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Brilliant_brown [7]
3 years ago
11

A telephone pole is tied to a stake with a

Mathematics
1 answer:
Harman [31]3 years ago
6 0
I think it will be d because i think that’s a good guess
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15y+3x-10=0 intercept form
storchak [24]
Hai! So, I have all the math and explanations in the picture attached.
I forgot to add on there(and you'll see it) wen i subtract 3x and add 10 to the other side it cancels each other out. -3 + 3= 0 and -10 + 10 = 0. Etc.  
Other than that, if you have any other questions, comment them down below. :) 

5 0
3 years ago
A rectangle has an area of 102 cm2. The length of the rectangle is 17 cm.
Aliun [14]
The area formula is length times width, and since the length is 17 cm in this case, 17 times width = 102, so we must divide 102 divided by 17, and you will get 6. The perimeter formula is 2 times the length plus 2 times the width, so
2(17) + 2(6), which gives us 46, so the perimiter is 46.
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3 years ago
Is 5x-7y=3 standard form?
barxatty [35]
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7 0
4 years ago
You can use a calculator
kolezko [41]

Answer:

238, 000

Step-by-step explanation:

hope this helps </3

3 0
3 years ago
Answer the following questions about the function whose derivative is f primef′​(x)equals=x Superscript negative three fifths Ba
Elden [556K]

Answer:

(a) The critical points of f are x=0 and x=3.

(b)f is decreasing on (0,3) and f is decreasing on (3,\infty).

(c) Therefore the local minimum of f is at x=3

Step-by-step explanation:

Given function is

f'(x)= x^{-\frac35}(x-3)

(a)

To find the critical point set f'(x)=0

\therefore  x^{-\frac35}(x-3)=0

\Rightarrow x=0,3

The critical points of f are 0,3.

(b)

The interval are (0,3) and (3,\infty).

To find the increasing or decreasing, taking two points one point from the interval (0,3) and another point (3,\infty).

Assume 1 and 4.

Now f'(1)=(1)^{-\frac35}(1-3)

and f'(4)=(4)^{-\frac35}(4-3)>0

Since 1∈(0,3) , f'(x)<0  and 4∈(3,\infty) , f'(x)>0

∴f is decreasing on (0,3) and f is decreasing on (3,\infty).

(c)

f'(x)= x^{-\frac35}(x-3)

Differentiating with respect to x

f''(x)=-\frac35x^{-\frac 85}(x-3)+x^{-\frac35}

Now

f''(0)=-\frac35(0)^{-\frac 85}(0-3)+(0)^{-\frac35}=0

and

f''(3)=-\frac35(3)^{-\frac 85}(3-3)+3^{-\frac35}

        =0.517>0

Since f''(x)>0 at x=3

Therefore the local minimum of f is at x=3

8 0
3 years ago
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