what is the name of the line that the function approaches?

a. y-axis
b. there is no such line
c. the unknown line
d. asymptote

Answer:

B. Multiply by 2; 64, 128.

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].

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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.

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

1. 10 cm

2. 15 mm

3. 2 cm

Hope this helps even though it's pretty late!

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.

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.

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%

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.

Answer:

The value of the equation for n is [tex]n=\frac{d +2}{4}[/tex].

Step-by-step explanation:

Consider the provided equation.

[tex]d = 4n - 2[/tex]

We need to solve the equation for n.

Add 2 from both the sides.

[tex]d +2= 4n - 2+2[/tex]

[tex]d +2= 4n [/tex]

Divide both side of the equation with 4.

[tex]\frac{d +2}{4}= \frac{4n}{4} [/tex]

[tex]\frac{d +2}{4}= n [/tex]

Hence, the value of the equation for n is [tex]n=\frac{d +2}{4}[/tex].

Answer:

$166.50

Step-by-step explanation:

The recursive formula that could be used for measuring the total amount of money is f(n+1)=f(n)+20.

Given that,

Based on the above information, we can see that there are 20 increments in every next number.

So, the equation should be f(n+1)=f(n)+20

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Solution 1:

If |a| = |b|, then either a = b or a = -b. Hence, if |2x - 1| = |3x + 5|, then either 2x - 1 = 3x + 5 or 2x - 1 = -(3x + 5).

If 2x - 1 = 3x + 5, then x = -6. If 2x - 1 = -(3x + 5), then x = -4/5. Therefore, the solutions are x = \boxed{-6, -4/5}.

Solution 2:

Another approach is to square both sides. Squaring both sides, we get 4x^2 - 4x + 1 = 9x^2 + 30x + 25 (because |a|^2 = a^2 for all a). This simplifies to 5x^2 + 34x + 24 = 0, which factors as (x + 6)(5x + 4) = 0, so the solutions are x = -6 and x = -4/5, as before. You must be careful when using this approach because squaring both sides of an equation can introduce false solutions. Thus, we need to check that both of these "solutions" work in the original expression before declaring them correct.