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wolverine [178]
4 years ago
12

How do you write 15/1000in scientific notation​

Mathematics
1 answer:
Amanda [17]4 years ago
6 0

Answer: 15 * 10^-3

Step-by-step explanation:

Divide 15/1000

= 0.015

=15 * 10^-3

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The diameter of a circle is __________.
Orlov [11]

Answer:

The answer is C

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Can yall tell me what B is? b-7/4=-2/3
Rashid [163]

Answer:

1 1/12

Step-by-step explanation:

b-7/4=-2/3

First, in an equation like this, you add 7/4 to each side which will look like this:

b-7/4=-2/3

 +7/4=+7/4

We add 7/4 to each side of the equation to balance the numbers.

b=+7/4-2/3 (because the minus symbol is in front, we put that)

Before adding we have to find the least common denominator known as LCD.

We find that by first looking at the denominators or the fractions which are 4,3

4,8,\12/,16,20,24.

3,6,9,\12/,15,18,21,24.

We can also find the LCD by multiplying 3 and 4. As you can see in the list, there are 12's in each list of numbers. We use that to change our fractions.

From the number 4 we find out how many times (x) we have to add 4 to get 12. We do 12/4 which is 3. And 12/3 which is 4.

7 x 3       2 x 4

-------   -   -------

4 x 3       3 x 4

which equals 21/12-8/12. Now that we have a common denominator, we can subtract.

21                  8                    13

---        -        ---        =         -----     =    1  \frac{1}{12}

12                 12                    12

I hope this helps, please let me know if I have any incorrections. :D

8 0
3 years ago
What are the solutions of x^2-2x+5=0
sweet [91]

The solution of x^{2}-2 x+5=0 are 1 + 2i and 1 – 2i

<u>Solution:</u>

Given, equation is x^{2}-2 x+5=0

We have to find the roots of the given quadratic equation

Now, let us use the quadratic formula

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}  --- (1)

<em><u>Let us determine the nature of roots:</u></em>

Here in x^{2}-2 x+5=0 a = 1 ; b = -2 ; c = 5

b^2 - 4ac = 2^2 - 4(1)(5) = 4 - 20 = -16

Since b^2 - 4ac < 0 , the roots obtained will be complex conjugates.

Now plug in values in eqn 1, we get,

x=\frac{-(-2) \pm \sqrt{(-2)^{2}-4 \times 1 \times 5}}{2 \times 1}

On solving we get,

x=\frac{2 \pm \sqrt{4-20}}{2}

x=\frac{2 \pm \sqrt{-16}}{2}

x=\frac{2 \pm \sqrt{16} \times \sqrt{-1}}{2}

we know that square root of -1 is "i" which is a complex number

\begin{array}{l}{\mathrm{x}=\frac{2 \pm 4 i}{2}} \\\\ {\mathrm{x}=1 \pm 2 i}\end{array}

Hence, the roots of the given quadratic equation are 1 + 2i and 1 – 2i

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I think the answer is A. 0
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Amar makes 8 dollars an hour for raking leaves
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