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3 years ago

Can someone help me ASAP !

Mathematics

2 answers:

babymother [125]3 years ago

5 0

Answer:

All of them you need answered?

Step-by-step explanation:

Novay_Z [31]3 years ago

4 0

Answer:

the first questions looks ike it reflects over all of them

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Help with this please

Alik [6]

Answer:

Yes, the given parallelogram is a rectangle.

Step-by-step explanation:

The vertices of parallelogram are J(-5,0), K(1,4), L(3,1) and M(-3,-3).

The slope formula is

$m=\frac{y_2-y_1}{x_2-x_1}$

$JK=\frac{4-0}{1-(-5)}=\frac{4}{6}=\frac{2}{3}$

$KL=\frac{1-4}{3-1}=\frac{-3}{2}$

$LM=\frac{-3-1}{-3-3}=\frac{-4}{-6}=\frac{2}{3}$

$JM=\frac{-3-0}{-3-(-5)}=\frac{-3}{2}$

The slopes of opposites sides are same it means they are parallel to each other.

The product of slopes of two consecutive sides is

$\frac{2}{3}\times \frac{-3}{2}=-1$

Since the product of slopes of two consecutive sides is -1, therefore the consecutive sides are perpendicular to each other.

Yes, the given parallelogram is a rectangle.

7 0

3 years 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 $n \geq 0$ ,

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

Observe that, because $n \geq 2$ and is an integer,

$n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5$

In concusion, the statement is true if and only if n is a non negative integer such that $n\leq 77$

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute $(4n)^4- \sum^{n}_{i=0} (2i)^4$ for 77 and 78 you will obtain:

$(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064$

$(4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992$

7 0

3 years ago

Help plz explain how to use discributive property

DiKsa [7]

1.Identify the fractions. Using the distributive property, you’ll eventually turn them into integers.

2.For all fractions, find the lowest common multiple (LCM) -- the smallest number that both denominators can fit neatly into. This will allow you to add fractions.

3.Multiply every term in the equation by the LCM.

4.Isolate variables adding or subtracting like terms on both sides of the equals sign.

5.Combine like terms.

6.Solve the equation and simplify, if needed.

5 0

3 years ago

Read 2 more answers

In 2004 a football team scored 27, 19, 12, 20, 20, 26, 23, 29, 20, 17, 61, 30, and 31 points in their first 13 games. Find the m

miv72 [106K]

After adding all the numbers together then dividing the sum by 13 I got 25.7 rounded to 26

8 0

3 years ago

a proportional relationship is represented on a graph with a line that had a slope of 65. what is the unit rate of the relations

erik [133]

Answer:

Step-by-step explanation:

8 0

3 years ago