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joja [24]
3 years ago
14

Where would square root 113 be located on a number line

Mathematics
1 answer:
Amanda [17]3 years ago
5 0

Answer: It's between 10 and 11

It's closer to 11 than it is to 10.

This is because \sqrt{113} \approx 10.63 after using a calculator.

You could also look at the list of perfect squares

{1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144,...}

The terms in bold are where 113 fits in between. So we can say,

100 < 113 < 121\\\\\sqrt{100} < \sqrt{113} < \sqrt{121}\\\\10 < \sqrt{113} < 11\\\\

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HELP RIGHT NOW AND ILL GIVE BRAINLY IF CORRECT
Stella [2.4K]

Answer:

-9

Step-by-step explanation:

3^{-36}= 27^{-12}

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3 years ago
If the demand function for a commodity is given by the equation
Margarita [4]

Answer:

Equilibrium quantity = 26.92

Equilibrium price is $31.13

Step-by-step explanation:

Given :Demand function : p^2 + 16q = 1400

           Supply function : 700 -p^2 + 10q = 0

To Find : find the equilibrium quantity and equilibrium price.

Solution:

Demand function : p^2 + 16q = 1400  --A

Supply function : p^2-10q=700 ---B

Now to find the equilibrium quantity and equilibrium price.

Solve A and B

Subtract B from A

p^2-10q -p^2-16q=700-1400

-26q=-700

26q=700

q=\frac{700}{26}

q=26.92

So, equilibrium quantity = 26.92

Substitute the value of q in A

p^2 + 16(26.92) = 1400

p^2 + 430.72 = 1400

p^2 = 1400- 430.72

p^2 = 969.28

p = \sqrt{969.28}

p = 31.13

So, equilibrium price is $31.13

7 0
3 years ago
*square root symbol* y^3 multiplied by *square root symbol* y^3
san4es73 [151]
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3 years ago
Read 2 more answers
2/3 x -9/8 x -4/5 x -1
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3 years ago
An urn contains three red balls and four blue balls. Draw two balls at random from the urn, without replacement. Compute the exp
mezya [45]

Step-by-step explanation:

in total we have 3+4 = 7 balls.

when we draw the first ball, the probability to draw a red ball is 3/7, and a blue ball 4/7.

when we draw the second ball, we have now only 6 balls in total.

the probabilty to draw a red back now depends also on the result of the first draw.

if the first ball was already red, then we have only a chance now of 2 out of 6.

if the first ball was blue, then we have now a chance of 3 out of 6.

so, the probably to draw at least 1 red ball in 2 draws is the probability of drawing one on the first draw plus the probability of drawing one on the second :

1 - probability to see 2 blue balls

1 - 4/7 × 3/6 = 1 - 12/42 = 30/42 = 0.714285714...

the expected number of red balls in 2 draws is

1 red in first red in first red in second

but not second and second but not first

1×(3/7 × 4/6) + 2×(3/7 × 2/6) + 1×(4/7 × 3/6) = 12/42 + 12/42 + 12/42 = 36/42 = 6/7 = 0.857142857 ≈ 0.8571

7 0
2 years ago
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