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devlian [24]
3 years ago
9

Please helpp will mark brainlyist

Mathematics
1 answer:
chubhunter [2.5K]3 years ago
6 0
I’m pretty sure it’s (3.63)


3.631923997916374
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The surface area of a cube is represented by the expression below, where s is the side length of the cube. Find the surface area
Dmitry [639]

Answer:

486 centimeters³

Step-by-step explanation:

Surface area of a cube = 6s²

Where,

s = side length

Find the surface area of a cube-shaped jewelry gift box with a side length of 9 centimeters.

s = 9 centimeters

Surface area of a cube = 6s²

= 6 × 9²

= 6 × 81

= 486 centimeters³

Surface area of a cube-shaped jewelry gift box with a side length of 9 centimeters 486 centimeters³

8 0
2 years ago
HELP ITS FOR 25 POINTS, i really needa finish
SSSSS [86.1K]

Answer:

Which gas creates the longest wavelength?

→Argon gas

Which gas creates the shortest wavelength?

→ Fluorine gas

Explaination:

1.0×10^-6 > 1.0×10^-7 > 1.0×10^-10 (a^-n =1/a^n)

8 0
2 years ago
100*.005<br><img src="https://tex.z-dn.net/?f=100%20%5Ctimes%20.005" id="TexFormula1" title="100 \times .005" alt="100 \times .0
andrew11 [14]
Answer is .5
move decimal place 2 places right
7 0
3 years ago
Verify the basic identity. What is the domain of validity? cot theta = cos theta csc theta
Kay [80]
Cot x = cos x csc x
1/tan x = cosx (1/sin x)
1/(sin x / cos x) = cos x / sin x
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The domain of validity is all real number values of x except for sin x = 0.
6 0
2 years ago
Read 2 more answers
Suppose a parabola has an axis of symmetry at x=-5, a maximum height of 9, and passes through the point (-7,1). Write the equati
Nutka1998 [239]

the parabola has maximum at 9, meaning is a vertical parabola and it opens downwards.

it has a symmetry at x = -5, namely its vertex's x-coordinate is -5.

check the picture below.

so then, we can pretty much tell its vertex is at (-5 , 9), and we also know it passes through (-7, 1)


\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} y=a(x- h)^2+ k\qquad \leftarrow \textit{using this one}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=-5\\ k=9 \end{cases}\implies y=a[x-(-5)]^2+9\implies y=a(x+5)^2+9


\bf \textit{we also know that } \begin{cases} x=-7\\ y=1 \end{cases}\implies 1=a(-7+5)^2+9 \\\\\\ -8=a(-2)^2\implies -8=4a\implies \cfrac{-8}{4}=a\implies -2=a \\\\[-0.35em] ~\dotfill\\\\ ~\hfill y=-2(x+5)^2+9~\hfill

7 0
3 years ago
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