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disa [49]
3 years ago
8

The length width and height of a rectangular solid are 8,4,1 respectively what is the length of the longest line segment whose e

ndpoints are two vertices of this solid
Mathematics
1 answer:
LUCKY_DIMON [66]3 years ago
3 0

Answer:

9 units

Step-by-step explanation:

Given

Length = 8

Width = 4

Height = 1

Required

Determine the length of the longest segment

To do this, we use:

Longest = \sqrt{Length^2 + Width^2 + Height^2}

So, we have:

Longest = \sqrt{8^2 + 4^2 + 1^2}

Longest = \sqrt{64 + 16 + 1}

Longest = \sqrt{81}

Take positive square root

Longest = 9

<em>Hence, the length of the longest segment is 9 units</em>

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Georgia kassem installs fencing. she is paid $7.50 per hour for an 8-hour day and time and a half for overtime for any work over
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Georgia worked overtime on 4 of the 5 days. Her total overtime hours were

... (12-8) +(9 -8) +(10 -8) +(10 -8) = 4 +1 +2 +2 = 9

Then her pay for the week is

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Cross Multiply by 'x - 5'
Hence
y(x - 5) = 4 -3x
xy - 5y = 4 - 3x

Move all the terms containing an 'x' to one side of the equals and all the others to the opposite side.
xy + 3x = 4 + 5y
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8 0
3 years ago
Find the smallest zero of f(x+5)
podryga [215]

ANSWER

1. k=13

2. x=-10

EXPLANATION

The given function is

f(x) =  {x}^{2}   + 3x  - 10

To find f(x+5), plug in (x+5) wherever you see x.

This implies that:

f(x) =  {(x + 5)}^{2}   + 3(x + 5) - 10

Expand:

f(x) =   {x}^{2}    + 10x + 25+ 3x + 15- 10

Simplify to obtain

f(x) =   {x}^{2}    + 13x + 30

We now compare with,

f(x) =   {x}^{2}    + kx + 30

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k = 13

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{x}^{2}    + 13x + 30 = 0

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(x + 3)(x + 10) = 0

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