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3 years ago

10

3. Consider the expressions sq root of 15, sq root of 97 and sq root of 40 Locate the approximate value of each on a number line

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Then, estimate each square root to the nearest tenth.

1 answer:

Crank3 years ago

5 0

Answer:

Square root of 15

√ 15 = 3.9 (Rounding to the nearest tenth)

Square root of 97

√97 = 9.9 (Rounding to the nearest tenth)

Square root of 40

√40 = 6.3 (Rounding to the nearest tenth)

Step-by-step explanation:

Square root of 15

√ 15 = 3.9 (Rounding to the nearest tenth)

Square root of 97

√97 = 9.9 (Rounding to the nearest tenth)

Square root of 40

√40 = 6.3 (Rounding to the nearest tenth)

Location on a number line

√ 15, between + 3 and + 4, very close to +4

√40, between + 6 and + 7, close to +6

√97, between + 9 and + 10, very close to +10

You might be interested in

Among all pairs of numbers whose difference is 20, find a pair whose product is as small as possible. what is the minimum produc

Elza [17]

Answer:

-100

-10, 10

Step-by-step explanation:

Let the smaller number be x.

Then the larger number is x + 20.

The product is x(x + 20).

Now you can write the function

y = x(x + 20)

y = x^2 + 20x

Take the first derivative of y with respect to x.

y' = 2x + 20

Set the first derivative equal to zero to find the x value for the minimum value of the function.

2x + 20 = 0

2x = -20

x = -10

The minimum value of the function occurs at x = -10. -10 is one of the two numbers.

y = x^2 + 20x

For x = -10,

y = (-10)^2 + 20(-10)

y = 100 - 200

y = -100

The minimum value of the product is -100.

x = -10

x + 20 = -10 + 20 = 10

The numbers are -10 and 10.

If you have not learned derivatives yet, then plot the function

y = x^2 + 20x

Now look at the graph and find the minimum y value and the x value at which it occurs.

The minimum y value is -100. That is the minimum product you are looking for.

The x value of the minimum function value is x = -10.

Then x + 20 = -10 + 20 = 10.

The numbers are -10 and 10.

4 0

3 years ago

frozen [14]

Answer:

the answer is 3/100

Step-by-step explanation:

the probability=

pr(blue)*pr(black)

=3/10*1/10

=3/100

hope this helps

please give brainest

4 0

2 years ago

Need help in this please

Alona [7]

Hmmm this is a little hard I believe c

8 0

3 years ago

If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,

julsineya [31]

<h2>Hello!</h2>

The answer is:

The first option, the amount dumped after 5 days is 0.166 tons.

<h2>Why?</h2>

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

$\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)$

$(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166$

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

5 0

3 years ago

In which quadrant is the point (–3, 5) located?

LuckyWell [14K]

Answer:

Step-by-step explanation:

7 0

3 years ago