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Charra [1.4K]
4 years ago
6

A rectangular lawn is 48 feet long and 61 feet wide. A path extends diagonally between 2 of the corners of the lawn, what is the

length of the path to the nearest 10th of a foot?

Mathematics
1 answer:
lana66690 [7]4 years ago
5 0
This is a question that will require Pythagoras' theorem as you want to find the length of a diagonal of a triangle and are given the measurements of two sides. I have attached a picture of how to solve this problem.
To the nearest 10th of a foot, the answer is 77.6ft

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Solve each system of equations using matrices.<br> -2x = -6y + 8<br> 2y = x + 1
Elden [556K]

Answer:    1.) x - 2y = 4 , add 2y to both sides. ---> answer is, X = 2y + 4

2.) 2x - 6y = 8, Add 6y to both sides, then divide both sides by 2. --> answer is, X = 3y + 4

6 0
3 years ago
Max and Sharifa are both saving to buy the same model of DVD player, which costs $99, including tax. Max already has $31.00 and
gogolik [260]

Answer:

Sharifa can pay for the DVD player first.

Step-by-step explanation:

<u>Max:</u>

99 - 31 = $68 remaining

68 / 8.5 = <em>8 weeks</em>

<u>Sharifa:</u>

99 - 25.5 = $73.50 remaining

73.5 / 10.5 = <em>7 weeks</em>

7 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
Ppppppppeeeelelelelele help
kirill115 [55]
11/3*6/11=66/33=2
6 24/20-2 15/20=2 9/20
5 0
3 years ago
Please help hurry!!!!! This is due in a couple of minutes!!!!
kumpel [21]

Answer:

200 in.

Step-by-step explanation:

1250/75=5000/300=50/3 simplification

50*12=600 conversion to inches

600/3=200 simplification

5 0
4 years ago
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