Verify the identity. quantity one minus sine of x divided by cosine of x equals cosine of x divided by quantity one plus sine of x

The pulse corresponding to one pixel is approximately 1.27 microseconds.

In this case, we can calculate the duration of the pulse corresponding to one pixel by dividing the time required to redraw the display by the total number of pixels.

Given that the display is redrawn 75 times a second, and the display has a resolution of 1024x768 pixels, the total number of pixels is 1024 * 768 = 786,432 pixels.

To find the duration of the pulse corresponding to one pixel, we can divide the time required to redraw the display (1/75 seconds) by the total number of pixels (786,432 pixels).

Therefore, the pulse corresponding to one pixel is approximately 1.27 microseconds.

s =1/2(u+v)t

u = 16, v=16, t=90

s = 1/2(16+16)90

s = 1/2(36)90

s = 16*90

s = 1440

Answer:  The required simplified expression is -2c.

Step-by-step explanation:  We are given to simplify the following expression :

[tex]E=\dfrac{1}{8}(-8c+16)-\dfrac{1}{3}(6+3c)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]

We will be using the following distributive property :

[tex]a(b+c)=ab+ac.[/tex]

From expression (i), we get

[tex]E\\\\\\=\dfrac{1}{8}(-8c+16)-\dfrac{1}{3}(6+3c)\\\\\\=\dfrac{1}{8}\times(-8c)+\dfrac{1}{8}\times16-\dfrac{1}{3}\times6-\dfrac{1}{3}\times3c\\\\=-c+2-2-c\\\\=-2c.[/tex]

Thus, the required simplified expression is -2c.

Answer:

$166.50

Step-by-step explanation:

Answer:

B. Multiply by 2; 64, 128.

A) 265,000* 1.023 = $271,095

B) 42,000,000*0.06 = 2,520,000*0.20 = $504,000

C) 504000 - 468000 = 36000/504000 = 0.0714 = 7.1%

1. 10 cm

2. 15 mm

3. 2 cm

Hope this helps even though it's pretty late!

Answer:  The correct option is

(C) [tex]4x^2+2x+6.[/tex]

Step-by-step explanation:  We are given the functions f(x) and g(x) as follows :

[tex]f(x)=2x+6,\\\\g(x)=4x^2.[/tex]

We are to find the value of (f+g)(x).

We know that

for any two functions p(x) and q(x), we have

[tex](p+q)(x)=p(x)+q(x).[/tex]

So, we get

[tex](f+g)(x)\\\\=f(x)+g(x)\\\\=2x+6+4x^2\\\\=4x^2+2x+6.[/tex]

Thus, the required value is [tex](4x^2+2x+6).[/tex]

Option (C) is CORRECT.

Final answer:

To solve the literal equation for y in the given equation a=9y-3yx, we can rearrange the terms and isolate y on one side. The solution for y is (a - 9y) / (-3x).

Explanation:

To solve the literal equation for y, we need to isolate y on one side of the equation.

Given the equation a = 9y - 3yx, we can start by rearranging the terms.

Subtract 9y from both sides:

a - 9y = -3yx

Divide both sides by -3x:

(a - 9y) / (-3x) = y

So, the solution for y is (a - 9y) / (-3x).

Final answer:

To solve the given literal equation for y, factor out y and then divide by (9 - 3x) to obtain the equation y = a / (9 - 3x), which allows y to be expressed in terms of a and x.

Explanation:

To solve the literal equation for y given as a = 9y - 3yx, we want to isolate y on one side of the equation. To do this, we first factor out y from the terms on the right side of the equation:

a = y(9 - 3x)

Next, we divide both sides of the equation by (9 - 3x) to get y by itself:

y = a / (9 - 3x)

This equation represents the value of y in terms of a and x. In the context of a linear equation, such as y = 9 + 3x, the number 9 is the y-intercept (denoted as the b term), and the number 3 is the slope (denoted as the m term).

Answer:

B. [tex]8\sqrt2+2\sqrt{10}[/tex] units

Step-by-step explanation:

We are given that a parallelogram PQRS in which point P(-2,5), Q(2,1),R(1,-2) and S(-3,2).

We have to find the perimeter of parallelogram PQRS.

Distance formula : [tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

Using the formula

[tex]SR=\sqrt{(1+3)^2+(-2-2)^2}=4\sqrt2[/tex]units

[tex]QR=\sqrt{(1-2)^2+(-2-1)^2}=\sqrt{10} units[/tex]

Now, the perimeter of parallelogram =2(SR+QR)

The perimeter of parallelogram =[tex]2(4\sqrt2+\sqrt{10})=8\sqrt2+2\sqrt{10} units[/tex]

Answer:B. [tex]8\sqrt2+2\sqrt{10}[/tex] units

There is a solution. Starting from C−1(A+X)B−1=In C−1(A+X)B−1=In, multiply both of the sides of the equation, by C and multiply both sides of the equation by B.

