A²x2=100 What’s the answer to my question??

Mathematics

2 answers:

pishuonlain [190]3 years ago

7 0

Answer: 25

Step-by-step explanation:

Give this person brainliest ^^ the deserve it

Lelechka [254]3 years ago

3 0

The previous answer is wrong.
The answer is the square root of 50. Because if your work backwards then you do 100/2 =50. And then you have to do the opposite of squaring so you do the square root of 50 which is √50 which can be simplified to
5 √3

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plz help whats the answer to this question !!! calculate the mass of a gold bar that is 18.00cm long, 9.21cm wide, and 4.45cm hi

erik [133]

We have to find the mass of the gold bar.

We have gold bar in the shape of a rectangular prism.

The length, width, and the height of the gold bar is 18.00 centimeters, 9.21 centimeters, and 4.45 centimeters respectively.

First of all we will find the volume of the gold bar which is given by the volume of rectangular prism:

Volume of the gold bar $= length \times width\times height$

Plugging the values in the equation we get,

Volume of the gold bar $=18.00 \times 9.21 \times 4.45= 737.721cm^3$

Now that we have the volume we can find the mass by using the formula,

$Mass= density \times volume$

The density of the gold is 19.32 grams per cubic centimeter. Plugging in the values of density and volume we get:

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3 years ago

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ExtremeBDS [4]

Answer: Yes

Step-by-step explanation:

It is a relation because each person will most likely have a different name and you can have plenty of the same outputs just as long as the input doesn't repeat at all.

5 0

2 years ago

Using a linear approximation, estimate f(2.1), given that f(2) = 5 and f'(x) = √3x-1.

algol [13]

Answer:

$f\left( {2.1} \right) \approx 5.22360$ .

Step-by-step explanation:

The linear approximation is given by the equation

${f\left( x \right) \approx L\left( x \right) }={ f\left( a \right) + f^\prime\left( a \right)\left( {x - a} \right).}$

Linear approximation is a good way to approximate values of $f(x)$ as long as you stay close to the point $x= a$ , but the farther you get from $x=a$ , the worse your approximation.