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3 years ago

HOW DO I SOLVE THIS???!!

Mathematics

1 answer:

aliina [53]3 years ago

5 0

Answer:

K=32

Step-by-step ex

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A wildlife conservancy maintains a ratio of 2 squirrels for every 8 birds how many birds would there be if there were 15 squirre

GrogVix [38]

Answer: 16 birds

Step-by-step explanation:

Acording to the info we know that: for every 1 squirrel there are 4 birds

=> the number of the birds if there were 15 squirrels: 15*4=60 (birds)

7 0

3 years ago

What is the vertex of the graph of y = x^2+ 4x?

xxTIMURxx [149]

Answer:

(-2, -4)

Step-by-step explanation:

You can complete the square of the equation to get

y+(4/2)^2 = x^2+4x+(4/2)^2

y+4 = x^2 + 4x + 4

y+4 = (x+2)^2

y = (x+2)^2 - 4

This gives the form y = a(x-h)^2 + k where (h, k) is the vertex of the equation. You can also arrive at the same conclusion by making some observations of the equation. (x+2)^2 minimum value is going to be 0 since and negative values resulting from x+2 is going to become positive because of the square. So the minimum value is when x+2 is 0 or when x is equal to -2 and when it's at that minimum value of 0 it's going to have 4 subtracted from it which gives the vertex of (-2, -4)

6 0

2 years ago

Rx+9/5=h, Solve for the value of x

muminat

rx+(9/5)=h

rx=h-(9/5)

x=(h-(9/5))/r

5 0

3 years ago

What is 64×34+8-3? Solve.

borishaifa [10]

The answer is 2,181 because 64x34=2176+8=2184-3=2181

6 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