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SOVA2 [1]
3 years ago
5

Veveref7fois7dgfl eeufhwieguwegfwgwegf

Mathematics
1 answer:
Alchen [17]3 years ago
6 0
The answer to your question is obviously bdjanabjmamajdhf jsjsju hshs2+2is4-thats3quickmafsna. duh....
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What algebraic property is demonstrated in the equation below?<br> 5(8x-4)-10=40x-20-10
grandymaker [24]
Distributive Property
3 0
3 years ago
During a 4​-year ​period, ​65% of snowstorms in a certain city caused power outages. 14 of the snowstorms did not cause power ou
Colt1911 [192]

Answer:

40

Step-by-step explanation:

S x .35 = 14

S = 14/.35

S = 40

Verify:

40 x 35% = 14

3 0
3 years ago
Solve for the inverse g(x) = 2x^2-6/9​
oksian1 [2.3K]

Answer:f(x)=x2−6x−9 f ( x ) = x 2 - 6 x - 9. Replace f(x) f ( x ) with y y . y=x2−6x−9 y = x 2 - 6 x - 9. Interchange the variables. x=y2−6y−9 x = y 2 - 6 y - 9.

Step-by-step explanation:

4 0
3 years ago
Please help im so confused!!??
victus00 [196]
The formula for the surface area of a cone is SA=(pi)(r^2)+(pi)(r)(l)

L is the slant height of the cone.
Because pi is in the answer choices, there is no need to multiply either by pi so the new formula becomes
SA=r^2+(r)(l)
Now you just plug in what you know
Sa=(8^2) + (8)(15)
Sa=64+120
Sa=184(pi)

*dont forget to put the pi back in since you took it out in the beginning of the equation*

Your answer is A.

Hope this helps :)
8 0
3 years ago
Consider the following. Cube roots of −343 (a) Use the formula zk = n r cos θ + 2πk n + i sin θ + 2πk n to find the indicated ro
Lapatulllka [165]

Answer:

Z0 = 7 ( cos 60° + isin60°)

Z1 = 7( cos180° + isin180° )

Z2 = 7 ( cos300° + isin300°)

Step-by-step explanation:

Given that;

cube root of  -343

so, z = -343 + 0i

∴ r = √ (( -343)² + (0)²)  = 343

so tan∅ = y/x ⇒ tan∅ = 0/-343

∅ = tan⁻¹ (0/-343)

= 0 or 180°

but we are going yo make use of 180° since -343 is negative x-axis

Zk = ∛343 ( cos 180/3 + 360K/3) + isin(180/3 + 360k/3)

here k = 0, 1, 2, 3 .........

SO z0 = z1 = z2 = ???

k=0

Z0 =  ∛343 ( cos 180/3 + isin180/3)

=  ∛343 ( cos 60° + isin60°)

Z0 = 7 ( cos 60° + isin60°)

K=1

Z1 = ∛343 ( cos 180/3 + 360×1 / 3 + isin180/3 + 360×1 / 3 )

= ∛343 ( cos180° + isin180°)

Z1 = 7( cos180° + isin180° )

K=2

Z2 = ∛343 ( cos 180/3 + 360×2 / 3 + isin180/3 + 360×2 / 3 )

= ∛343 ( cos300° + isiN300°)

Z2 = 7 ( cos300° + isin300°)

6 0
3 years ago
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