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3 years ago

When two numbers are not equal, and what is a sentence that uses the symbols

Mathematics

1 answer:

aivan3 [116]3 years ago

8 0

6 and 7 are not equal because 7>6. 7 is greater than 6

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Problem Solving<br> Find the exact solution of each equation for 0 < theta < 2 pi

horrorfan [7]

Answer:

C) $\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$

Step-by-step explanation:

$sin^2\theta=\frac{1}{2},\: 0\leq\theta\leq2\pi\\ \\\frac{1-cos(2\theta)}{2}=\frac{1}{2}\\ \\1-cos(2\theta)=1\\\\cos(2\theta)=0\\\\2\theta=\frac{\pi}{2}+\pi n\\ \\\theta=\frac{\pi}{4}+\frac{\pi}{2}n\\ \\ \theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$

Recall the identity $sin^2\theta=\frac{1-cos(2\theta)}{2}$

5 0

2 years ago

I need help thank you

IRISSAK [1]

Answer:

1) k = 3

2) 3, 4, 5

3) k + (k + 2) + (k + 4) = 4k

4) 6, 8, 10

Step-by-step explanation:

Jared used the equation;

k + (k + 1) + (k + 2) = 4k

Let's find k.

3k + 3 = 4k

4k - 3k = 3

k = 3

Thus,the integers will be;

3, (3 + 1) and (3 + 2)

This is; 3, 4, 5

He is looking for consecutive even integers but yet got 3 and 5 which are odd numbers. However, he should have used the format;

k, (k + 2), (k + 4) as the 3 consecutive even integers.

Thus, rewriting the main equation gives;

k + k + 2 + k + 4 = 4k

3k + 6 = 4k

4k - 3k = 6

k = 6

Thus,the integers are;

6, (6 + 2), (6 + 4)

This is; 6, 8, 10

6 0

3 years ago

jenna has a rope that is 8 7/8 feet long. she cuts the rope into peices that are 1 3/4 feet long. How many pieces will she have

Mandarinka [93]

Answer:20 pieces

Step-by-step explanation:

8 7/8=71/8

71/8 *4/7(reciprocal of 1 3/4)= 284/56= 20.2867

7 0

3 years ago

A male elephant weighs 6,549 pounds. A female elephant weighs 5,701 pounds. To the nearest

GrogVix [38]

Answer:

12200

Step-by-step explanation:

6549 plus 5701 equals 12250 but to nearest 100 12200

4 0

3 years ago

Read 2 more answers

Evaluate 1^3 + 2^3 +3^3 +.......+ n^3

Molodets [167]

Notice that

$(n+1)^4-n^4=4n^3+6n^2+4n+1$

so that

$\displaystyle\sum_{i=1}^n((n+1)^4-n^4)=\sum_{i=1}^n(4i^3+6i^2+4i+1)$

We have

$\displaystyle\sum_{i=1}^n((i+1)^4-i^4)=(2^4-1^4)+(3^4-2^4)+(4^4-3^4)+\cdots+((n+1)^4-n^4)$

$\implies\displaystyle\sum_{i=1}^n((i+1)^4-i^4)=(n+1)^4-1$

so that

$\displaystyle(n+1)^4-1=\sum_{i=1}^n(4i^3+6i^2+4i+1)$

You might already know that

$\displaystyle\sum_{i=1}^n1=n$

$\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2$

$\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6$

so from these formulas we get

$\displaystyle(n+1)^4-1=4\sum_{i=1}^ni^3+n(n+1)(2n+1)+2n(n+1)+n$

$\implies\displaystyle\sum_{i=1}^ni^3=\frac{(n+1)^4-1-n(n+1)(2n+1)-2n(n+1)-n}4$

$\implies\boxed{\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4}$

If you don't know the formulas mentioned above:

The first one should be obvious; if you add $n$ copies of 1 together, you end up with $n$ .
The second one is easily derived: If $S=1+2+3+\cdots+n$ , then $S=n+(n-1)+(n-2)+\cdots+1$ , so that $2S=n(n+1)$ or $S=\dfrac{n(n+1)}2$ .
The third can be derived using a similar strategy to the one used here. Consider the expression $(n+1)^3-n^3=3n^2+3n+1$ , and so on.

7 0

4 years ago