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2 years ago

PLEASE HELP!!!

Mathematics

1 answer:

Jobisdone [24]2 years ago

6 0

The answer it 1. is A.
The answer to 2. is B.

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Tju [1.3M]

See the attached picture:

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3 years ago

PLEASE ANSWER QUICK PLEASE

frozen [14]

1/32 maybe................

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2 years ago

A rectangular storage container with a lid is to have a volume of 2 m3. The length of its base is twice the width. Material for

Scilla [17]

Answer:

Dimensions are 2 m by 1 meter by 1 meter,

Minimum cost is $ 18.

Step-by-step explanation:

Let w be the width ( in meters ) of the container,

Since, the length is twice of the width,

So, length of the container = 2w,

Now, if h be the height of the container,

Volume = length × width × height

2 = 2w × w × h

1 = w² × h

$\implies h=\frac{1}{w^2}$

Since, the area of the base = l × w = 2w × w = 2w²,

Area of the lid = l × w = 2w²,

While the area of the sides = 2hw + 2hl

= 2h( w + l)

$= 2\times \frac{1}{w^2}(w+2w)$

$=\frac{6w}{w^2}$

$=\frac{6}{w}$

Since, Material for the base costs $1 per m². Material for the sides and lid costs $2 per m²,

So, the total cost,

$C(w) = 1\times 2w^2+2\times 2w^2 + 2\times \frac{6}{w}$

$=2w^2+4w^2+\frac{12}{w}$

$=6w^2+\frac{12}{w}$

Differentiating with respect to w,

$C'(w) = 12w -\frac{12}{w^2}$

Again differentiating with respect to w,

$C''(w) = 12 + \frac{24}{w^3}$

For maxima or minima,

C'(w) = 0

$\implies 12w -\frac{12}{w^2}=0$

$\implies 12w^3 - 12=0$

$w^3-1=0\implies w = 1$

For w = 1, C''(w) = positive,

Hence, for width 1 m the cost is minimum,

Therefore, the minimum cost is C(1) = 6(1)²+12 = $ 18,

And, the dimension for which the cost is minimum is,

2 m by 1 meter by 1 meter.

7 0

3 years ago

Pls answer this my friend says i cant do it i cant but i wanna prove him wrong

Vladimir [108]

Answer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

A newly discovered planet orbits a distant star with the same mass as the Sun at an average distance of 101 million kilometers.

Elden [556K]

Kepler's third law described the relation between semi-major axis (or average distance to the star) and the orbital period (how long it takes to complete one lap) as follows:
a^3 / p^2 = constant
In the case of our Solar system the constant is 1
This means that, for this problem:
a^3 / p^2 = 1
p^2 = a^3
p = a^(3/2)
The semi major axis is given as 101 million km. We need to convert this into AU where 1 AU is approximately 150 million Km
101 million Km = (101x1) / 150 = 0.67 AU

Now, we substitute in the equation to get the orbital period as follows:
p = (0.67)^(3/2) = 0.548 earth years

3 0

2 years ago