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2 years ago

What is the volume of a cube

Mathematics

2 answers:

Vesna [10]2 years ago

6 0

You would use v=lwh (volume= length x width x height)

ddd [48]2 years ago

5 0

The volume of a cube: V= a^3

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Solve quadratic equations by factoring<br> x cubed +3x squared - 4x= 0

Papessa [141]

<span>x cubed +3x squared - 4x= 0
= x^3 + 3x^2 -4x = 0
=x(x^2 +3x -4) = 0
= x (x+4)(x-1) = 0

x = 0
x+ 4 = 0 then x = -4
x - 1 = 0 then x = 1
answer: x = 0, x = -4 and x = 1</span>

8 0

2 years ago

Read 2 more answers

In a study relating the degree of warping, in mm, of a copper plate (y) to temperature in °C (x), the following summary statisti

elena-14-01-66 [18.8K]

Answer:

y = 0.3035 + 0.0082x

0.6315 mm

x = 23.9634 °C

Step-by-step explanation:

a. Compute the least-squares line for predicting warping from temperature. Round the answers to four decimal places.

We need to find an equation of the form

y = b + mx

where m is the slope and b the Y-intercept.

The slope m can be computed with the formula

$\bf m=\displaystyle\frac{\displaystyle\sum_{i=1}^n(x_i-\bar x)(y_i-\bar y)}{\displaystyle\sum_{i=1}^n(x_i-\bar x)^2}$

Replacing the values in our formula (we will round at the end of the calculations)

$\bf m=\displaystyle\frac{806.94}{98,775}=0.008169476$

the Y-intercept b is computed with the formula

$\bf b=\bar y-m\bar x$

therefore we have

$\bf b=0.5188-0.008169476*26.36=0.303452613$

and the least-squares line rounded to 4 decimals would be

y = 0.3035 + 0.0082x

b. Predict the warping at a temperature of 40°C. Round the answer to three decimal places.

We simply replace x with 40 to get

y = 0.3035 + 0.0082*40 = 0.6315 mm

c. At what temperature will we predict the warping to be 0.5 mm? Round the answer to two decimal places

Here, we replace y with 0.5 and solve for x

0.5 = 0.3035 + 0.0082x ===> x = (0.5-0.3035)/0.0082 ===>

x = 23.9634 °C

6 0

3 years ago

Suppose the half-life of a material is 10 days. You have one 1 kg of the material today. How much of the material would you have

boyakko [2]

Answer:

0.50 kg of the material would be left after 10 days.

0.25 kg of the material would be left after 20 days.

Step-by-step explanation:

We have been given that the half-life of a material is 10 days. You have one 1 kg of the material today. We are asked to find the amount of material left after 10 days and 20 days, respectively.

We will use half life formula.

$A=a\cdot(\frac{1}{2})^{\frac{t}{h}}$ , where,

A = Amount left after t units of time,

a = Initial amount,

t = Time,

h = Half-life.

$A=1\text{ kg}\cdot(\frac{1}{2})^{\frac{10}{10}}$

$A=1\text{ kg}\cdot(\frac{1}{2})^{1}$

$A=1\text{ kg}\cdot\frac{1}{2}$

$A=0.5\text{ kg}$

Therefore, amount of the material left after 10 days would be 0.5 kg.

$A=1\text{ kg}\cdot(\frac{1}{2})^{\frac{20}{10}}$

$A=1\text{ kg}\cdot(\frac{1}{2})^{2}$

$A=1\text{ kg}\cdot\frac{1^2}{2^2}$

$A=1\text{ kg}\cdot\frac{1}{4}$

$A=0.25\text{ kg}$

Therefore, amount of the material left after 20 days would be 0.25 kg.

6 0

2 years ago

(Mhanifa) please solve this question thank you!

Dima020 [189]

Answer:

Option D.

Step-by-step explanation:

The angles forming a straight angle are supplementary.

We have angles APC and DPB forming a straight angle with adjacent angle to both (either angle APD or CPB).

It means the correct choice is D.

8 0

1 year ago

The graph of y = 4 – is shown.

Ratling [72]

Rather than try to select the correct answer without doing work, I'd rather do the work first and the matching later.

Given y = 4 - x, or y = 4 - 1x,

1) the y-intercept is (0,4).
2) this being a linear function, it has neither maximum nor minimum.
3) the x-intercept is found by letting y = 0 and solving for x: 0 = 4 - x; x - 4. The x-intercept is (4,0).

Which of the given statements is true?

7 0

2 years ago