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notka56 [123]
4 years ago
13

Which statement correctly describes this expression 2m3-11 A. Twice the cube of the number subtracted from 11 B.the difference o

f twice a number and 11 cubes
Mathematics
1 answer:
Ad libitum [116K]4 years ago
5 0

A number is :

m

_________________________________

Cube of it equals :

{m}^{3}

_________________________________

Twice the cube equals :

2 {m}^{3}

_________________________________

Subtract it from 11 equals :

2 {m}^{3} - 11

Which is the expression we have.

_________________________________

So the correct answer is A.

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

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Solve by completing the square.
IRINA_888 [86]

Answer:

-23.18, -0.82

Step-by-step explanation:

  • h² + 24h + 19 = 0
  • h²+2h*12+12²-125=0
  • (h+12)²-√125²=0
  • (h+12+11.18)(h+12-11.18)=0
  • (h+23.18)(h+0.82)=0
  • h+23.18=0 ⇒ h= -23.18
  • h+0.82=0 ⇒ h= -0.82
7 0
3 years ago
Read 2 more answers
Subtraction with renaming 9 3/8 - 8 7/8 <br> please help me
BaLLatris [955]
Converting 9 3/8 to an improper fraction is 75/8. Converting 8 7/8 to an improper fraction is 71/8. Subtracting these, we get 75/8 - 71/8 = 4/8 = 1/2.
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Hope this helps!
7 0
3 years ago
The chief chemist for a major oil and gasoline production company claims that the regular unleaded gasoline produced by the comp
Svetradugi [14.3K]

Answer:

The probability of finding an average in excess of 4.3 ounces of this ingredient from 100 randomly inspected 1-gallon samples of regular unleaded gasoline = P(x > 4.3) = 0.00621

Step-by-step explanation:

This is a normal distribution problem

The mean of the sample = The population mean

μₓ = μ = 4 ounces

But the standard deviation of the sample is related to the standard deviation of the population through the relation

σₓ = σ/√n

where n = Sample size = 100

σₓ = 1.2/√100

σₓ = 0.12

The probability of finding an average in excess of 4.3 ounces of this ingredient from 100 randomly inspected 1-gallon samples of regular unleaded gasoline = P(x > 4.3)

To do this, we first normalize/standardize the 4.3 ounces

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (4.3 - 4)/0.12 = 2.5

To determine the probability of finding an average in excess of 4.3 ounces of this ingredient from 100 randomly inspected 1-gallon samples of regular unleaded gasoline = P(x > 4.3) = P(z > 2.5)

We'll use data from the normal probability table for these probabilities

P(x > 4.3) = P(z > 2.5) = 1 - P(z ≤ 2.5) = 1 - 0.99379 = 0.00621

5 0
4 years ago
HELP MahHH PLEaSE ASAP
tatyana61 [14]

Amelia can make 9 bracelets in 25 mins

Step-by-step explanation:

45 - 20 = 25

5 0
3 years ago
Compute the differential of surface area for the surface S described by the given parametrization.
AysviL [449]

With S parameterized by

\vec r(u,v)=\langle e^u\cos v,e^u\sin v,uv\rangle

the surface element \mathrm dS is

\mathrm dS=\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|\,\mathrm du\,\mathrm dv

We have

\dfrac{\partial\vec r}{\partial u}=\langle e^u\cos v,e^u\sin v,v\rangle

\dfrac{\partial\vec r}{\partial v}=\langle -e^u\sin v,e^u\cos v,u\rangle

with cross product

\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\langle ue^u\sin v-ve^u\cos v,-ve^u\sin v-ue^u\cos v,e^{2u}\cos^2v+e^{2u}\sin^2v\rangle

\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\langle e^u(u\sin v-v\cos v),-e^u(v\sin v+u\cos v),e^{2u}\rangle

with magnitude

\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=\sqrt{e^{2u}(u\sin v-v\cos v)^2+e^{2u}(v\sin v+u\cos v)^2+e^{4u}}

\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=e^u\sqrt{u^2+v^2+e^{2u}}

So we have

\mathrm dS=\boxed{e^u\sqrt{u^2+v^2+e^{2u}}\,\mathrm du\,\mathrm dv}

8 0
3 years ago
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