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Over [174]
4 years ago
10

Can you help me please. ​

Mathematics
1 answer:
KIM [24]4 years ago
6 0

Step-by-step explanation:

a) \\ (8 \times  {10}^{ - 3} ) \times (2 \times  {10}^{ - 4} ) \\    = 8 \times 2 \times  {10}^{ - 3}  \times {10}^{ - 4}  \\  = 16 \times  {10}^{ - 3 - 4} \\  = 16 \times  {10}^{ - 7} \\  = 1.6 \times  {10}^{ - 6}  \\  \\ b) \\ (6 \times  {10}^{ 2} )  \div  (3\times  {10}^{ - 5} ) \\  \\     =  \frac{6 \times  {10}^{ 2}}{3\times  {10}^{ - 5}}  \\  \\  =  \frac{2\times  {10}^{ 2}}{ {10}^{ - 5}}   \\  \\  = 2 \times {10}^{ 2 + 5} \\  \\  = 2 \times {10}^{7} \\

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It would be awesome if I had some help please. An answer is fine, but an explanation would be much appreciated.
frutty [35]

Answer:

868x

Step-by-step explanation:

4 0
3 years ago
The sum of a number and its additive inverse is equal to ___.
serious [3.7K]
The sum of a number and its additive inverse is equal to zero. E.g. an additive inverse of 2 is -2.

3 0
3 years ago
Read 2 more answers
Explain the answer please
valentina_108 [34]
The area of a square is the square of the length of its side. Here, we're told that the side of each square is equal to the radius (r) of the circle. Then the area of each square is
.. Asquare = r^2
There are 3 of them, so their total area is
.. Aall_squares = 3*r^2

The area of a circle is given by the formula
.. Acircle = π*r^2 . . . . . where r represents the radius of the circle

Fernie wants to compare the area of the 3 squares to that of the circle. We know that the value of π is about 3.1416, a little more than 3, so we have
.. Aall_squares = 3*r^2
.. Acircle ≈ 3.1416*r^2

We notice that 3.1416 is more than 3, so the area of the circle is greater than the area of Fernie's 3 squares.


___
It is not clear to me that Fernie's drawing will explain the formula A = π*r^2, unless it can somehow be used to show that the parts of each square that are outside the circle add up to an amount that is slightly less than the uncovered part of the circle.
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PLEASE HELP! 25 PTS!!
zubka84 [21]

Answer:

Solution given:

f(x)=\frac{x-16}{x^2+6x-40}

g(x)=\frac{1}{x+10}

now

f(x)+g(x)=\frac{x-16}{x^2+6x-40}+\frac{1}{x+10}....(1)

now

factoring x²+6x-40

we get

x²+10x-4x-40

x(x+10)-4(x+10)

(x+10)(x-4)

now substituting in equation 1 ,we get

f(x)+g(x)=\frac{x-16}{(x+10)(x-4)}+\frac{1}{x+10}

taking l.c.m

=\frac{(x-16)+(x-4)}{(x-10)(x-4)}

=now

opening bracket

\frac{x-16+x-4}{x²-10x-4x+40}

=\frac{2x-20}{x²+6x-40}

So

answer is :

B. \bold b\frac{2x-20}{x^2+6x-40}

6 0
3 years ago
Slope Application, I need help ASAP.
JulsSmile [24]

Answer:

89/3x

Step-by-step explanation:

Use Marhway for these types of problems

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