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3 years ago

Choose the digit that shows 3.096 rounded to the nearest hundred

2 answers:

6 0

The round it off 3.096 because 0i is between the 3 or 0 the nearest hundred is zero but you rounded the number higher the number is for you to rounded I will keep has the same

Eddi Din [679]3 years ago

4 0

The 0 because the 3 is thousand and the 0 is hundred and th 9 nearest tenth and 6 nearest ones

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A sno-cone machine priced at $139 is on sale for 20% off. What is the price of the sno-cone machine after the discount? *

lawyer [7]

So percents are just written as 20% = 0.2 so keep that in mind

139 divided by 0.2 is 95 , which would be the price after the discount :)

hope this helps!
give me brainliest

7 0

3 years ago

The radioactive element carbon-14 has half of 5750 years. A scientist determined that the bones from a mastodon has lost 56.6% o

IRISSAK [1]

Answer:7881 years

Step-by-step explanation:

You can use the equation

fraction remaining = 0.5n where n is # of half lives elapsed

n = ?

fraction remaining = 1 - 0.612 = 0.388

0.388 = 0.5n

n = 1.37 = the # of half lives

1.37 x 5750 yrs = 7881 years

8 0

3 years ago

In an experiment, a ball is drawn from an urn containing 11 orange balls and 8 blue balls. If the ball is orange, three coins ar

Sophie [7]

Answer:

a) 4

b) 12

Step-by-step explanation:

Since, when an orange ball is drawn, three coins are tossed. So, sample points are OHHH, OHHT, OHTH, OHTT. OTHH, 0THT, OTTH, OTTT

So, clearly, 8 sample points will have an orange ball.

When a blue ball is drawn, two coins are tossed. So, sample points are BHH,BHT,BTH,BTT.

So, clearly, 4 samples points will when a blue ball is drawn.

Total samples points=4+8=12

4 0

4 years ago

Describe a situation that the expression -15÷(-15) can represent

FrozenT [24]

-15 so that is negative. so that is negative 15÷15 so 15 ÷15 is 1 so it would be -15÷-15=-1

7 0

3 years ago

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$

So we have

$g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}$

and we can simplify this by factoring out $18(36-x^2)^{-3/2}$ to end up with

$g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}$

5 0

3 years ago