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gtnhenbr [62]
2 years ago
14

Somebody help this assignment is due tonight for a grade

Mathematics
1 answer:
Law Incorporation [45]2 years ago
6 0
1.) 24 pack for $6.88
2.) 12 oz box for $2.15
3.) 16 pound turkey for $20.00
4.) $1.60 per notebook
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julia spends 2.25 on gas for her lawn mower.she earns 15.00 mowing her neighbors yard.what is julias profit?
Eduardwww [97]
This one is easy all you have to do is subtract the 2.25 by the 15 to get 12.75.
8 0
3 years ago
Read 2 more answers
A figure with 6 edges.<br> How many edges does this figure have?<br><br> It has <br> edges.
Anestetic [448]

Hexagon has six edges. The figure you're trying to figure out is hexagon I guess.

3 0
2 years ago
Abbys school has a track it takes eight laps around the track to run 1 mile.
insens350 [35]
The correct answer to the following questions is this:

<span>Abbys school has a track it takes eight laps around the track to run 1 mile. 
This means that you have to run 8 laps in order to complete a 1 mile. 
1 lap = 1/8 mile

A. What fraction of a mile is two laps around the track? Show your work or explain how you know
Since 1 lap = 1/8 mile, then 2 laps would be 2/8 mile or 1/4 mile.

Abby ran 3 1/8 miles on Monday and 1 3/8 miles on Tuesday.
B. What was the total number of miles every right on Monday and Tuesday? show your work or explain how you know
On Monday, she ran 3 1/8 miles, 1 3/8 miles on Tuesday. The total number of miles is 3 1/8 + 1 3/8 = (3+1) (1/8 + 3/8) = 4 4/8 = 4 1/2 miles.

C. How many total lasted gym run on Monday and Tuesday? Show your work or explain how you know
This seems not clear regarding the gym.
</span>
7 0
3 years ago
Solve the following proportion for X
mixer [17]
Answer: 3.7 (to nearest tenth)

Working:
7/17 = x/9
63 = 17x
x = 63/17
x = 3.7 (to nearest tenth)
8 0
3 years ago
I need help with my algebra 2. I also want to know how to solve this.
Lisa [10]

Add 1 to both sides:

\sqrt{x+3} = x+1

In cases like this, we have to remember that a root is always positive, so we can square both sides only assuming that

x+1 \geq 0 \iff x \geq -1

Under this assumption, we square both sides and we have

x+3 = (x+1)^2 \iff x+3 = x^2+2x+1 \iff x^2+x-2 = 0

The solutions to this equation are

x = -2,\ x=1

But since we can only accept solutions greater than -1, we discard x=-2 and accept x=1.

In fact, we have

x=-2 \implies \sqrt{-2+3}-1=0\neq -2

and

x=1 \implies \sqrt{4}-1=1

which is the only solution.

4 0
3 years ago
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