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MissTica
3 years ago
11

Please Hurry!!

Mathematics
2 answers:
sertanlavr [38]3 years ago
8 0
6/3 = 2
10^4/10^-5 = 10^9

so it's 2 x 10^9

answer
<span>A 2 × 10^9</span>
const2013 [10]3 years ago
4 0
The answer is A. What you do first is divide 6 by 3 to get 2 then you bring down your 10-----> 2*10. Then you subtract -5 from 4 ---> 4-(-5) (if the equation looks like this and only this then you just add the 4 and the 5, you can remember this because the subtraction sign, parenthese, and negative sign looks like a big addition sign.) Your answer will be 9 so just bring that sown too! Your final answer is 2*10^9
You might be interested in
Which ordered pairs are solutions to the inequality 2x-y&gt;1?
RideAnS [48]

Answer:

Any points in the shaded region including (2,-2) and (-3,-8)

Step-by-step explanation:

Convert the line into slope intercept form and graph it.

2x-y > 1 becomes -y>1-2x. Divide both sides by -1 and you get y<2x-1. Graph it with the shaded area on the right and a dashed line.

Any point which falls within the shaded red of the graph is a solution. No points on the line since it is not equal to (its dashed) are solutions. Check the location of your points to verify that they fall within this area.

(-3, -8)  ---Yes

(-1, -3)  ---No

(0, 5)  --- No

(1, 6)  --- No

(2, -2) ---Yes

8 0
3 years ago
The mean annual income for adult women in one city is $28,520 and the standard deviation of the incomes is $5600. The distributi
pshichka [43]

Answer:

E(\bar{x}) = 28520

Standard error of mean = 689

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = $28,520

Standard Deviation, σ = $5600

Mean of sampling distribution =

E(\bar{x}) = \mu = 28520

As per Central Limit Theorem, if the sample size is large enough, then the sampling distribution of the sample means follow approximately a normal distribution.

Sample size, n = 66

Since the sample size is large, we can use normal distribution for approximation.

Standard error of mean =

\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{5600}{\sqrt{66}} = 689.31 \approx 689

7 0
3 years ago
I have a pile of 20 coins that includes dimes and quarters. The coins total is $2.75. How many quarters do I have?
Ilia_Sergeevich [38]
You have 5 quarters which equals 25 cents each. Which in total comes to $1.25.
You have 15 dimes which is worth 10 cents each.
Which in total comes to $1.50.
$1.50
+$1.25
=$2.75
5 0
3 years ago
PLEASE HELP ME WITH THIS! IM BEHIND
jolli1 [7]

Answer:40

Step-by-step explanation: i may be wrong but i subtracted 136 from 95 and i got 40 im sorry if it is wrong

5 0
2 years ago
Find dy/dx if f(x) = (x + 8)^3x.
vlabodo [156]
This is quite a complex problem. I wrote out a really nice solution but I can't work out how to put it on the website as the app is very poorly made. Still, I'll just have to type it all in...

Okay so you need to use a technique called logarithmic differentiation. It seems quite unnatural to start with but the result is very impressive.

Let y = (x+8)^(3x)

Take the natural log of both sides:

ln(y) = ln((x+8)^(3x))

By laws of logarithms, this can be rearranged:

ln(y) = 3xln(x+8)

Next, differentiate both sides. By implicit differentiation:

d/dx(ln(y)) = 1/y dy/dx

The right hand side is harder to differentiate. Using the substitution u = 3x and v = ln(x+8):

d/dx(3xln(x+8)) = d/dx(uv)

du/dx = 3

Finding dv/dx is harder, and involves the chain rule. Let a = x+ 8:

v = ln(a)
da/dx = 1
dv/da = 1/a

By chain rule:

dv/dx = dv/da * da/dx = 1/a = 1/(x+8)

Finally, use the product rule:

d/dx(uv) = u * dv/dx + v * du/dx = 3x/(x+8) + 3ln(x+8)

This overall produces the equation:

1/y * dy/dx = 3x/(x+8) + 3ln(x+8)

We want to solve for dy/dx, achievable by multiplying both sides by y:

dy/dx = y(3x/(x+8) + 3ln(x+8))

Since we know y = (x+8)^(3x):

dy/dx = ((x+8)^(3x))(3x/(x+8) + 3ln(x+8))

Neatening this up a bit, we factorise out 3/(x+8):

dy/dx = (3(x+8)^(3x-1))(x + (x+8)ln(x+8))

Well wasn't that a marathon? It's a nightmare typing that in, I hope you can follow all the steps.

I hope this helped you :)
8 0
3 years ago
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