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kondor19780726 [428]
3 years ago
15

3,350 to the nearest ten.

Mathematics
2 answers:
oksian1 [2.3K]3 years ago
7 0

Answer:

Step-by-step explanation:

3350 to the nearest ten is 3350

lisabon 2012 [21]3 years ago
4 0
Your answer will be 3350
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Solve for X and solve for Y?
Llana [10]

Answer:

x=3\sqrt 5\\

y = \sqrt {126}

Step-by-step explanation:

By geometric mean property:

x =  \sqrt{9 \times 5}  =  \sqrt{45} = 3 \sqrt{5}   \\  \\ by \: pythagoras \: theorem \\  \\  {y}^{2}  =  {9}^{2}  +  {x}^{2}  \\  \\  {y}^{2}  =  {9}^{2}  +  {(3 \sqrt{5}) }^{2}  \\  \\  {y}^{2}  = 81 + 45 \\  \\  {y}^{2}  = 126 \\  \\ y =  \sqrt{126}

7 0
2 years ago
What is the answer for this question:
Sergio [31]
There are thirty seven seats in total.
5 0
3 years ago
A linear model for the situation below passes through the origin. Find the missing value. Round to the nearest ten if necessary.
BARSIC [14]
17.55 to buy 13 apples
4 0
2 years ago
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Can someone help me out with this question?
goblinko [34]

Answer:

The third option

Step-by-step explanation:

The third option represents 65 over 100, meaning 65% of whatever of it would the 65% of 90 be

6 0
3 years ago
A tank contains 300 liters of fluid in which 40 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu
bonufazy [111]

Answer:

A(t) = 300 -260e^(-t/50)

Step-by-step explanation:

The rate of change of A(t) is ...

A'(t) = 6 -6/300·A(t)

Rewriting, we have ...

A'(t) +(1/50)A(t) = 6

This has solution ...

A(t) = p + qe^-(t/50)

We need to find the values of p and q. Using the differential equation, we ahve ...

A'(t) = -q/50e^-(t/50) = 6 - (p +qe^-(t/50))/50

0 = 6 -p/50

p = 300

From the initial condition, ...

A(0) = 300 +q = 40

q = -260

So, the complete solution is ...

A(t) = 300 -260e^(-t/50)

___

The salt in the tank increases in exponentially decaying fashion from 40 grams to 300 grams with a time constant of 50 minutes.

6 0
3 years ago
Read 2 more answers
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