All categories

3 years ago

BRAINLIESTTT ASAP! PLEASE HELP ME :)

Mathematics

1 answer:

Tasya [4]3 years ago

8 0

$\displaystyle y = Atan(Bx - C) + D$

According to this trigonometric function, −C gives you the OPPOSITE terms of what they really are, so be EXTREMELY CAREFUL:

$\displaystyle Phase\:[Horisontal]\:Shift → \frac{-\frac{π}{6}}{1} = -\frac{π}{6} \\ Period → \frac{π}{1} = π$

Therefore we have our answer.

Extended Information on the trigonometric function

$\displaystyle Vertical\:Shift → D \\ Phase\:[Horisontal]\:Shift → \frac{C}{B} \\ Period → \frac{π}{B} \\ Amplitude → |A|$

NOTE: Sometimes, your vertical shift might tell you to extend the troughs on each end of your graphs, beyond the midline.

* All tangent functions have NO AMPLITUDE.

I am joyous to assist you anytime.

You might be interested in

Surface area for a net

dolphi86 [110]

Finding the areas of each of the rectangles and squares of the net of a rectangular prism and adding up those areas gives the surface area or total surface area of the prism. For example, if the length of one side of the cube 4 units then the area of one its face is 4 × 4 = 16 square units.

8 0

3 years ago

Find the area of a rectangle whose perimeter is 60in and width is 10in.

Dafna1 [17]

Answer:

250 sq. inch

Step-by-step explanation:

Perimeter: 60 inch

: 2 X L+B

: 2 X ? + 10

Length: 60-10

: 50/2

: 25 inch

Area: L X B

: 25 X 10

: 250 sq.inch

Hope it helps ;)

Please tell if it's correct

Please mark as brainliest!

3 0

2 years ago

A rancher wishes to build a fence to enclose a 2250 square yard rectangular field. Along one side the fence is to be made of hea

Bess [88]

Answer:

The least cost of fencing for the rancher is $1200

Step-by-step explanation:

Let x be the width and y the length of the rectangular field.

Let C the total cost of the rectangular field.

The side made of heavy duty material of length of x costs 16 dollars a yard. The three sides not made of heavy duty material cost $4 per yard, their side lengths are x, y, y. Thus

$C=4x+4y+4y+16x\\C=20x+8y$

We know that the total area of rectangular field should be 2250 square yards,

$x\cdot y=2250$

We can say that $y=\frac{2250}{x}$

Substituting into the total cost of the rectangular field, we get

$C=20x+8(\frac{2250}{x})\\\\C=20x+\frac{18000}{x}$

We have to figure out where the function is increasing and decreasing. Differentiating,

$\frac{d}{dx}C=\frac{d}{dx}\left(20x+\frac{18000}{x}\right)\\\\C'=20-\frac{18000}{x^2}$

Next, we find the critical points of the derivative

$20-\frac{18000}{x^2}=0\\\\20x^2-\frac{18000}{x^2}x^2=0\cdot \:x^2\\\\20x^2-18000=0\\\\20x^2-18000+18000=0+18000\\\\20x^2=18000\\\\\frac{20x^2}{20}=\frac{18000}{20}\\\\x^2=900\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{900},\:x=-\sqrt{900}\\\\x=30,\:x=-30$

Because the length is always positive the only point we take is $x=30$ . We thus test the intervals $(0, 30)$ and $(30, \infty)$

$C'(20)=20-\frac{18000}{20^2} = -25 < 0\\\\C'(40)= 20-\frac{18000}{20^2} = 8.75 >0$

we see that total cost function is decreasing on $(0, 30)$ and increasing on $(30, \infty)$ . Therefore, the minimum is attained at $x=30$ , so the minimal cost is

$C(30)=20(30)+\frac{18000}{30}\\C(30)=1200$