Ask question

Ask question

All categories

3 years ago

13

What is the volume of a certain cube? (1) The sum of the lengths of the edges of the cube is 36. (2) The surface area of the cub

e is 54.

1 answer:

german3 years ago

8 0

Answer:

(1) The sum of the lengths of the edges of the cube is 36.

A cube has 12 equal edges. Sum = 36. Length of each edge = 36/12 = 3

Volume = 3*3*3 = 27

(2) The surface area of the cube is 54.

A cube has 6 identical faces. Area of each face = s^2 (s is the length of the side)

6s^2 = 54

s = 3

Volume = 3*3*3 = 27

Step-by-step explanation:

All you need to uniquely define a cube is any one measurement - length of a side/edge, area of a surface, volume etc. If you have any one of them, you can uniquely determine the others. So each statement alone is sufficient here.

To show how,

(1) The sum of the lengths of the edges of the cube is 36.

A cube has 12 equal edges. Sum = 36. Length of each edge = 36/12 = 3

Volume = 3*3*3 = 27

(2) The surface area of the cube is 54.

A cube has 6 identical faces. Area of each face = s^2 (s is the length of the side)

6s^2 = 54

s = 3

Volume = 3*3*3 = 27

You might be interested in

I don’t know what I’m doing wrong. I need help understanding this problem please...

denpristay [2]

Answer:

Step-by-step explanation:

Cost (C) depends on the number of miles traveled (x).

So C is the dependent variable.

Or C(x) if they want you to be more specific.

x is your independent variable.

Miles driven do not depend on anything.

5 0

3 years ago

PLS HELP ME d + 7 2d-5 10d+ 1 Write and simplify an expression

tester [92]

answer: D = 1

Isolate the variable by dividing each side by factors that don't contain the variable. 7 (2d - 1) + 5 (2 - 3d) = 2d

8 0

2 years ago

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

melamori03 [73]

Answer:

$6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}$

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs $AB=3$ , $BC=4$ , and $AC=5$ . The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base $b$ and height $h$ is given by $A=\frac{1}{2}bh$ . Therefore, the area of this right triangle is:

$A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}$

The other triangle is a bit trickier. Triangle $\triangle ADC$ is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

$A=\sqrt{s(s-a)(s-b)(s-c)}$ , where $a$ , $b$ , and $c$ are three sides of the triangle and $s$ is the semi-perimeter ( $s=\frac{a+b+c}{2}$ ).

The semi-perimeter, $s$ , is:

$s=\frac{5+5+4}{2}=\frac{14}{2}=7$

Therefore, the area of the isosceles triangle is:

$A=\sqrt{7(7-5)(7-5)(7-4)},\\A=\sqrt{7\cdot 2\cdot 2\cdot 3},\\A=\sqrt{84}, \\A=2\sqrt{21}\:\mathrm{cm^2}$

Thus, the area of the quadrilateral is:

$6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}$

4 0

3 years ago

An instructor who taught two sections of Math 161A, the first with 20 students and the second with 30 students, gave a midterm e

uranmaximum [27]

Answer:

<em>The answers are for option (a) 0.2070 (b)0.3798 (c) 0.3938 </em>

Step-by-step explanation:

<em>Given:</em>

<em>Here Section 1 students = 20 </em>

<em> Section 2 students = 30 </em>

<em> Here there are 15 graded exam papers. </em>

<em> (a )Here Pr(10 are from second section) = ²⁰C₅ * ³⁰C₁₀/⁵⁰C₁₅= 0.2070 </em>

<em> (b) Here if x is the number of students copies of section 2 out of 15 exam papers. </em>

<em> here the distribution is hyper-geometric one, where N = 50, K = 30 ; n = 15 </em>

<em>Then, </em>

<em> Pr( x ≥ 10 ; 15; 30 ; 50) = 0.3798 </em>

<em> (c) Here we have to find that at least 10 are from the same section that means if x ≥ 10 (at least 10 from section B) or x ≤ 5 (at least 10 from section 1) </em>

<em> so, </em>

<em> Pr(at least 10 of these are from the same section) = Pr(x ≤ 5 or x ≥ 10 ; 15 ; 30 ; 50) = Pr(x ≤ 5 ; 15 ; 30 ; 50) + Pr(x ≥ 10 ; 15 ; 30 ; 50) = 0.0140 + 0.3798 = 0.3938 </em>

<em> Note : Here the given distribution is Hyper-geometric distribution </em>

<em> where f(x) = kCₓ)(N-K)C(n-x)/ NCK in that way all these above values can be calculated.</em>

7 0

3 years ago

What is the area of the rectangle? shoe your work! do not round anything until the very end and round your final answer to the t

weeeeeb [17]

Answer:

The area of the rectangle is 42 units^2

Step-by-step explanation:

we know that

The area of rectangle is equal to

$A=LW$

In this problem

AB=DC

BC=AD

see the attached figure with letters to better understand the problem

we have that

$L=BC\\W=AB$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have the points

$A(2,-1),B(5,2),C(12,-5),D(9,-8)$

<em>Find out the distance BC</em>

we have

$B(5,2),C(12,-5)$

substitute in the formula

$d=\sqrt{(-5-2)^{2}+(12-5)^{2}}$

$d=\sqrt{(-7)^{2}+(7)^{2}}$

$d_B_C=\sqrt{98}\ units$

<em>Find out the distance AB</em>

we have

$A(2,-1),B(5,2)$

substitute in the formula

$d=\sqrt{(2+1)^{2}+(5-2)^{2}}$

$d=\sqrt{(3)^{2}+(3)^{2}}$

$d_A_B=\sqrt{18}\ units$

<em>Find out the area</em>

$A=(L)(W)$

we have

$L=d_B_C=\sqrt{98}\ units$

$W=d_A_B=\sqrt{18}\ units$

substitute

$A=(\sqrt{98})(\sqrt{18})=42.0\ units^2$

4 0

3 years ago