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3 years ago

13

Cable hangs between two poles of equal height and 37 feet apart. at a point on the ground directly under the cable and x feet fr

om the point on the ground halfway between the poles the height of the cable in feet is

1 answer:

Leya [2.2K]3 years ago

5 0

Weight= (38.55 ft)(11.7 lb/ft) = 451 pounds

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Can anyone help find the area pls ty

AlexFokin [52]

Answer:

See below ~

Step-by-step explanation:

<u>Question #8</u>

Area = 10 x 11.7
Area = <u>117 m²</u>

<u></u>

<u>Question #10</u>

Area = 11 x 11
Area = <u>121 m²</u>

8 0

2 years ago

Determine the domain of the function (fog) (x) where f(x) = 3x-1/x-4 and g(x)=x+1/x

Mrac [35]

Answer:

$(-\infty,0)\cup(0, \frac{1}{3})\cup (\frac{1}{3},\infty)$

Step-by-step explanation:

The given functions are $f(x)=\frac{3x-1}{x-4}$ and $g(x)=\frac{x+1}{x}$ .

We now composed the two functions to find:

$(f\circ g)(x)=f(g(x))$

$\implies (f\circ g)(x)=f(\frac{x+1}{x})$

$\implies (f\circ g)(x)=\frac{3(\frac{x+1}{x})+1}{\frac{x+1}{x}-4}$

$\implies (f\circ g)(x)=\frac{4x+3}{1-3x}$

This function is defined if the denominator is not zero.

$1-3x\ne0$

$x\ne\frac{1}{3}$

We write this in interval notation as:

$(-\infty,\frac{1}{3})\cup (\frac{1}{3},\infty)$

We need to be cautious here as x=0 is not in the domain of g(x).

Therefore the domain of $(f\circ g)(x)$ is

$(-\infty,0)\cup(0, \frac{1}{3})\cup (\frac{1}{3},\infty)$

4 0

3 years ago

Ed's new car has a value of $23,500, but it is expected to depreciate at a rate of 7.5% per year. What is the value of Ed's car

Kazeer [188]

Answer:

$13,616

Step-by-step explanation:

For each year we will have to constantly decrease it by 7.5%

23,500 - 7.5%

= 21,737.5

= 21,737.5 - 7.5%

= 20,107.1875

= 20,107.1875 - 7.5%

= 18599.1484375

= 18599.1484375 - 7.5%

= 17204.2123047

= 17204.2123047 - 7.5%

= 15913.8963818

= 15913.8963818 - 7.5%

= 14720.3541532

= 14720.3541532 - 7.5%

= 13616.3275917

Definitively we will get the answer $13,616

8 0

3 years ago

Slav-nsk [51]

The answer is a, because the absolute value of x(the actual weight of the product) minus the target weight of 18.25 needs to be greater that the target weight difference of .36 so that it is outside the range. So basically, any number outside of the range will be correct.You can plug in number that you know are in/ out of the acceptable range of numbers for this problem in.
|18.62-18.25|>.36
.37>.36
18.62 is not in range.

7 0

3 years ago

Read 2 more answers

Using (x + a) (x + b) = x² + (a + b) x + ab, evaluate 75 x 77

Gre4nikov [31]

(x + a) (x + b) = x² + (a + b) x + ab

75 * 77=

=(70+5)(70+7)

=70²+(5+7)*70+5*7

=4900+12*70+35

=4900+840+35

=5775

4 0

3 years ago