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wel
3 years ago
14

Find the value of given expression

tle=" \sqrt{35 + 1} " alt=" \sqrt{35 + 1} " align="absmiddle" class="latex-formula">
​
Mathematics
2 answers:
aliya0001 [1]3 years ago
6 0
Answer is 6 if you do 35+1 that gives you 36, the square root of 36 is 6 .6•6=36 so that’s your answer 6
Simora [160]3 years ago
4 0

Answer and Step-by-step explanation:

35 + 1 = 36

\sqrt{36} (The square root of 36) is equal to 6.

6 is the answer.

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

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3 0
3 years ago
A number has the digits 1, 9 and 5 to the nearest 100 the number rounds to 600. What is the number?
mars1129 [50]

Answer:

<u><em>591</em></u>

Step-by-step explanation:

the closest number to 600 out of every combo

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591

4 0
3 years ago
Read 2 more answers
Prove that for all n in the naturals, n(n+1)(n+2) is divisible by 3
DerKrebs [107]
By induction:

It's true for n=1, since 1\cdot2\cdot3 clearly contains a factor of 3.

Suppose it's true for n=k, that k(k+1)(k+2) is divisible by 3. Then

(k+1)(k+2)(k+3)=\dfrac{k(k+1)(k+2)(k+3)}k=\dfrac{3m(k+3)}k

where m is an integer. This reduces to

\dfrac{3m(k+3)}k=3m+9\dfrac mk

and both terms are clearly multiples of 3. We know that \dfrac mk is an integer since we had set m=k(k+1)(k+2) previously, which implies m is a multiple of k. So the statement is true.
6 0
3 years ago
Solve 47 (math operation)
Alchen [17]

Answer:

  t ≈ -2.014 or 3.647

Step-by-step explanation:

Add the opposite of the expression on the right side of the equal sign to put the equation into standard form.

  4.9t² -8t -36 = 0

You can divide by 4.9 to make this a little easier to solve.

  t² -(8/4.9)t -36/4.9 = 0

Now, add and subtract the square of half the x-coefficient to "complete the square."

  t² -(8/4.9)t +(4/4.9)² -36/4.9 -(4/4.9)² = 0

  (t -4/4.9)² -192.4/4.9² = 0 . . . . simplify

Add the constant term, then take the square root.

  (t -4/4.9)² = 192.4/4.9²

  t -4/4.9 = ±(√192.4)/4.9

  t = (4 ± √192.4)/4.9

  t ≈ {-2.014, 3.647}

8 0
3 years ago
If
likoan [24]

It's easy to show that 7\tan(4x) is strictly increasing on x\in\left[0,\frac\pi8\right]. This means

M = \max \left\{7\tan(4x) \mid \dfrac\pi{16} \le x \le \dfrac\pi{12}\right\} = 7\tan(4x) \bigg|_{x=\pi/12} = 7\sqrt3

and

m = \min \left\{7\tan(4x) \mid \dfrac\pi{16} \le x \le \dfrac\pi{12}\right\} = 7\tan(4x) \bigg|_{x=\pi/16} = 7

Then the integral is bounded by

\displaystyle 7\left(\frac\pi{12} - \frac\pi{16}\right) \le \int_{\pi/16}^{\pi/12} 7\tan(4x) \, dx \le 7\sqrt3 \left(\frac\pi{12} - \frac\pi{16}\right)

\implies \displaystyle \boxed{\frac{7\pi}{48}} \le \int_{\pi/16}^{\pi/12} 7\tan(4x) \, dx \le \boxed{\frac{7\sqrt3\,\pi}{48}}

7 0
2 years ago
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