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3 years ago

PLS HELPP pls answer all! pls show work

Mathematics

1 answer:

ICE Princess25 [194]3 years ago

7 0

Answer: middle one = 4/4 + 1 + 2 =3

Step-by-step explanation:

4/4 also equals one so x=4 1+2=3

4/4=1

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If 50 percent of x=66 then x =

sasho [114]

Answer:

132

Step-by-step explanation:

50 % means half

so 66 is half of whatever x is.

to find x you have to multiply 66 by 2 or add 66 to 66 which will give you 132 either way.

<em>Hope this helps you! have an amazing day :)</em>

7 0

3 years ago

Read 2 more answers

The can is 12cm high and 8cm in diameter. Using formula V=TT r2h, how much is in the can

scoray [572]

The formula you have given here is for the volume of a cylinder. It should read $V = \pi r^{2} h$ .
First, to find r, the radius, divide the diameter by two (since the radius is the distance from the circumference to the centre and the diameter is the distance from any point on the circumference to the opposite point (through the centre))

r = 8/2 = 4

Now we can calculate this.
$V = \pi *4^{2}*12 = \pi * 16 * 12 = 603.185... \\ 
V = 603 cm^{3}$

If your answer needs to be in litres, one cubic centimetre = one millilitre.

603 cm3 = 603mL = 0.603 L

4 0

3 years ago

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

3 years ago

Union City Bakery uses a chocolate chip muffin recipe that yields 1/2 cup of milk for every batch of 12 chocolate chip muffins.

nikitadnepr [17]

Answer:

Grease bottoms only of 12 muffin cups or line with baking cups. ... I made some changes to the recipe;

Step-by-step explanation:

I added 1 cup milk, 1 cup chips, 1 cup ... was way too much batter for 12 muffins; I ended up making two trays and they were done in 18 minutes. I give it four stars only because of all the changes I had to make

6 0

3 years ago

Question 5 (Multiple Choice Worth 4 points)

loris [4]

(-2,2) (0,1)
slope = y1-y2/x1-x2 .... 2-1/-2-0 = -1/2 the answer is C

7 0

3 years ago

Read 2 more answers

PLS HELPP pls answer all! pls show work ​

PLS HELPP pls answer all! pls show work