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3 years ago

What number is between 2/7 and 4/11 CHECK MY ANSWER PLEASE?!?!

Mathematics

1 answer:

-BARSIC- [3]3 years ago

3 0

Yes, that's correct. All you do is add the two numbers together and then divide by 2!

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A town is fencing a rectangular field that is 200 feet long and 150 feet wide. At $10 per foot, how much will it cost to fence t

labwork [276]

Perimeter of a rectangle is

2*width + 2*length

then multiply the perimeter by the cost per foot

2*(150ft) + 2*(200ft)

300ft + 400ft

700ft

700ft * $10/ft

$7000

6 0

3 years ago

Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$

Viktor [21]

When $n=0$ , we have

$5^{3^0} + 1 = 5^1 + 1 = 6$

$3^{0 + 1} = 3^1 = 3$

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for $n=k$ , that

$3^{k + 1} \mid 5^{3^k} + 1$

Now for $n=k+1$ , we have

$5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)$

so we know the left side is at least divisible by $3^{k+1}$ by our assumption.

It remains to show that

$3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1$

which is easily done with Fermat's little theorem. It says

$a^p \equiv a \pmod p$

where $p$ is prime and $a$ is any integer. Then for any positive integer $x$ ,

$5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3$

Furthermore,

$5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3$

which goes all the way down to

$5^{3^k} \equiv 5 \pmod 3$

So, we find that

$\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3$

QED

5 0

2 years ago

Prove that 1+cosA/sinA + sinA/1+cosA=2cosecA

Alla [95]

$\frac{1+cos\alpha}{sin\alpha}+\frac{sin\alpha}{1+cos\alpha}=2cosec\alpha\\\\L=\frac{(1+cos\alpha)(1+cos\alpha)+sin\alpha\cdot sin\alpha}{sin\alpha(1+cos\alpha)}=\frac{1+2cos\alpha+cos^2\alpha+sin^2\alpha}{sin\alpha(1+cos\alpha)}\\\\=\frac{1+2cos\alpha+1}{sin\alpha(1+cos\alpha)}=\frac{2+2cos\alpha}{sin\alpha(1+cos\alpha)}=\frac{2(1+cos\alpha)}{sin\alpha(1+cos\alpha)}\\\\=\frac{2}{sin\alpha}=2\cdot\frac{1}{sin\alpha}=2cosec\alpha=R$

5 0

3 years ago

A circle has a radius of 7/4 units and is centered at <br> (−2.5,−4.4)

lana66690 [7]

Are you asking for the expression?

If so, you plug into the format (x - h)² + (y - k)² = r² where (h, k) is the center and r is the radius.

(x + 2.5)² + (y + 4.4)² = (7/4)²

Simplify to get your equation and answer:

(x + 2.5)² + (y + 4.4)² = 49/16

6 0

3 years ago

Read 2 more answers

Tell whether the triangles are similar. Explain.

Nata [24]

Answer:

Similar

Step-by-step explanation:

They are similar because they are the same shape

3 0

3 years ago

Read 2 more answers