All categories

3 years ago

The area of a square is 16 square units. Since 42 = 16 and (-4)2 = 16, both 4 and -4

2 answers:

3 0

It is true that (-4)^2 = 16; however, it doesn't make sense to have a square with a negative side lengths. Lengths and distances must be positive numbers.

So in many problems, we'll just ignore the negative solution to equations like x^2 = 16. It makes sense to only consider x = 4 for situations like this.

Vikentia [17]3 years ago

3 0

67572882737478282847474738374

You might be interested in

Romashka [77]

9.42 Units

Step-by-step explanation:

Since the angle shown is 90 degrees we can deduce that the rest of the arc is 270 degrees therefore we can make the fraction:

270/360; 360 being the total angle measurement of a circle

270/360 simplifies to 3/4

The fraction represents how much of the circle that arc covers

Now we have to find the circumference which would be

2(3.14)r

r = 2 therefore: 2(3.14)2 = 12.56

Now that we have the circumference of the WHOLE circle we multiply it by 3/4 to find out the arc length

Which gives us 9.42

8 0

3 years ago

What is the value of 4(25-5)?

sweet [91]

First you subtract 25 from 5 which equals 20 then you multipliy 4 to 20 and you get 80...so 80 is your solution.

7 0

3 years ago

Alexia ran three laps around her neighborhood. Each lap is 1 and StartFraction 3 over 8 EndFraction miles. Which is the best est

Rama09 [41]

Answer:

4 1/2

Step-by-step explanation:

each lap is 1 3/8 times that by 3

3/8 * 3 = 1.125

1* 3 = 3

so would be 4.125

7 0

3 years ago

Read 2 more answers

PLS HURRY, I neeeedd helppppp

Sedaia [141]

Answer:

third option

r > 11

hope it helps

4 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.

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

4 years ago