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Nonamiya [84]
3 years ago
5

Define a variable and write an inequality for the statement:

Mathematics
2 answers:
Rashid [163]3 years ago
8 0
Algebra course is the defined variable and the inequality is y=70x+b
BartSMP [9]3 years ago
6 0
The dependent variable is the test grade. the constant or control variable is the algebra A1 course. the independent variable is if the test score changes.
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Consider the quadratic function:
storchak [24]
X = -b/2a = 8/2 = 4 

y = 16 -8*4 -9 = 16 - 32 -9 = -25

the wertex is (4,-25)
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4 years ago
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A car traveled 210 miles in 3 hours. Find the unit rate in miles per hour
Airida [17]
70 miles per hour, just divide 210 between 3
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3 years ago
Ms. Green tells you that a right triangle has a hypotenuse of 12 and a leg of 5. She asks you to find the other leg of the trian
Anit [1.1K]

let the other leg be x cm.

(5)^2+(x)^2=(12)

25+(x)^2=144

(x)^2=144-25

x=√119

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3 years ago
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What is 24.051 rounded to the nearest hundredth? How did you solve it?
4vir4ik [10]
This is like late elementary/early middle school stuff where you learn this.
First off, what number is in the position of the hundereth place? 0? No. 1? No. 5 yes! Now how i learned how to round is you take  the number to the right of whatever you are being asked about and round it up or down. Now if the number (in this case it is 1) is lower than 5, you round down (or keep it the same) but if it is 5 or higher you round up. since 1 is lower then 5, that 5 in the hundereths place stays the same. 
7 0
3 years ago
Joe bikes at the speed of 30 km/h from his home toward his work. If Joe's wife leaves home 5 mins later by car, how fast should
GuDViN [60]

Answer:

Joe's wife must drive at a rate of 45km/hour.

Step-by-step explanation:

We are given that Joe leaves home and bikes at a speed of 30km/hour. Joe's wife leaves home five minutes later by car, and we want to determine her speed in order for her to catch up to Joe in 10 minutes.

Since Joe bikes at a speed of 30km/hour, he bikes at the equivalent rate of 0.5km/min.

Then after five minutes, when his wife leaves, Joe is 5(0.5) or 2.5 km from the house. He will still be traveling at a rate of 0.5km/min, so his distance from the house can be given by:

2.5+0.5t

Where <em>t</em> represents the time in minutes after his wife left the house.

And since we want to catch up in 10 minutes, Joe's distance from the house 10 minutes after his wife left will be:

2.5+0.5(10)=7.5\text{ km}

Let <em>s</em> represent the wife's speed in km/min. So, her speed times 10 minutes must total 7.5 km:

10s=7.5

Solve for <em>s: </em>

<em />\displaystye s=0.75\text{ km/min}<em />

Thus, Joe's wife must drive at a rate of 0.75km/min, or 45km/hour.

3 0
3 years ago
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