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3 years ago

If _ angles are then lines are parallel

1 answer:

7 0

If two lines are cut by a transversal and the alternate interior angles are congruent,the lines are parallel.

I hope this helps.

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Use the bar diagram to write an equation. Then solve. Drag numbers to complete the equations.

N76 [4]

Answer/Step-by-step explanation:

The value of x can be solved as follows, take note of the numbers you will need to drag to complete the equations:

4x + 12x = 320 (segment addition postulate)

16x = 320 (combining like terms)

x = 20 (dividing both sides by 16)

We would end up with the value of x, which equals 20.

Numbers that we end up using are: 12, 16, 20, and 320.

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3 years ago

How would you lable a math bar model

Anastasy [175]

Well,. you could say bees in hive + bees in garden = bees in total.

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3 years ago

How is Pascal's triangle used in binomial theorem?.

Scorpion4ik [409]

To simplify the process of expanding a binomial of the type (a+b) n (a + b) n, use Pascal's triangle. The same numbered row in Pascal's triangle will match the power of n that the binomial is being raised to.

A triangular array of binomial coefficients known as Pascal's triangle can be found in algebra, combinatorics, and probability theory. Even though other mathematicians studied it centuries before him in India, Persia, China, Germany, and Italy, it is called after the French mathematician Blaise Pascal in a large portion of the Western world. Traditionally, the rows of Pascal's triangle are listed from row =0 at the top (the 0th row). Each row's entries are numbered starting at k=0 on the left and are often staggered in relation to the numbers in the next rows. The triangle could be created in the manner shown below: The top row of the table, row 0, contains one unique nonzero entry.

Learn more about triangle here

brainly.com/question/2773823

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7 0

1 year ago

Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe

ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

As we can see, $\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4$ and $(4n)^4=256n^4$ . Then,

$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

Now, we will use the formula for the sum of the first 4th powers:

$\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$

Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$