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2 years ago

9

Solve for x. Round your answer to the nearest tenth.

1 answer:

kipiarov [429]2 years ago

8 0

25 because you do 19 squared plus 16 squared equals 617, you do 617 rooted. It’s called pythagorus.

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HELP WHAT IS THE SLOPE OF THIS GRAPH!!! DUE TODAY ALMOST LATE!!!!!

Nady [450]

Answer:

wait what?

how is something meant to be due late, like b r u h you can't make a task due late. LOL :)

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

There are 45 new houses being built in a neighborhood. Last month 1/3 of them were sold. This month 2/5 of the remaining houses

polet [3.4K]

Answer:

total=45

1/3+2/5=11/15

remaining fraction of houses=1-11/15

=4/15

reamaimg houses=4/15×45=12houses

7 0

3 years ago

Red ink pens are sold at $1.50 each black ink pens are sold at $30 which color is the cheaper buy?

Lena [83]

Answer:

Red Ink pens.

Step-by-step explanation:

Red ink pens are sold at $1.50 while Black ink pens are sold at $30. One black ink pens cost more than one red ink pen by $28.50.

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3 years ago

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

2 years ago

Find three consecutive odd integers such that the sum of 7 times the smallest and twice the largest is -91.

ollegr [7]

Find three consecutive odd integral such that the sum of seven times the smallest and twice the largest is -91

3 0

2 years ago