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slavikrds [6]
3 years ago
9

Evaluate the expression x^2 - y^2, if x = -3, and y= 2

Mathematics
1 answer:
Darina [25.2K]3 years ago
4 0

\\ \rm\longmapsto x^2-y^2

\\ \rm\longmapsto (-3)^2-(2)^2

\\ \rm\longmapsto 9-4

\\ \rm\longmapsto 5

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30 Points! please help!
SVEN [57.7K]

Answer:

18) 4 * 10 ^ -10

19) t ^ 9

Step-by-step explanation:

18. When multiplying in scientific notation, just multiply the regular numbers together and add the exponents for the tens.

So it'll be 40 * 10^-11

However, the beginning number should have the decimal right after the 4, not any place after (for example, it's 3.056, not 30.56 or 3056.)

To fix this, move the decimal place forward one space from 40. to 4.0, and add 1 to the -11 power.

So the answer is now 4 * 10^-10

19. When dividing exponents, if the base number is the same, you can divide. (Basically, you can not divide x^2 and y^3 because x and y aren't the same)

Since in this case it's t, you can divide.

Dividing exponents is simple: just subtract them.

14 - 5 = 9

so the answer is t ^ 9

3 0
4 years ago
Rob coloured 1/4 of a rectangle .what is another way to name 1/4?
mixas84 [53]

another way to name 1/4 is a quarter because 1/4 of 100 is 25 %, so that's why it is called a quarter.

hopefully this helps you :)

5 0
3 years ago
Round the number to the place of the underlined digit.<br><br> 4.327 (The underlined Digit is 3)
Genrish500 [490]

Answer:4.3

You round to the tenths place

5 0
3 years ago
Let f(x) = x2 − 3x − 7. Find f(−3).
MrRa [10]
Answer is:
f(-3) = 11
5 0
3 years ago
Read 2 more answers
Solve the following equation: lim x-&gt;0 sin3x/5x^3-4x.
goblinko [34]
Usual limit of sin is sinX/X--->1, when X--->0

sin3x/5x^3-4x=0/0?, sin3x/3x--->1 when x --->0, so sin3x/5x^3-4x=                    [3x. sin3x / 3x] /(5x^3-4x)=(sin3x / 3x) . (3x/5x^3-4x)
   =(sin3x / 3x) . (3/5x^2- 4)
finally lim  sin3x/5x^3-4x=lim (sin3x / 3x) .(3/5x^2- 4)=1x(3/-4)= - 3/4
                 x----->0            x---->0
3 0
3 years ago
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