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suter [353]
3 years ago
12

If k(x) = 5x - 6, which expression is equivalent to (k+ k)(4)?

Mathematics
1 answer:
erik [133]3 years ago
8 0

Answer:

3h33j333jj3

Step-by-step explanation:b3n3n3nn3n33

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Find the volume of a right circular cone that has a height of 17.8 ft and a base with a radius of 17.8 ft. Round your answer to
Alona [7]

Answer:

5905.9 ft  

Step-by-step explanation:

5 0
3 years ago
What is 0.0592 in scientific notation
DiKsa [7]

Answer:

5.92 times 10 to the 2nd power.

Step-by-step explanation:

So funny I just finished my unit on scientific notation! So what you do is you count the distance from the first non-zero number (in this case, it would be after 5), and count how many places there are until the decimal point. I hope this helped! :D

7 0
3 years ago
Read 2 more answers
100 POINT QUESTION!!!!!!!!!!!!!!!!!!!!!!!!!!!!
babunello [35]

Answer:

Ok I don't know waht you mean but the answer is

Step-by-step explanation:

Answer:

x=3/2±(√11)2

x=1.5+1.65831i

x=1.5−1.65831i

Find the Solution for

2x^22−6x+10=0

using the Quadratic Formula where

a = 2, b = -6, and c = 10

x=(−b±√(b^2−4ac))/2a

x=(−(−6)±√((−6)2−4(2)(10)))2(2)

x=(6±√(36−80))/4

x=(6±√−44)/4

The discriminant b^2−4ac<0

so, there are two complex roots.

Simplify the Radical:

x=(6±2√11 i)/4

x=6/4±(2√11 i)4

Simplify fractions and/or signs:

x=3/2±(√11)2

which becomes

x=1.5+1.65831i

x=1.5−1.65831i

4 0
2 years ago
Use what you know about tangent lines to circles to solve for x
tigry1 [53]
X=5
Since both lines are tanget meaning the only touch the circle once, both sides will be the same we can set up an equation 33=x^2+8 which is 25=x^2 and square root both sides to get 5 equals x
3 0
3 years ago
I really need help, please dont put something silly.
Brut [27]

Answer:

1) 2.92405063291 (put it in the correct amount of significant figures)

2) 6.4688 (put it in the correct amount of significant figures)

Step-by-step explanation:

i dont know how many significant figures to put it into but theres the full answers

7 0
3 years ago
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