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3 years ago

9

Find two consecutive odd integers such that 28 more than the lesser is three times the greater.

1 answer:

Elan Coil [88]3 years ago

8 0

Answer:

lesser odd integer =11

Greater odd integer=13

Step-by-step explanation:

let the consecutive odd integers be x,x+2

and x<x+2

3(x+2)=x+28

3x+6=x+28

3x-x=28-6

2x=22

x=11

x+2=11+3=13

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Fernando followed two diagonal paths, Paths 1 and 2, to get from his house, F, to a neighborhood corner store, C, as shown below

aleksandrvk [35]

Answer:

42 meters

Step-by-step explanation:

Given

See attachment for complete question

Required

Calculate distance FC

FC is calculated as:

$FC = Path\ 1 + Path\ 2$

Where

$Path\ 1 = FB$

$Path\ 2 = BC$

Considering $\triangle FEB$ , we have:

$FB^2 = FE^2 + BE^2$ --- Pythagoras theorem

Where

$FE = FD - BA$

$FE = 22m - 15m$

$FE = 7m$

and

$BE = CD - CA$

$BE = 32m - 8m$

$BE = 24m$

So:

$FB^2 = FE^2 + BE^2$

$FB^2 = 7^2 +24^2$

$FB^2 = 49 +576$

$FB^2 = 625$

Take square roots of both sides

$FB = 25$

So:

$Path\ 1 = 25$

Considering $\triangle BAC$ , we have:

$BC^2 = BA^2 + AC^2$ --- Pythagoras theorem

Where:

$BA = 15$ and $AC = 8$

So, we have:

$BC^2 = 15^2 + 8^2$

$BC^2 = 225 + 64$

$BC^2 = 289$

Take square roots of both sides

$BC = 17$

So;

$Path\ 2= 17$

Recall that:

$FC = Path\ 1 + Path\ 2$

$FC = 25 + 17$

$FC = 42$

6 0

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What is the relationship between 5 in these two numbers 8,523 and 9,852

babymother [125]

Answer:

The 5 in 8523 is 10x greater than the 5 in 9852.

Step-by-step explanation:

Note the place value of the two number's digit 5:

8<u>5</u>23: Hundreds place value.

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10 x 10 = 100 ∴ The 5 in 8523 (hundreds place value) is 10x greater than the 5 in 9852 (tens place value).

~

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