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3 years ago

The edge length of a cube is 17 cm. Identify the length of its diagonal. Round to the nearest tenth, if necessary. HELP PLEASE!!

Mathematics

1 answer:

lara31 [8.8K]3 years ago

3 0

Answer:

29.4 cm

Step-by-step explanation:

The length of the space diagonal can be found to be the root of the squares of the three orthogonal edge lengths. For a cube, those edge lengths are all the same, so the diagonal length is ...

d = √(17^2 + 17^2 +17^2) = 17√3 ≈ 29.4 . . . . cm

_____

Consider a rectangular prism with edge lengths a, b, c. Then the face diagonal of the face perpendicular to edge "a" has length ...

(face diagonal)^2 = (b^2 +c^2)

and the space diagonal has length ...

(space diagonal)^2 = a^2 + (face diagonal)^2 = a^2 +b^2 +c^2

So, the length of the space diagonal is ...

space diagonal = √(a^2 +b^2 +c^2)

when the prism is a cube, these are all the same (a=b=c). This is the formula we used above.

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fomenos

Answer:

Therefore, the solutions of the quadratic equations are:

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The graph is also attached.

Step-by-step explanation:

The solution of the graph could be obtained by finding the x-intercept.

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Finding the x-intercept by substituting the value y = 0

$y=\left(x-\:4\right)^2-1$

$\:0\:=\:\left(x\:-\:4\right)^2\:-\:1$ ∵ y = 0

$\left(x-4\right)^2-1=0$

$\left(x-4\right)^2-1+1=0+1$

$\left(x-4\right)^2=1$

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$\mathrm{Solve\:}\:x-4=\sqrt{1}$

$\mathrm{Apply\:rule}\:\sqrt{1}=1$

$x-4=1$

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$\mathrm{Solve\:}\:x-4=-\sqrt{1}$

$\mathrm{Apply\:rule}\:\sqrt{1}=1$

$x-4=-1$

$x=3$

So, when y = 0, then x values are 3, and 5.

Therefore, the solutions of the quadratic equations are:

$x=5,\:x=3$

The graph is also attached. As the graph is a Parabola. It is visible from the graph that the values of y = 0 at x = 5 and x = 3. As the graph is a Parabola.

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