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Arisa [49]
3 years ago
11

198round to the nearest tens and hundreds

Mathematics
1 answer:
Xelga [282]3 years ago
8 0
198 round to the nearest hundreds is 200
198 round to the nearest tens is 200 too
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3x+3y+2(2x-y) equivalent fractions to this problem
Anastaziya [24]
Distribute the 2 into the 2x and -y

Then add “like terms” which are 3x and 4x... 3y and -2y

Which then becomes 7x+y

There are no more “like terms” so that is your final answer

6 0
3 years ago
Help me out please and thanks
omeli [17]

Answer:

Step-by-step explanation:

If you are trying to <u>solve the system of equations</u> then its

(x,y)=(-1,2)

If you are trying to <u>rewrite the system of equations</u> then its

x+6y=11

x-2y=-5

3 0
2 years ago
Simplify the expression using properties of operations. (−r−5)−(−2r−4)
Pachacha [2.7K]

Answer:r-1

Step-by-step explanation:

(-r-5)-(-2r-4) Or (-r-5)-(-2r-4)

-r+-5+2r+4. -r-5+2r+4

r-1 r-1

You bring down the first Parentheses Then have to multiply the (implied 1) -1•-2=2 and -1•-4=4 and do the math regularly

8 0
3 years ago
Can someone help me with this. Will Mark brainliest.
Anon25 [30]

Answer:

the distance between the two points is √130.

Step-by-step explanation:

a^2 + b^2 = c^2

a^2 = 9^2 = 81

b^2 = 7^2 = 49

c^2 = 81 + 49 = 130

c = √130

4 0
3 years ago
Please help!!
sveta [45]
For this case we have the following equation:
 d =  \sqrt{\frac{3h}{2}}
 Where,
 d: the distance they can see in thousands
 h: their eye-level height in feet
 For Kaylib:
 d = \sqrt{\frac{3(48)}{2}}
 d=\sqrt{3(24)}
 d=\sqrt{72}
 d=6\sqrt{2}
 For Addison:
 d = \sqrt{\frac{3(85\frac{1}{3})}{2}}
 d = \sqrt{\frac{256}{2}}
 d=\sqrt{128}
 d=8\sqrt{2}
 Subtracting both distances we have:
 8\sqrt{2} - 6\sqrt{2} = 2\sqrt{2}
 Answer:
 
2\sqrt{2}
 B. 2√2 mi
6 0
3 years ago
Read 2 more answers
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