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andreev551 [17]
4 years ago
8

60,000 is equal to how many ones

Mathematics
2 answers:
Diano4ka-milaya [45]4 years ago
8 0
60,000 ones
60,000=60,000 ones
maksim [4K]4 years ago
5 0
60,000 ones equals 60 thousands.
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4 years ago
What is the area of the figure?<br><br> A.234 in.^2<br> B.315 in.^2<br> C.225 in.^2<br> D.252 in.^2
saul85 [17]

Let's imagine we enclosed the entire shape in a 15 inch wide, 21 inch tall rectangle, area 315 sq inches.

From that rectangle we have to take away the 3x6 bottom left corner, and a strip 21-6=15 inches long and 3 inches wide on the right.

A = 15×21 - 3×6 - 3×15 = 315 - 63 = 252 sq in

Answer: D.   252 inches squared, last choice

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3 years ago
Pls pls help me I will mark you brainest
Musya8 [376]

Answer:

It would be 125m^2

Step-by-step explanation:

17*10=170

170/2=85

8*10=80

80/2=40

85+40=125

7 0
2 years ago
Read 2 more answers
Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE
EleoNora [17]

Answer:

0, 10

Step-by-step explanation:

The given function is:

g(y) = \frac{y-5}{y^2-3y+15}

According to the quotient rule:

d(\frac{f(y)}{h(y)})  = \frac{f(y)*h'(y)-h(y)*f'(y)}{h^2(y)}

Applying the quotient rule:

g(y) = \frac{y-5}{y^2-3y+15}\\g'(y)=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}

The values for which g'(y) are zero are the critical points:

g'(y)=0=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}\\(y-5)*(2y-3)-(y^2-3y+15)=0\\2y^2-3y-10y+15-y^2+3y-15\\y^2-10y=0\\y=\frac{10\pm \sqrt 100}{2}\\y_1=\frac{10-10}{2}= 0\\y_2=\frac{10+10}{2}=10

The critical values are y = 0 and y = 10.

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3 years ago
Please help me someone!
Elena-2011 [213]

Answer:

That looks hard but idk how to answer it

Step-by-step explanation:


8 0
3 years ago
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