Help me pls!!!!!!!!!!!!!!!!

What is 1 divided by 5^-2? Please give clear explanation. thanks.

Pani-rosa [81]

1 / 5^-2

First, let's work out the denominator.
5^ -2 = 1/25. When an exponent is negative, it means to put a 1 over the product. Like we did earlier.

Now we have 1 / (1/25).

1 / (1/25)
Divide 1 by 25.
1 / 0.04
Divide 1 by 0.04
25

The answer is 25!

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7 0

2 years ago

If k is a positive integer and n = k(k + 7), is n divisible by 6 ? (1) k is odd. (2) When k is divided by 3, the remainder is 2.

olganol [36]

Answer:

1.) Yes

2.) Yes

Step-by-step explanation:

Given that

n = k(k + 7)

If k is a positive integer and n = k(k + 7), is n divisible by 6 ?

(1) k is odd. Yes.

Let assume that k = 3

Then, n = 3(3 + 7)

n = 3 × 10

n = 30.

30 is divisible by 6.

(2) When k is divided by 3, the remainder is 2. That is,

Let k = 5

Then,

5/3 = 1 remainder 2

Substitute k into the equation

n = k(k + 7)

n = 5(5 + 7)

n = 5 × 12

n = 60

And 60 is divisible by 6.

Therefore, the answer to both questions is Yes.

5 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

2 years ago

Only answer if known and please show work

rusak2 [61]

Easy, the second one. Both figures seem congruent.

4 0

3 years ago

Please help equation below

qaws [65]

Answer:

3y √21 + 2 y√15

Step-by-step explanation:

To simply, we open up the bracket

We have this as;

√(189y^2) + √60y

= 3y √21 + 2 y√15

7 0

2 years ago