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alekssr [168]
3 years ago
9

Express 0.000023 in standard form . *​

Mathematics
1 answer:
sdas [7]3 years ago
5 0

Answer:

Step-by-step explanation:

2.3*10'-2

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How do you know the quotient of 639÷6 is greater than 100 before you actually divide
svp [43]
The 6 in the hundreds place tells you the quotient is at least 100. The non-zero digits in the 1s and 10s places tell you the quotient is more than 100.
4 0
3 years ago
Solve y = x2-5 for x.<br> O A. x = +/-5<br> B. X = y + 5<br> C. X = +/y+5<br> O D. x = y-5
OLEGan [10]
I got C b I’m not 100% sure
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Use the graph of the function to find, f(-2).<br><br> A. -2<br> B. -1<br> C. 1<br> D. 2
Misha Larkins [42]

Since f(-2) means that our x-value is -2, we can go to that position on the graph and slide up the y-axis to find the intersection this point has with the function graphed.

If I am correct the answer should be that the y value is 2!

7 0
3 years ago
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Pre-calc questions (trignometry)
miv72 [106K]

Answer:  (1a) 250°  (1b) 70°    (2) see below

<u>Step-by-step explanation:</u>

1a) -110 + 360 (one rotation clockwise) = 250°

1b) 430 - 360 (one rotation counterclockwise) = 70°

2) sec Ф = -8/5    in Quadrant III

Quadrant III identifies that both sin (y) and cos (x) are negative.

sec = r/x --> r = 8, x = -5, and y = -√39

(Use Pythagorean Theorem x² + y² = r² to solve for y)

\sin\theta=\dfrac{y}{r}=\dfrac{-\sqrt{39}}{8}              \csc\theta=\dfrac{r}{y}=\dfrac{-8}{\sqrt{39}}  rationalized = \dfrac{-8\sqrt{39}}{39}

\cos\theta=\dfrac{x}{r}=\dfrac{-5}{8}                  \sec\theta=\dfrac{r}{x}=\dfrac{-8}{5}  (GIVEN)

\tan\theta=\dfrac{y}{x}=\dfrac{\sqrt{39}}{5}              \cot\theta=\dfrac{x}{y}=\dfrac{-5}{-\sqrt{39}}  rationalized = =\dfrac{5\sqrt{39}}{39}

NOTE THAT YOU ARE ALLOWED A MAXIMUM OF 3 QUESTIONS PER POST.  Please repost #3 and #4 as a different question and I will answer them.

8 0
3 years ago
12 (3 + 4) =<br><br> 36 + 48<br> 0.84<br> 123 + 124<br> 15 + 16
Nitella [24]
36+48 is the only right answer here
3 0
3 years ago
Read 2 more answers
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