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ZanzabumX [31]
4 years ago
13

Ax) = 4x - 12f'(x) =​

Mathematics
1 answer:
Virty [35]4 years ago
7 0

Answer:

x = 8

Step-by-step explanation:

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Write (3a)^3 without exponents
makkiz [27]

Answer: 27a^3

It is 27a^3 because if you do 3×3=9 then 9×3= 27

7 0
3 years ago
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I've been stuck on this for about 2 hours but it still doesn't make any sense to me, please helpppppppppp
Mariana [72]
I really hope this helps. I’m sure the answer is c:)
5 0
3 years ago
Write an equation in slope-intercept form for the line with slope -6 and y-intercept 5​
algol [13]

Answer:

y=-6+5

Step-by-step explanation:

5 0
3 years ago
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One day at football practice, Darrell, the kicker, punted the ball so that its height in feet above the ground was given by the
nika2105 [10]

When the football is 81 feet high, it will be √  3 / 8 seconds

<h3>Function</h3>

Function relates input to output. Therefore,

  • h(t)  = - 16t² + 75

Therefore,

t = number of seconds

h(t) = - 16t² + 75

h(t) = 81 ft

Therefore,

81 = - 16t² + 75

- 16t²  = 81 - 75

- 16t²  = 6

divide both sides by -16

t² = 6 / -16

t² = - 3 / 8

square root both sides

t = √  3 / 8 seconds

learn more on function here: brainly.com/question/4418459

6 0
2 years ago
the equation a equals 1 half b h, where b represents base and h represents height, is used to find the area of a .
NikAS [45]

The formula for b from the given area is equal to (2A / h).

As given in the question,

Formula to find area of a triangle is equal to A = (1/2) b h

b represents the base of the triangle

h represents the height of the triangle

A = (1/2) b h

Multiply both the side by 2 in area formula

2A = b h

Divide both the side by h

b = (2A) / h

Therefore, the formula for b from the given area is equal to (2A / h).

The complete question is:

The formula A = (1/2)b h represents the area of a triangle where A represents the area, b is the base of the triangle and h is the height of the triangle. 1. Solve this formula for b.

Learn more about area here

brainly.com/question/27683633

#SPJ4

6 0
2 years ago
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