is <u>equal</u> to,
(3x + 15). "EQUAL" is very powerful. It means that wherever you see 'y',
you can put (3x + 15) in its place, and you won't change anything or
upset anything. One thing you can do is take that (3x + 15) from the <span>
2nd</span> equation, and put it right into the first equation in place of 'y'.
You'll see how that helps as soon as you do it.
First equation: <u>2x + 3y = 1</u>
Substitute for 'y' : 2x + 3(<em>3x + 15</em>) = 1
Remove parentheses: 2x + 3(3x) + 3(15) = 1
2x + 9x + 45 = 1
Combine the terms with 'x' in them: 11x + 45 = 1
Look what you have now ! An equation with only one variable in it !
Subtract 45 from each side: 11x = -44
Divide each side by 11 : <em> x = -4</em>
You're more than halfway there. Now you know what 'x' is,
and you can use it with either equation to find what 'y' is.
-- If you use it with the first equation: <u> 2x + 3y = 1</u>
Put in the value of 'x': 2(<em>-4</em>) + 3y = 1
Remove the parentheses: -8 + 3y = 1
Add 8 to each side: 3y = 9
Divide each side by 3 : <em> y = 3</em>
-- If you use it with the 2nd equation: <u>y = 3x + 15</u>
Put in the value of 'x' : y = 3(<em>-4</em>) + 15
Remove the parentheses: y = -12 + 15
Add numbers on the right side: <em> y = 3</em> (same as the other way)
So there's your solution for the system of two equations:
<em> x = -4</em>
<em> y = 3</em>