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2 years ago

15

Ned and Sam are playing a game. Ned has -3 points. Sam has -8 points. How many more points does Ned have than Sam? Please provid

e an equation and an answer.

1 answer:

9966 [12]2 years ago

4 0

Answer:

5 points

Step-by-step explanation:

Subtracting their scores, we get (-3)-(-8)=5

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You are given a 16-bit binary integer X= (1110 0100 0111 1010)2 . Convert X to decimal if:

podryga [215]

Step-by-step explanation:

a). The decimal numbers is equal to the sum of binary numbers $(b_n)$ times their powers of $2(2^n) .$

That is,

$b_n \times 2^n + b_{n-1} 2^{n-1}+ ..... + b_02^0$

$\therefore (1110 \ \ 0100 \ \ 0111 \ \ 1010)_2$

= $(0 \times 2^0) + (1 \times 2^1) + (0 \times 2^2) + (1 \times 2^3) + (1 \times 2^4) + (1 \times 2^5) + (1 \times 2^6) + ( 0 \times 2^7) + (0 \times 2^8) + (0 \times 2^9) + (1 \times 2^{10} )$ $+ (0 \times 2^{11}) + (0 \times 2^{12}) + (1 \times 2^{13}) + (1 \times 2^{14})+ (1 \times 2^{15})$

= 58490

b).

The left most bit of the number represents the sign of the number. If bit = 1, the number is negative else if the bit is 0, the number is positive.

All the remaining bits (i.e all the bits except the leftmost bit) represent the magnitude of the number.

Now, X = 1110 0100 0111 1010

left most bit = 1 => number is negative

remaining bits Y = 110 0100 0111 1010

converting Y to decimal => Y = 25722

c).

Check the leftmost bit(lmb) of the number

if leftmost bit is 0 => convert do the binary to decimal conversion

if leftmost bit is 1, follow the following steps

1. Flip the bits (convert the 0's into 1's and the vice versa)

2. convert the new number from binary to decimal.

3. Add a negative sign / multiply by -1

Now, X = 1110 0100 0111 1010

left most bit = 1;

flipping the bits => Y = 0001 1011 1000 0101

converting Y into decimal => Y = 7045

multiplying Y with -1 => Y = -7045

Therefore, one's complement of X to decimal is -7045

Thus, X = -25722 in decimal.

d).

Check the leftmost bit(lmb) of the number

if leftmost bit is 0 => convert do the binary to decimal conversion

if leftmost bit is 1, follow the following steps

1. Flip the bits (convert the 0's into 1's and the vice versa)

2. convert the new number from binary to decimal.

3. Add 1 to the converted number

4. Add a negative sign / multiply by -1

Now, X = 1110 0100 0111 1010

left most bit = 1;

flipping the bits => Y = 0001 1011 1000 0101

converting Y into decimal => Y = 7045

Adding 1 to Y => Y = 7046

multiplying Y with -1 => Y = -7046

Thus, two's complement of X to decimal is -7046

7 0

3 years ago

IF A SUBSCRIPTION IS $499 PLUS 8% TAX FOR 30 DAYS, BUT IS BEING PRORATED FOR 7 DAYS, PLUS THERE IS A $10 OFF COUPON, WHAT'S THE

natita [175]

Answer:

$528.12

Step-by-step explanation:

499-10

489×1.08= 528.12

6 0

3 years ago

Read 2 more answers

Suppose M is the midpoint of line FG.

alukav5142 [94]

$A)\\|FM|=5y+13;\ |MG|=5-3y\\\\|FG|=|FM|+|MG|\\\\|FG|=(5y+13)+(5-3y)=2y+18\\\\B)\\|FM|=8a+1;\ |FG|=42\\\\|FG|=2|FM|\\\\2(8a+1)=42\\16a+2=42\ \ \ \ \ |subtract\ 2\ from\ both\ sides\\16=40\ \ \ \ \ |divide\ both\ sides\ by\ 16\\a=\frac{40}{16}\\a=\frac{5}{2}\\a=2.5$

6 0

3 years ago

What is the answer to -t+9-4t=59?

kkurt [141]

Answer:

-10

Step-by-step explanation:

combine like terms -5x+9=59 subtract 9 from both side and get -5x=50 so x=-10

3 0

3 years ago

Read 2 more answers

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

-----------------------

Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

6 0

3 years ago