All categories

3 years ago

Can you please help me fill this out thanks

Mathematics

1 answer:

swat323 years ago

7 0

Answer: $\sqrt{2}$

Step-by-step explanation:

In a 45 45 90 triangle the two sides are 1 wile the hypothenuse is $\sqrt{2}$

You might be interested in

50 POINTS! WILL MARK BRAINLEST! NEED IN 10 - 15 MINUTES!

masha68 [24]

Answer:

(2,14)

14/2=7

14+7=21

21 is final answer

Erika had a 95% chance of guessing wrong

20 can be seen as 100%

the % will go down by 5's because of the original number

.5*20=

This will give you a 95% success rate.

7 0

3 years ago

SOMEONE PLEASE HELPPP PLEASE ASPAAP

Feliz [49]

<h3>Answer: 35</h3>

==========================================================

Explanation:

Check out the diagram below.

U is between T and V, and U is on segment TV. We can see that the smaller pieces TU and UV add back up to the largest segment TV

TU + UV = TV

17 + 18 = TV

35 = TV

TV = 35

Segment TV is 35 units long.

5 0

3 years ago

urgente ordenar de mas grande a mas pequeño estos numeros: 85,94 85,490 , 85,93, 85,035 ,85

Ilia_Sergeevich [38]

Largest to Smallest:-

First lets write all the numbers down. 45, 94, 85, 490, 85, 93, 85, 035, 85.

035 is 35.

So lets order them now.

The biggest number is 490 and smallest is 35 (035).

490 - 94 - 93 - 85 - 85 - 85 - 85 - 035 (35)

4 0

3 years ago

Read 2 more answers

Consider the sequence defined recursively by do = 0, an = an-1 + 3n – 1. a) Write out the first 5 terms of this sequence,

creativ13 [48]

Answer:

The first 5 terms of the sequence is 2,7,15,26,40.

Step-by-step explanation:

Given : Consider the sequence defined recursively by $a_0=0$ $a_n=a_{n-1}+3n-1$

To find : Write out the first 5 terms of this sequence ?

Solution :

$a_n=a_{n-1}+3n-1$ and $a_0=0$

The first five terms in the sequence is at n=1,2,3,4,5

For n=1,

$a_1=a_{1-1}+3(1)-1$

$a_1=a_{0}+3-1$

$a_1=0+2$

$a_1=2$

For n=2,

$a_2=a_{2-1}+3(2)-1$

$a_2=a_{1}+6-1$

$a_2=2+5$

$a_2=7$

For n=3,

$a_3=a_{3-1}+3(3)-1$

$a_3=a_{2}+9-1$

$a_3=7+8$

$a_3=15$

For n=4,

$a_4=a_{4-1}+3(4)-1$

$a_4=a_{3}+12-1$

$a_4=15+11$

$a_4=26$

For n=5,

$a_5=a_{5-1}+3(5)-1$

$a_5=a_{4}+15-1$

$a_5=26+14$

$a_5=40$

The first 5 terms of the sequence is 2,7,15,26,40.

5 0

3 years ago

2. Solve the system of equations. Write your answer in parametric vector notation. X, +5x2 + 4x3 + 3x4 +9xs = 18 2x, +6x2 + 4x3

ratelena [41]

Answer:

The system is inconsistent.

Step-by-step explanation:

The system is

$x_1+5x_2+4x_3+3x_4+9x_5=18\\2x_1+6x_2+4x_3+6x_4+6x_5=28\\3x_1+7x_2+4x_3+10x_4+x_5=41\\4x_1+6x_2+2x_3+7x_4+4x_5=29$

The associated matrix to the system is

$\left[\begin{array}{cccccc}1&5&4&3&9&18\\2&6&4&6&6&28\\3&7&4&10&1&41\\4&6&2&7&4&29\end{array}\right]$

Now we use row operations to find the echelon form of the matrix:

1. We substract to row 2, two times the row 1.

We substract to row 3, three times the row 1.

We substract to row 4, four times the row 1 and obtain the matrix

$\left[\begin{array}{cccccc}1&5&4&3&9&18\\0&-4&-4&0&-12&-8\\0&-8&-8&1&-26&-13\\0&-14&-14&-5&-32&-43\end{array}\right]$

2. We multiply the second row of the preview step by -1/4. We obtain the matrix

$\left[\begin{array}{cccccc}1&5&4&3&9&18\\0&1&1&0&3&2\\0&-8&-8&1&-26&-13\\0&-14&-14&-5&-32&-43\end{array}\right]$

We add to row 3, eight times the row 2.

We add to row 4, fourtheen times the row 2 and obtain the matrix

$\left[\begin{array}{cccccc}1&5&4&3&9&18\\0&1&1&0&3&2\\0&0&0&1&-2&3\\0&0&0&-5&10&-155\end{array}\right]$

4. We add to row 4 of the preview step, five times the row 3 and obtain the matrix

$\left[\begin{array}{cccccc}1&5&4&3&9&18\\0&1&1&0&3&2\\0&0&0&1&-2&3\\0&0&0&0&0&-140\end{array}\right]$

Using backward substitution we have that

$0x_5=-140$ , then $0=-140$ and this is absurd. Then The system is inconsistent.

4 0

3 years ago