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3 years ago

What is the additive inverse of 38

Mathematics

2 answers:

Mumz [18]3 years ago

4 0

-1/38 is your inverse of 38

hope this helps

Roman55 [17]3 years ago

4 0

-1/38

I hope this helps.

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One kilogram is 2.2 pounds. If i had 1 pound would that be 0.45 kilograms why or why not?

prisoha [69]

Step-by-step explanation:

1 kg = 2.2 pounds

0.45 kg = 1 pounds to see whether it's correct or not we can cross multiply the given equations

multiply 1 kg with 1 pounds and 0.45 kg with 2.2 pounds then check if they are equal

1 × 1 = 2.2 × 0.45

1 = 0.99 as you can see this is not an equality therefore the statement is wrong.

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4 years ago

Graph the circle (x-8)^2 + (y+5)^2 = 4

Kobotan [32]

Answer:

The answer to your question is See the attachment

Step-by-step explanation:

Equation

(x - 8)² + (y + 5)² = 4

Process

1.- Write the standard equation of the circle

(x - h)² + (y - k)² = r²

2.- Find the center

The center are the letters h and k

h = 8 k = -5

3.- Find the radius

r = 2

4.- Graph the circle ( i will use geogebra)

See below

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3 years ago

Evaluate tan 30° without using a calculator by using ratios in a reference triangle.

Sloan [31]

Answer: Root 3 / 3

Step-by-step explanation: Use unit circle trigonometry.

6 0

3 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,

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

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3 years ago

A spool contains 270 inches of wire. How many pieces of wire, each with the a length of 1 7/8 inches, can be cut from the spool?

prisoha [69]

You would be able to cut 144 pieces of wire from the spool because 270/1 7/8 (or 1.875) equals 144

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3 years ago