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3 years ago

Determine how to translate triangle ABC to triangle A'B'C'

Mathematics

1 answer:

vichka [17]3 years ago

8 0

Do you still need the answer to this or nah?

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Solving a right triangle means using given information to find all the angles and side lengths on

nikklg [1K]

All triangles add up to 180 degrees so I don’t get the just of the question but there you go.

4 0

3 years ago

All boxes with a square base, an open top, and a volume of 60 ft cubed have a surface area given by S(x)equalsx squared plus

Karo-lina-s [1.5K]

Answer:

The absolute minimum of the surface area function on the interval $(0,\infty)$ is $S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2$

The dimensions of the box with minimum surface area are: the base edge $x=2\sqrt[3]{15}\:ft$ and the height $h=\sqrt[3]{15} \:ft$

Step-by-step explanation:

We are given the surface area of a box $S(x)=x^2+\frac{240}{x}$ where x is the length of the sides of the base.

Our goal is to find the absolute minimum of the the surface area function on the interval $(0,\infty)$ and the dimensions of the box with minimum surface area.

1. To find the absolute minimum you must find the derivative of the surface area ( $S'(x)$ ) and find the critical points of the derivative ( $S'(x)=0$ ).

$\frac{d}{dx} S(x)=\frac{d}{dx}(x^2+\frac{240}{x})\\\\\frac{d}{dx} S(x)=\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(\frac{240}{x}\right)\\\\S'(x)=2x-\frac{240}{x^2}$

Next,

$2x-\frac{240}{x^2}=0\\2xx^2-\frac{240}{x^2}x^2=0\cdot \:x^2\\2x^3-240=0\\x^3=120$

There is a undefined solution $x=0$ and a real solution $x=2\sqrt[3]{15}$ . These point divide the number line into two intervals $(0,2\sqrt[3]{15})$ and $(2\sqrt[3]{15}, \infty)$

Evaluate S'(x) at each interval to see if it's positive or negative on that interval.

$\begin{array}{cccc}Interval&x-value&S'(x)&Verdict\\(0,2\sqrt[3]{15}) &2&-56&decreasing\\(2\sqrt[3]{15}, \infty)&6&\frac{16}{3}&increasing \end{array}$

An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We can see from the table that f(x) decreases before $x=2\sqrt[3]{15}$ , increases after it, and is defined at $x=2\sqrt[3]{15}$ . So f(x) has a relative minimum point at $x=2\sqrt[3]{15}$ .

To confirm that this is the point of an absolute minimum we need to find the second derivative of the surface area and show that is positive for $x=2\sqrt[3]{15}$ .

$\frac{d}{dx} S'(x)=\frac{d}{dx}(2x-\frac{240}{x^2})\\\\S''(x) =\frac{d}{dx}\left(2x\right)-\frac{d}{dx}\left(\frac{240}{x^2}\right)\\\\S''(x) =2+\frac{480}{x^3}$

and for $x=2\sqrt[3]{15}$ we get:

$2+\frac{480}{\left(2\sqrt[3]{15}\right)^3}\\\\\frac{480}{\left(2\sqrt[3]{15}\right)^3}=2^2\\\\2+4=6>0$

Therefore S(x) has a minimum at $x=2\sqrt[3]{15}$ which is:

$S(2\sqrt[3]{15})=(2\sqrt[3]{15})^2+\frac{240}{2\sqrt[3]{15}} \\\\2^2\cdot \:15^{\frac{2}{3}}+2^3\cdot \:15^{\frac{2}{3}}\\\\4\cdot \:15^{\frac{2}{3}}+8\cdot \:15^{\frac{2}{3}}\\\\S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2$

2. To find the third dimension of the box with minimum surface area:

We know that the volume is 60 $ft^3$ and the volume of a box with a square base is $V=x^2h$ , we solve for h

$h=\frac{V}{x^2}$

Substituting V = 60 $ft^3$ and $x=2\sqrt[3]{15}$

$h=\frac{60}{(2\sqrt[3]{15})^2}\\\\h=\frac{60}{2^2\cdot \:15^{\frac{2}{3}}}\\\\h=\sqrt[3]{15} \:ft$

The dimension are the base edge $x=2\sqrt[3]{15}\:ft$ and the height $h=\sqrt[3]{15} \:ft$

6 0

3 years ago

Jillian deposit 3/4 of his paycheck into the bank the deposit slip shows he deposit $46.50 how much was the amount of his payche

Lynna [10]

Do 46.50 divided by 3. That gives you 15.5. Now do 15.5 multiplied by 4 because it says he deposited 3/4 of his paycheck.
The answer is 62.
(Sorry if there were any mistakes or if i read the question wrong) <span />

4 0

3 years ago

HELP

Yakvenalex [24]

Answer:

5. 50.3 m²

6. 113.1 cm²

7. 102.1 yards²

8. 50.3 yards²

9. 452.2 ft²

10.379.8 ft²

Step-by-step explanation:

Area of a circle = nr²

The circumference of a circle = 2nr

n = pi = 22/7

r = radius

the diameter is the straight line that passes through the centre of a circle and touches the two edges of the circle.

A radius is half of the diameter

radius = diameter / 2

5. (22/7) x 4² = 50.3 m²

6. (22/7) x 6² =113.1 cm²

7. radius = 11.4/2 = 5.7 yards

(22/7) x 5.7² =102.1 yards²

8. radius = 8/2 = 4 yards

(22/7) x 4² = 50.3 yards²

9. 2nr = 75.4 yards

2 x (22/7) x r = 75.4

to find r, divide both sides of the equation by 2 and 7/22

r = 12 ft

Area = (22/7) x 12² = 452.2 ft²

10. 2nr = 69.1 yards

2 x (22/7) x r = 69.1

to find r, divide both sides of the equation by 2 and 7/22

r = 10.99 ft²

Area = (22/7) x 10.99² = 379.8 ft²

the tenth is the first number after the decimal place. To convert to the nearest tenth, look at the number after the tenth (the hundredth). If the number is greater or equal to 5, add 1 to the tenth figure. If this is not the case, add zero

8 0

3 years ago

Verify the identity

Igoryamba

Answer:

We have to prove

sin⁡(α+β)-sin⁡(α-β)=2 cos⁡ α sin ⁡β

We will take the left hand side to prove it equal to right hand side

So,

=sin⁡(α+β)-sin⁡(α-β) Eqn 1

We will use the following identities:

sin⁡(α+β)=sin⁡ α cos⁡ β+cos⁡ α sin⁡ β

and

sin⁡(α-β)=sin⁡ α cos ⁡β-cos ⁡α sin ⁡β

Putting the identities in eqn 1

=sin⁡(α+β)-sin⁡(α-β)

=[ sin⁡ α cos ⁡β+cos⁡ α sin⁡ β ]-[sin⁡ α cos ⁡β-cos ⁡α sin ⁡β ]

=sin⁡ α cos⁡ β+cos⁡ α sin ⁡β- sin⁡α cos⁡ β+cos ⁡α sin ⁡β

sin⁡α cos⁡β will be cancelled.

=cos⁡ α sin ⁡β+ cos ⁡α sin ⁡β

=2 cos⁡ α sin ⁡β

Hence,

sin⁡(α+β)-sin⁡(α-β)=2 cos ⁡α sin ⁡β

8 0

3 years ago