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Furkat [3]
3 years ago
14

A tree is 7 ft tall when it is planted and this type of tree grows two feet per year. How tall will the tree be in 6 years? How

long will it take for the tree to reach 44ft?
Mathematics
1 answer:
s344n2d4d5 [400]3 years ago
4 0

2x+7=?

X=6

2(6)+7=

12+7=19

In six years, the tree will be 19ft

2x+7=44

Minus seven on both sides

2x=37

Divide two from both sides

X=18.5

Thus, it will take 18 and a half years for the tree to reach 44ft

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What is the x value such that y= 15-3x and y=0?
enot [183]

Answer:

5

Step-by-step explanation:

y = 15 - 3x

0 = 15 - 3x

15 - 0 = 3x

x = 5

7 0
3 years ago
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Isaiah is having his house painted. He paid $525 for the required materials, and he will pay $30 an hour for the painters to com
alukav5142 [94]

Answer:

Step-by-step explanation:

x is the amount of hours it will take. Since he pays $30 per hour, your answer would be 30x + 525

7 0
3 years ago
Given f(x) = x -2, if the domain is 5, find the range
Mrrafil [7]

Answer:

Step-by-step explanation:

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3 0
3 years ago
Sergio ate 3.5 cookies. Each cookie contained 5.7 grams of sugar. How many grams of sugar did Sergio eat?
FrozenT [24]
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He ate 19.95 grams of sugar.

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5 0
3 years ago
Consider the following. f(x) = x5 − x3 + 6, −1 ≤ x ≤ 1 (a) Use a graph to find the absolute maximum and minimum values of the fu
kykrilka [37]

ANSWER

See below

EXPLANATION

Part a)

The given function is

f(x) =  {x}^{5}  -  {x}^{3}  + 6

From the graph, we can observe that, the absolute maximum occurs at (-0.7746,6.1859) and the absolute minimum occurs at (0.7746,5.8141).

b) Using calculus, we find the first derivative of the given function.

f'(x) = 5 {x}^{4} - 3 {x}^{2}

At turning point f'(x)=0.

5 {x}^{4} - 3 {x}^{2}  = 0

This implies that,

{x}^{2} (5 {x}^{2}  - 3) = 0

{x}^{2}  = 0 \: or \: 5 {x}^{2}  - 3 = 0

x =  - \frac{ \sqrt{15} }{5}   \: or \: x = 0 \:  \: or \: x =\frac{ \sqrt{15} }{5}

We plug this values into the original function to obtain the y-values of the turning points

(   -  \frac{ \sqrt{15} }{5}  , \frac{1}{125} ( 6 \sqrt{15}  +750)) \:and \:  (0, - 6) \: and\: (   \frac{ \sqrt{15} }{5}  , \frac{1}{125} ( - 6 \sqrt{15}  +750))

We now use the second derivative test to determine the absolute maximum minimum on the interval [-1,1]

f''(x) = 20 {x}^{3}  - 6x

f''( -  \frac{ \sqrt{15} }{5} ) \:   <  \: 0

Hence

(   -  \frac{ \sqrt{15} }{5}  , \frac{1}{125} ( 6 \sqrt{15}  + 750))

is a maximum point.

f''( \frac{ \sqrt{15} }{5} ) \:    >  \: 0

Hence

(     \frac{ \sqrt{15} }{5}  , \frac{1}{125} (- 6 \sqrt{15}  + 750))

is a minimum point.

f''(0) \: =\: 0

Hence (0,-6) is a point of inflexion

4 0
3 years ago
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