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VladimirAG [237]

2 years ago

11

Can you help me please I’ll give brainliest

1 answer:

kolezko [41]2 years ago

5 0

The answer is B I hope it’s right

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What’s the correct answer for this?

joja [24]

Answer:

MN = 48

Step-by-step explanation:

Since AB is a diameter , so it divides MN into 2 equal parts i.e. MO = NO

7x-4 = 6x

7x-6x = 4

x = 4

Now

MN = MO + NO

MN = 7x-4+6x

MN = 7(4)-4+6(4)

MN = 28-4+24

MN = 48

6 0

3 years ago

Read 2 more answers

Area of a trapezoid is 70 meters squared, and its bases are 5m and 9m, find its height

MrMuchimi

Answer:

10

Step-by-step explanation:

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3 years ago

What is the y-intercept of the function f(x)=0.65^x-2 +5. Round to two decimal places

lana [24]

The y-intercept is 5.

6 0

3 years ago

All angles are right angles this describes a parallelogram, rhombus, rectangle, or a square (there can be more than one answer

Sidana [21]

Rectangle and square

7 0

3 years ago

The base of an aquarium with given volume V is made of slate and the sides are made of glass. If the slate costs seven times as

lubasha [3.4K]

Answer:

$l = \sqrt[3]{\frac{2V}{7}}$ $b = \sqrt[3]{\frac{2V}{7}}$ $h = \sqrt[3]{\frac{49V}{4}}$

Step-by-step explanation:

Represent the volume of the box with V and the dimensions with l, b and h.

The volume (V) is:

$V = l * b * h$

Make h the subject of the formula

$h = \frac{V}{lb}$

The surface area (S) of the aquarium is:

$S = lb + 2(lh + bh)$

Where lb represents the area of the base (i.e. slate):

The cost (C) of the surface area is:

$C = 7 * lb + 1 * 2(lh + bh)$

$C = 7lb + 2(lh + bh)$

$C = 7lb + 2h(l + b)$

Substitute $\frac{V}{lb}$ for h in the above equation

$C = 7lb + 2*\frac{V}{lb}(l + b)$

$C = 7lb + \frac{2V}{lb}(l + b)$

$C = 7lb + \frac{2V}{b} + \frac{2V}{l}$

Differentiate with respect to l and with respect to b

$C_l=7b - \frac{2V}{l^2}$ $=0$

$C_b=7l - \frac{2V}{b^2}$ $=0$

To solve for b and l, we equate both equations and set l to b (to minimize the cost)

$7b - \frac{2V}{l^2}=7l - \frac{2V}{b^2}$

$7l - \frac{2V}{l^2}=7b - \frac{2V}{b^2}$

By comparison:

$l =b$

$C_l=7b - \frac{2V}{l^2}$ $=0$ becomes

$7l - \frac{2V}{l^2}=0$

$7l = \frac{2V}{l^2}$

Cross Multiply

$7l^3 = 2V$

Solve for l

$l^3 = \frac{2V}{7}$

$l = \sqrt[3]{\frac{2V}{7}}$

Recall that: $l =b$

$b = \sqrt[3]{\frac{2V}{7}}$

Also recall that:

$h = \frac{V}{lb}$

$h = \frac{V}{\sqrt[3]{\frac{2V}{7}}*\sqrt[3]{\frac{2V}{7}}}$

$h = \frac{V}{\sqrt[3]{\frac{4V^2}{49}}}$

Apply law of indices

$h = \sqrt[3]{\frac{49V^3}{4V^2}}$

$h = \sqrt[3]{\frac{49V}{4}}$

The dimension that minimizes the cost of material of the aquarium is:

$l = \sqrt[3]{\frac{2V}{7}}$ $b = \sqrt[3]{\frac{2V}{7}}$ $h = \sqrt[3]{\frac{49V}{4}}$

8 0

3 years ago