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4 years ago

Whats 2x4 because i dont really know that answer

Mathematics

2 answers:

labwork [276]4 years ago

6 0

8 is the answer ik u know lol

saw5 [17]4 years ago

3 0

Answer:8 .......................

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PLEASE ANSWER ASAP!!!!!!!!!!

IgorLugansk [536]

The equation of the sin function is y = 3sin(πx/2) + 1, and the graph of the function is shown in the picture.

<h3>What is sin function?</h3>

It is defined as a function that is sinusoidal in nature, and it has a domain of all real numbers and lies between the [a, a]where is the amplitude of the function.

As we know the sin function is represented by:

y = Asin(Bx) + C

Here,

A = The amplitude = 3

C = The vertical shift = 1

B - The period in terms of π

B = 2π/amplitude

B = 2π/4

B = π/2

The becomes:

$\rm y = 3sin(\dfrac{\pi }{2}x)+1$

Thus, the equation of the sin function is y = 3sin(πx/2) + 1, and the graph of the function is shown in the picture.

Learn more about the sin function here:

brainly.com/question/14397255

#SPJ1

3 0

2 years ago

Anderson has $50 in his savings account. He deposits $5 every week. His father also deposits $20 into the account every time And

Sav [38]

Part A: 5 is the coefficient, w is the variable, 50 is the constant

Part B: 10 x 5 = 50 + 2 x 20 = 40+ 50 = $140

Part C: the constant,50,would change and become 85. Hope this helps :)

8 0

3 years ago

Read 2 more answers

Will mark brainliest.

snow_tiger [21]

The ordered pair (- 3, 1 ) is a solution to the second equation because it makes the second equation true

To test , substitute the coordinates of (- 3, 1) into the equations and if both sides are true then it is a solution to the equation.

x - 4y = - 3 - 4 = - 7 ≠ 6 ← not true

(3 × - 3) + 1 = - 9 + 1 = - 8 ← true

Thus (- 3, 1 ) is a solution to the second equation

3 0

4 years ago

Whats the area of 5 and 16

icang [17]

Answer:

The area of a shape that has a length of 5 and a width of 16 is 80.

Step-by-step explanation:

Just multiply 5 by 16.

5 0

3 years ago

Find the linear approximating polynomial for the following function centered at the given point a.point a. b. Find the quadratic

ZanzabumX [31]

Answer:

$(a)L(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})\\(b)Q(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})+\frac{\sqrt{2} }{4}(x+\frac{\pi}{4})^2\\(c)L(-0.23\pi)=-0.6626\\Q(-0.23\pi)=-0.6613$

Step-by-step explanation:

Given the function:

$f(x)=sin x, a=-\frac{\pi}{4}$

(a)Linear approximating polynomial

$L(x)=f(a)+f'(a)(x-a)\\f(a)=sin(-\frac{\pi}{4})=-\frac{\sqrt{2} }{2}\\f'(x)=cos x, a=-\frac{\pi}{4}\\f'(a)=cos (-\frac{\pi}{4})=\frac{\sqrt{2} }{2} \\Therefore:\\L(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x-(-\frac{\pi}{4}))\\L(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})$

(b)Quadratic approximating polynomial

$Q(x)=L(x)+\frac{1}{2}f''(a)(x-a)^2\\f''(x)=-sin(x), \\f''(a)=-sin(-\frac{\pi}{4})=\frac{\sqrt{2} }{2}\\Q(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})+\frac{\sqrt{2} }{4}(x+\frac{\pi}{4})^2$

(c)When $x=-0.23\pi$

Using Linear Approximation polynomial

$L(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})\\L(-0.23\pi)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(-0.23\pi+\frac{\pi}{4})\\L(-0.23\pi)=-0.6626$

Using the Quadratic approximating polynomial

$Q(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})+\frac{\sqrt{2} }{4}(x+\frac{\pi}{4})^2\\Q(-0.23\pi)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(-0.23\pi+\frac{\pi}{4})+\frac{\sqrt{2} }{4}(-0.23\pi+\frac{\pi}{4})^2=-0.6613$

4 0

3 years ago