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3 years ago

Evaluate√0.792using tables of Square roots

Mathematics

2 answers:

garik1379 [7]3 years ago

8 0

Answer:

0.28142

Step-by-step explanation:

Hope this helps you

photoshop1234 [79]3 years ago

5 0

Step-by-step explanation:

0.28142

$\sqrt{0.792 \div 2 = 0.28142$

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Error Analysis! Answer both parts of the question.

Maurinko [17]

Answer:

Kaitlin's answer is wrong because she is supposed to by dividing. When trying to get rid of a number from a variable you have to divide the number by itself and whatever you do to one side you have to do to the other side

so you would do 5/5 - cancels out

15/5=3

y=3

Step-by-step explanation:

have a great day :D

8 0

2 years ago

PLEASE HELP ASAP 25 PTS + BRAINLIEST TO RIGHT/BEST ANSWER

Alex777 [14]

Answer:

Its D

Step-by-step explanation:

x = 1

x = -1

x= 0.0000 - 3.0000 i

x= 0.0000 + 3.0000 i

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "x2" was replaced by "x^2". 4 more similar replacement(s).

Step by step solution :

Step 1 :

Equation at the end of step 1 :

((((((x6)-(2•(x5)))+(9•(x4)))-24x3)-x2)+18x)-9 = 0

Step 2 :

Equation at the end of step 2 :

(x6)-(2•(x5)+(32x4)-24x3)-x2)+18x)-9 = 0

Step 3 :

Equation at the end of step 3 :

(x6)-2x5)+32x4)-24x3)-x2)+18x)-9 = 0

4 0

2 years ago

How do you find the volume of the solid generated by revolving the region bounded by the graphs

d1i1m1o1n [39]

Answer:

About the x axis

$V = 4\pi[ \frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}$

About the y axis

$V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}$

About the line y=8

$V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}$

About the line x=2

$V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi$

Step-by-step explanation:

For this case we have the following functions:

$y = 2x^2 , y=0, X=2$

About the x axis

Our zone of interest is on the figure attached, we see that the limit son x are from 0 to 2 and on y from 0 to 8.

We can find the area like this:

$A = \pi r^2 = \pi (2x^2)^2 = 4 \pi x^4$

And we can find the volume with this formula:

$V = \int_{a}^b A(x) dx$

$V= 4\pi \int_{0}^2 x^4 dx$

$V = 4\pi [\frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}$

About the y axis

For this case we need to find the function in terms of x like this:

$x^2 = \frac{y}{2}$

$x = \pm \sqrt{\frac{y}{2}}$ but on this case we are just interested on the + part $x=\sqrt{\frac{y}{2}}$ as we can see on the second figure attached.