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disa [49]
3 years ago
14

what is three and sixty-tow thousandths rounded to the nearest hundredth and to the nearest whole number

Mathematics
2 answers:
vitfil [10]3 years ago
7 0
Round 2 from the 6 and then your answer will  be 360,000.
Alchen [17]3 years ago
5 0
Rounded to nearest whole number = 4
Rounded to nearest hundredth =3.62
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O 12 units<br> O 15 units<br> O 20 units<br> O 25 units
nadezda [96]

Answer:

12 units

Step-by-step explanation:

Because the two triangles are similar:

\dfrac{x}{9}=\dfrac{16}{x} \\\\\\\dfrac{x^2}{9}=16 \\\\\\x^2=144 \\\\\\x=12

Hope this helps!

3 0
3 years ago
Which of the following equations could be used to solve the given equation?
Pepsi [2]
If we simplify like terms on left and right sides we gwt

16x + 9 =  4x

Its B
8 0
3 years ago
Read 2 more answers
A school wants to put on a dance recital Middle School dancers have short programs that run 4 minutes long High School performer
sp2606 [1]

Answer:  The answer is \textup{M}+\textup{H}\leq 10~~\textup{and}4\textup{M}+7\textup{H}\leq 90.


Step-by-step explanation: Given that a school wants to put on a dance recital. The programs of Middle school dancer will be short of 4 minutes each and programmes of High school performers will be long of 7 minutes each.

Number of middle school dances is "M" and number od high school dances is "H".

Since there cannot be more than 10 dances all together, so fisrt inequality can be written as

\textup{M}+\textup{H}\leq 10.

Also, the entire recital must take maximum 90 minutes, so the second inequality can be written as

4\textup{M}+7\textup{H}\leq 90.

Thus, the pair of inequalities is

\textup{M}+\textup{H}\leq 10,\\\\4\textup{M}+7\textup{H}\leq 90.


6 0
3 years ago
1/2x+2 = x/2x^2+8<br><br> Show work plz
Igoryamba

Answer:

x=24

Step-by-step explanation:

1/2x+2 = 1/2x^2+8

1/2x^2=1/4x

1/2x+2=1/4x+8

1/2x -1/4x =0.25x

0.25x+2=8

8-2=6

0.25x=6

0.25/6 = 24

x=24

7 0
3 years ago
Read 2 more answers
In a pair of similar polygons, corresponding angles are congruent.
Brilliant_brown [7]
This is true based on the theorem corresponding parts of congruent figures are congruent.
3 0
2 years ago
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