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3 years ago

What is the length of the 9 centimeter ribbon in millimeters?

Mathematics

2 answers:

Dahasolnce [82]3 years ago

8 0

Answer:

90 millimeters

Step-by-step explanation:

multiply the length value by 10

miskamm [114]3 years ago

4 0

Answer:

i got 90 Millimeters

Step-by-step explanation:

The source that i used was https://mayarts.com/size-chart/

Hope this helped have a great day!

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Based on the piecewise graph, which is the domain?

joja [24]

The answer is 0 < x < infinity

5 0

2 years ago

Find area of the composite figure..

saveliy_v [14]

Answer:

627square inches

Step-by-step explanation:

Area of the composite figure.. = area of triangle + area of parallelogram

Since the triangle is an equilateral triangle;

Area of the triangle = r²sintheta

Area of the triangle = 18²sin60

Area of the triangle = 324sin60

Area of the triangle = 324(0.8860)

Area of the triangle = 280.584 square in

Area of the triangle = 281 square inches

Area of parallelogram = absintheta

Area of parallelogram = 20(20)sin60

Area of parallelogram = 400(0.8660)

Area of parallelogram = 346.4square inches

Area of parallelogram = 346 square inches

Area of the figure = 346 + 281

Area of the figure = 627square inches

3 0

2 years ago

Help me please, thank you!

elena-s [515]

A is 3 square root 2
Since 3*6 is 18
Find the factors of 18
So we have 9 and 2
9 is a perfect square so it’s 3 with 2 inside radical

7 0

2 years ago

Read 2 more answers

The mean radius of Main Street Park is 15 km and the mean radius of Uptown Park is 8 km. What is the approximate difference in t

Temka [501]

Step-by-step explanation:

circumference= 3.142*diameter

Diameter of main Street park is 30

Diameter of uptown park is 16

3.142*30=94.26

3.142*16=30.272

94.26-30.272=43.988

round off to tenth =44.0km

5 0

2 years ago

A box designer has been charged with the task of determining the surface area of various open boxes (no lid) that can be constru

Viktor [21]

Answer:

1) $S = 2\cdot w\cdot l - 8\cdot x^{2}$ , 2) The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$ , 3) $S = 176\,in^{2}$ , 4) $x \approx 4.528\,in$ , 5) $S = 164.830\,in^{2}$

Step-by-step explanation:

1) The function of the box is:

$S = 2\cdot (w - 2\cdot x)\cdot x + 2\cdot (l-2\cdot x)\cdot x +(w-2\cdot x)\cdot (l-2\cdot x)$

$S = 2\cdot w\cdot x - 4\cdot x^{2} + 2\cdot l\cdot x - 4\cdot x^{2} + w\cdot l -2\cdot (l + w)\cdot x + l\cdot w$

$S = 2\cdot (w+l)\cdot x - 8\cdpt x^{2} + 2\cdot w \cdot l - 2\cdot (l+w)\cdot x$

$S = 2\cdot w\cdot l - 8\cdot x^{2}$

2) The maximum cutout is:

$2\cdot w \cdot l - 8\cdot x^{2} = 0$

$w\cdot l - 4\cdot x^{2} = 0$

$4\cdot x^{2} = w\cdot l$

$x = \frac{\sqrt{w\cdot l}}{2}$

The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$

3) The surface area when a 1'' x 1'' square is cut out is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1\,in)^{2}$

$S = 176\,in^{2}$

4) The size is found by solving the following second-order polynomial:

$20\,in^{2} = 2 \cdot (8\,in)\cdot (11.5\,in)-8\cdot x^{2}$

$20\,in^{2} = 184\,in^{2} - 8\cdot x^{2}$

$8\cdot x^{2} - 164\,in^{2} = 0$

$x \approx 4.528\,in$

5) The equation of the box volume is:

$V = (w-2\cdot x)\cdot (l-2\cdot x) \cdot x$

$V = [w\cdot l -2\cdot (w+l)\cdot x + 4\cdot x^{2}]\cdot x$

$V = w\cdot l \cdot x - 2\cdot (w+l)\cdot x^{2} + 4\cdot x^{3}$

$V = (8\,in)\cdot (11.5\,in)\cdot x - 2\cdot (19.5\,in)\cdot x^{2} + 4\cdot x^{3}$

$V = (92\,in^{2})\cdot x - (39\,in)\cdot x^{2} + 4\cdot x^{3}$

The first derivative of the function is:

$V' = 92\,in^{2} - (78\,in)\cdot x + 12\cdot x^{2}$

The critical points are determined by equalizing the derivative to zero:

$12\cdot x^{2}-(78\,in)\cdot x + 92\,in^{2} = 0$

$x_{1} \approx 4.952\,in$

$x_{2}\approx 1.548\,in$

The second derivative is found afterwards:

$V'' = 24\cdot x - 78\,in$

After evaluating each critical point, it follows that $x_{1}$ is an absolute minimum and $x_{2}$ is an absolute maximum. Hence, the value of the cutoff so that volume is maximized is:

$x \approx 1.548\,in$

The surface area of the box is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1.548\,in)^{2}$

$S = 164.830\,in^{2}$

4 0

2 years ago