In other terms, this is the solution:

Given: C^(-1)(A + X)B^(-1) = In

= CC^(-1)(A + X)B^(-1)B = CInB

= In(A + X)In = CB

= AInIn + XInIn = CB


= A + X = CB


X = CB – A

The final answer Is X = CB – A

The type of rice which is most expensive is [tex]\boxed{\text{\bf type B}}[/tex].

Further details:

It is given that the rice seller mixes three types of rice and the total number of mixture is three.

The first mixture consists [tex]1\text{ kg}[/tex] rice of type [tex]\text{A}[/tex], [tex]2\text{ kg}[/tex] rice of type [tex]\text{B}[/tex] and [tex]4\text{ kg}[/tex] rice of type [tex]\text{C}[/tex].

The selling price of the first mixture is [tex]19500[/tex].

In equation form it can be written as follows:

[tex]\boxed{\text{A}+2\text{B}+4\text{C}=19500}[/tex]      .......(1)

The second mixture consists [tex]2\text{ kg}[/tex] rice of type [tex]\text{A}[/tex] and [tex]3\text{ kg}[/tex] rice of type [tex]\text{B}[/tex].

The selling price of the second mixture is [tex]19000[/tex].

In equation form it can be written as follows:

[tex]\boxed{2\text{A}+3\text{B}=19000}[/tex]   ......(2)

The third mixture consists [tex]1\text{ kg}[/tex] rice of type [tex]\text{B}[/tex] and [tex]1\text{ kg}[/tex] rice of type [tex]\text{C}[/tex].

The selling price of the third mixture is [tex]6250[/tex].

In equation form it can be written as follows:

[tex]\boxed{\text{B}+\text{C}=6250}[/tex] ........(3)

Rearrange the equation (3) to obtain the value of [tex]\text{B}[/tex] as follows:

[tex]\text{B}=6250-\text{C}[/tex]            ...........(4)

Substitute this value of [tex]\text{B}[/tex] in equation (2).

[tex]\begin{aligned}2\text{A}+3(6250-\text{C})&=19000\\2A+(3\cdot 6250)-3\text{C}&=19000\\2\text{A}-3\text{C}+18750&=19000\\2\text{A}-3\text{C}&=19000-18750\\2\text{A}-3\text{C}&=250\end{aligned}[/tex]  

Further solve the above equation,

[tex]\text{A}-1.5\text{C}=125[/tex]   ......(5)

Now substitute the value of [tex]\text{B}[/tex] from equation (4) in equation (1) to obtain an equation in the form of [tex]\text{A}[/tex] and [tex]\text{C}[/tex] as follows,

[tex]\begin{aligned}\text{A}+2(6250-\text{C})+4\text{C}&=19500\\ \text{A}+(2\cdot 6250)-2\text{C}+4\text{C}&=19500\\ \text{A}+12500+2\text{C}&=19500\\ \tex{A}+2\text{C}&=19500-12500\end{aligned}[/tex]

Further the above equation can be simplifies as follows,

[tex]\text{A}+2\text{C}=7000[/tex]            .......(6)

Subtract the equation (4) from (5) to obtain the value of [tex]\text{A}[/tex] and [tex]\text{C}[/tex].  

[tex]\begin{aligned}\text{A}+2\text{C}-\text{A}+1.5\text{C}&=7000-125\\3.5\text{C}&=6875\\ \text{C}&=1964.3\end{aligned}[/tex]

Substitute this value of [tex]\text{C}[/tex] in equation (5) to obtain the value of [tex]\text{A}[/tex] as follows,

[tex]\begin{aligned}\text{A}+(2\cdot 1964.3)&=7000\\ \text{A}&=7000-3928.6\\ \text{A}&=3071.4\end{aligned}[/tex]

Again substitute [tex]\text{C}=1964.3[/tex] in equation (4) to obtain the value of [tex]\text{B}[/tex].

[tex]\begin{aligned}\text{B}&=6250-1964.3\\&=4285.7\end{aligned}[/tex]  

Therefore, the most expensive type of rice is [tex]\boxed{\text{\bf type B}}[/tex].

Learn more:

1.  Problem on percentage in a survey: https://brainly.com/question/3724002

2. Problem on unit conversion factor: https://brainly.com/question/5009365

3. Problem on to find the value in pounds: https://brainly.com/question/4837736

Answer details:

Grade: Middle School

Subject: Mathematics

Chapter: Mixtures

Keywords:  Rice seller, mixture, sold, expensive, type A, type B, type C, simplification, consist, 19500, 1 kg,addition, substitution, multiplication.