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Shkiper50 [21]
3 years ago
12

Brainly sucks d ï ç k no one answers on here​

Mathematics
1 answer:
Hunter-Best [27]3 years ago
8 0

Answer:

bet

Step-by-step explanation:

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The ratio of the heights of two similar cylinders is 1:3. If the volume of the smaller cylinder is 67cm^3, find the volume of th
Darina [25.2K]

Answer:

  • 1809 cm³

Step-by-step explanation:

<u>Given the scale factor:</u>

  • k = h/H = 1/3

Since the cylinders are similar, the other dimensions should have same scale factor of 1/3.

<u>The volumes are:</u>

  • v = 67 cm³
  • V = x

<u>Ratio of volumes is the cube of the scale factor, as the volume is the product of 3 dimensions:</u>

  • v/x = k³
  • 67/x = (1/3)³
  • x = 67*27
  • x = 1809 cm³

7 0
3 years ago
Read 2 more answers
Solve the equation x³ - 5x²+2x+8 = 0 given that - 1 is a zero of f(x)= x3 - 5x²+2x+8.
exis [7]

Answer:

added in the picture

Step-by-step explanation:

added in the picture

3 0
2 years ago
For any perfect square trinomial (quadratic), the constant term (last term) must be positive.
Ronch [10]

Answer:

For the perfect square trinomial (quadratic) i.e. \left(x\:+\:3\right)^2, the constant term (last term) is positive.

Step-by-step explanation:

"Perfect square trinomials" are termed as the quadratics that are the outcomes of squaring binomials.

For example:

\left(x\:+\:3\right)^2

\mathrm{Apply\:Perfect\:Square\:Formula}:\quad \left(a+b\right)^2=a^2+2ab+b^2

a=x,\:\:b=3

=x^2+2x\cdot \:3+3^2

=x^2+6x+9

Therefore, for the perfect square trinomial (quadratic) i.e. \left(x\:+\:3\right)^2, the constant term (last term) is positive.

5 0
3 years ago
Find y when x=14 if y varies directly with x2 and y=72 when x=6
ICE Princess25 [194]
This should help you:


8 0
3 years ago
You are working for a company that designs boxes, bottles and other containers. You are currently working on a design for a milk
ra1l [238]

Volume is a measure of the <u>quantity </u>of <em>substance</em> a given <u>object</u> can contain. The required answers are:

1.1  The <u>volume</u> of each <u>milk</u> carton is 360 cm^{3}.

1.2  The area of <em>cardboard</em> required to make a single <u>milk</u> carton is  332.6 cm^{2}.

1.3  Each <u>carton</u> can hold 0.36 liters of <u>milk</u>.

1.4  The <em>cost</em> of filling the 200 <u>cartons</u> is R 86.40.

The <u>volume</u> of a given <u>shape</u> is the amount of <em>substance</em> that it can contain in a 3-dimensional <em>plane</em>. Examples of <u>shapes</u> with volume include cubes, cuboids, spheres, etc.

The <u>area</u> of a given <u>shape</u> is the amount of <em>space</em> that it would cover on a 2-dimensional <em>plane</em>. Examples of <u>shapes</u> to be considered when dealing with the area include triangle, square, rectangle, trapezium, etc.

The box to be considered in the question is a <u>cuboid</u>. So that;

<u>Volume</u> of <u>cuboid</u> = length x width x height

Thus,

1.1 The <u>volume</u> of each <u>milk</u> carton = length x width x height

                                                         = 6 x 6 x 10

                                                        = 360

The <u>volume</u> of each <u>milk</u> carton is 360 cm^{3}.

1.2 The <em>total area</em> of<em> cardboard </em>required to make a single<u> milk</u> carton can be determined as follows:

i. <u>Area</u> of the <u>rectangular</u> surface = length x width

                                                    = 6 x 10

                                                    = 60

Total <u>area</u> of the <u>rectangular</u> surfaces = 4 x 60

                                                     = 240 cm^{2}

ii. <u>Area</u> of the <u>square</u> surface = side x side = s²

                                                   = 6 x 6  

 <u>Area</u> of the <u>square</u> surface = 36 cm^{2}

iii. There are four <em>semicircular</em> <u>surfaces</u>, this implies a total of 2 <u>circles</u>.

<em>Area</em> of a <u>circle</u> = \pi r^{2}

where r is the <u>radius</u> of the <u>circle</u>.

Total <u>area</u> of the <em>semicircular</em> surfaces = 2 \pi r^{2}

                                        = 2 x \frac{22}{7} x (3)^{2}

                                        = 56.57

Total <u>area</u> of the <em>semicircular</em> surfaces = 56.6 cm^{2}

Therefore, total area of  <em>cardboard</em> required = 240 + 36 + 56.6

                                                            = 332.6 cm^{2}

The <u>area</u> of <em>cardboard</em> required to make a single <em>milk carton</em> is  332.6 cm^{2}.

1.3 Since,

  1 cm^{3}  = 0.001 Liter

Then,

360 cm^{3} = x

x = 360 x  0.001

  = 0.36 Liters

Thus each<em> carton</em> can hold 0.36 liters of <u>milk</u>.

1.4 total cartons = 200

<em>Total volume</em> of <u>milk </u>required = 200 x 0.36

                                                 = 72 litres

But, 1 kiloliter costs R1 200. Thus

<em>Total volume</em> in kiloliters = \frac{72}{1000}

                                         = 0.072 kiloliters

The <u>cost</u> of filling the 200 cartons = R1200 x 0.072

                                         = R 86.40

The <u>cost</u> of filling the 200 <u>cartons</u> is R 86.40.

For more clarifications on the volume of a cuboid, visit: brainly.com/question/20463446

#SPJ1

4 0
2 years ago
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