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3 years ago

What is nbr2 at the end of the program segment execution? nbr2 = 0 calc = 1 x = 7 while x >=3 calc = calc * x x = x - 2

Mathematics

1 answer:

Rashid [163]3 years ago

5 0

Answer:

nbr2 = 0

Step-by-step explanation:

Given

The following program segment

nbr2 = 0

calc = 1

x= 7

while x >=3:

calc = calc * x

x = x - 2

Required

The value of nbr2

In the first line, nbr2 was initialized to 0 i.e.

nbr2 = 0

From the second line of the program till the last, no operation was carried out on nbr2

This means that nbr2 will maintain its original value (which is 0)

Hence,

The value of nbr2 is 0

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Solution -9-8(1+4h)= -17

geniusboy [140]

Step 1. Add 9 to both sides

 $-8(1+4h)=-17+9$

Step 2. Simplify $-17+9$ to $-8$

$-8(1+4h)=-8$

Step 3. Divide both sides by $-8$

$1+4h=1$

Step 4. Cancel 1 on both sides

$4h=0$
Step 5. Divide both sides by 4

$h=0$

Done! :) Hope this helps! :)

To check $-9-8(1+4h)=-17$ follow these steps below:

Step 1. Let h = 0

$-9-8*(1+4*0)=-17$

Step 2. Simplify 4 * 0 to 0

$-9-8*(1+0)=-17$

Step 3. Simplify 1 + 0 to 1

$-9-8*1=-17$

Step 4. Simplify 8 * 1 to 8

$-9-8=-17$

Step 5. Simplify -9 -8 to -17

 $-17=-17$

Done! :)

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3 years ago

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asambeis [7]

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