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2 years ago

Find the value of b. Enter as an exact, simplified value:

Mathematics

1 answer:

zubka84 [21]2 years ago

5 0

Already answered here: brainly.com/question/26835844

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5/8 of the cats perferred drinking with there tongue but 3/16 wanted a straw if there was 39 cats that gave one of these 2 answe

AveGali [126]

Answer:

208cats

Step-by-step explanation:

⅝=tongue

³/16=straw=39cats

⅝+³/16=13/16

meaning total number of cats is 16/16

If ³/16=39cats

16/16=?

39×16/3

=208cats

6 0

3 years ago

Can you please explain to me how I would solve this problem? Can you please explain/and show all the steps?

Lena [83]

22 = z + (-9) Switch the sides
z = 22 - (-9) Solve
z = 31 So, z = 31

7 0

2 years ago

The value of a piece of land is $500 an acre. The price per acre is expected to increase 7% per year. Which function models the

kobusy [5.1K]

To set the equation,we can first find how much price would be raised within a year:

=500×(1+7%)

=500×1.07

As the years go by the increase would accumulate,and the equation would be:

$500 \times {1.07}^{t}$
Hope it helps!

4 0

2 years ago

I have a challenge. All of you that read this, comment or make an answer. After you are done, do the same thing! Make a free bra

padilas [110]

Answer:

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

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

2 years ago