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3 years ago

Find the length of the unknown leg.

Mathematics

1 answer:

Yuri [45]3 years ago

3 0

Answer:

The length of leg <em>b</em> is 21

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A large rectangular area is to be fenced off ( a large rectangle divided into 2 smaller rectangles). The fence used to divide th

Anvisha [2.4K]

Answer:

Length of area is 857.14 ft

Width of area is 857.14 ft

Length of dividing fence is 857.14 ft

Step-by-step explanation:

Here we have

Area of rectangle = Length × Width

The dimension, of the rectangle with the largest area is the dimension of a square, hence we have;

Length of rectangle = Width of rectangle = x

Hence, the perimeter of the area = 4·x, while the width of the dividing fence = x

Therefore, since the we have;

Cost of the perimeter fence = $15/foot

Cost of the dividing fence = $10/foot

Then;

4·x × 15 + x × 10 = 60000

60·x + 10·x = 60000

x = 60000/70 = 857.14 ft

Which gives the following dimensions;

Length of area = 857.14 ft

Width of area = 857.14 ft

Length of dividing fence = 857.14 ft.

3 0

3 years ago

What is the standard notation of 2.934x10^8

bogdanovich [222]

Answer:

293,400,000

Step-by-step explanation:

Just move the decimal 8 places to the right --->

293,400,000 is = 2.934 x 10^8

7 0

4 years ago

What are the coordinates of the x- intercept for the graph shown below?

3241004551 [841]

C the horizontal line represents x and it goes to -1 you place your coordinates as (x,y)

6 0

3 years ago

Find the area of the figure A) 14 cm B) 50 cm C) 60 cm D) 64 cm

aleksandrvk [35]

Answer:

The answer would be 64 cm.

Step-by-step explanation:

Step 1: Identify which shapes you will do to find the area
(I did 2 and 12 and 8 and 5)
Step 2: Do 12x2 (24)
Step 3: Do 8x5 (40)
Step 4: Add 40+24 (64)
Step 5: Label
I hope this helps!

3 0

3 years ago

The U.S. Fish and Wildlife Service reported that the mean length of six-year-old rainbow trout in the Arolik River in Alaska is

Anettt [7]

Answer:

The size limit for 15% of of six-year-old trout is 438.52 mm.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 481 millimeters

Standard Deviation, σ = 41 millimeters

We are given that the distribution of mean length of six-year-old rainbow trout is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.15

$P( X < x) = P( z < \displaystyle\frac{x - 481}{41})=0.15$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 481}{41} = -1.036\\\\x = 438.524$

Thus, the size limit for 15% of of six-year-old trout is 438.52 mm.

5 0

3 years ago