How do I solve questions 1,2 and 6?

Answer:

=14.69km²

Step-by-step explanation:

We can use the Hero's formula to calculate the area

A= √(s(s-a)(s-b)(s-c))

s is obtained by adding the lengths of the three sides of the triangle and then dividing by 2, a, b and c are the ides of the triangle.

S=(5+6+7)/2

=9

A=√(9(9-6)(9-5)(9-7))

=√(9×3×4×2)

=√216

=14.69km²

Answer:

Option D.The decrease in the value of the car, which is 8%

Step-by-step explanation:

we have a exponential function of the form

[tex]f(x)=a(b)^{x}[/tex]

where

y is the value of the car

x is the time in years

a is the initial value

b is the base

r is the rate of decrease

b=1+r

In this problem we have

a=$24,000 initial value of the car

b=0.92

so

0.92=1+r

r=0.92-1=-0.08=-8%-----> is negative because is a rate of decrease

Answer:

D.The decrease in the value of the car, which is 8%​

Step-by-step explanation:

Since, in the exponential function,

[tex]f(x)=ab^x[/tex]

a is the initial value,

b is the growth ( if > 1 ) or decay factor ( if between 0 and 1 ),

Here, the given equation that shows the value of car after x years,

[tex]f(x)=24000(0.92)^x[/tex]

By comparing,

b = 0.92 < 1

Thus, 0.92 is the decay factor that shows the decrease in the value of car,

∵ Decay rate = 1 - decay factor

= 1 - 0.92

= 0.08

= 8%

Hence, the value of car is decreasing with the rate of 8%.

Option 'D' is correct.

Answer with explanation:

Size of the sample = n =225

Mean[\text] \mu[/text]=48.5

Standard deviation [\text] \sigma[/text]= 1.8

[tex]Z_{90 \text{Percent}}=Z_{0.09}=0.5359\\\\Z_{score}=\frac{\Bar X -\mu}{\frac{\sigma}{\sqrt{\text{Sample size}}}}\\\\0.5359=\frac{\Bar X -48.5}{\frac{1.8}{\sqrt{225}}}\\\\0.5359=15 \times \frac{\Bar X -48.5}{1.8}\\\\0.5359 \times 1.8=15 \times (\Bar X -48.5)\\\\0.97=15 \Bar X-727.5\\\\727.5+0.97=15 \Bar X\\\\728.47=15 \Bar X\\\\ \Bar X=\frac{728.47}{15}\\\\\Bar X=48.57[/tex]

→Given Confidence Interval of Mean =48.8

→Calculated Mean of Sample =48.57 < 48.8

So, the value of Sample mean lies within the confidence interval.

Answer:

sample answer

Step-by-step explanation:

To find the margin of error, multiply the z-score by the standard deviation, then divide by the square root of the sample size.

The z*-score for a 90% confidence level is 1.645.

The margin of error is 0.20.

The confidence interval is 48.3 to 48.7.

48.8 is not within the confidence interval.

For this case we have that by definition, the equation of the line in slope-intersection form is given by:

[tex]y = mx + b[/tex]

Where:

m: It's the slope

b: It is the cutoff point with the y axis

We have:

[tex](x1, y1): (- 8,11)\\(x2, y2): (4,3.5)[/tex]

[tex]m = \frac {y2-y1} {x2-x1} = \frac {3.5-11} {4 - (- 8)} = \frac {-7.5} {4 + 8} = \frac {-7.5} {12 } = - \frac {\frac {15} {2}} {12} = - \frac {15} {24} = - \frac {5} {8}[/tex]

Thus, the equation will be given by:

[tex]y = - \frac {5} {8} x + b[/tex]

We substitute a point to find "b":

[tex]11 = - \frac {5} {8} (- 8) + b\\11 = 5 + b\\b = 11-5\\b = 6[/tex]

Finally:

[tex]y = - \frac {5} {8} x + 6[/tex]

Answer:

[tex]y = - \frac {5} {8} x + 6[/tex]

Answer:

So our answers could be any of these depending on the form wanted*:

[tex]y=\frac{-5}{8}x+6[/tex]

[tex]5x+8y=48[/tex]

[tex]y-11=\frac{-5}{8}(x+8)[/tex]

[tex]y-\frac{7}{2}=\frac{-5}{8}(x-4)[/tex]

*There are other ways to write this equation.

Step-by-step explanation:

So we are given two points on a line: (-8,11) and (4,7/2).

We can find the slope by using the formula [tex]\frac{y_2-y_1}{x_2-x_1} \text{ where } (x_1,y_1) \text{ and } (x_2,y+2) \text{ is on the line}[/tex].

So to do this, I'm going to line up my points vertically and then subtract vertically, then put 2nd difference over 1st difference:

( 4  ,  7/2)

-(-8 ,    11)

----------------

12      -7.5

So the slope is -7.5/12 or -0.625 (If you type -7.5 division sign 12 in your calculator).

-0.625 as a fraction is -5/8 (just use the f<->d button to have your calculator convert your decimal to a fraction).

Anyways the equation of a line in slope-intercept form is y=mx+b where m is the slope and b is y-intercept.

We have m=-5/8 since that is the slope.

So plugging this into y=mx+b gives us y=(-5/8)x+b.

So now we need to find b. Pick one of the points given to you (just one).

Plug it into y=(-5/8)x+b and solve for b.

y=(-5/8)x   +b with (-8,11)

11=(-5/8)(-8)+b

11=5+b

11-5=b

6=b

So the equation of the line in slope-intercept form is y=(-5/8)x+6.

We can also put in standard form which is ax+by=c where a,b,c are integers.

y=(-5/8)x+6

First step: We want to get rid of the fraction by multiplying both sides by 8:

8y=-5x+48

Second step: Add 5x on both sides:

5x+8y=48 (This is standard form.)

Now you can also out the line point-slope form, [tex]y-y_1=m(x-x_1) \text{ where } m \text{ is the slope and } (x_1,y_1) \text{ is a point on the line }[/tex]

So you can say either is correct:

[tex]y-11=\frac{-5}{8}(x-(-8))[/tex]

or after simplifying:

[tex]y-11=\frac{-5}{8}(x+8)[/tex]

Someone might have decided to use the other point; that is fine:

[tex]y-\frac{7}{2}=\frac{-5}{8}(x-4)[/tex]

So our answers could be any of these depending on the form wanted*:

[tex]y=\frac{-5}{8}x+6[/tex]

[tex]5x+8y=48[/tex]

[tex]y-11=\frac{-5}{8}(x+8)[/tex]

[tex]y-\frac{7}{2}=\frac{-5}{8}(x-4)[/tex]

Ah, all you have to do is combine 2/5m and 3/5m.

In this case:

=2/5m - 4/5 - 3/5m

=-1/5m - 4/5

=-m/5-4/5

Answer:

[tex]\frac{-1}{5}m-\frac{4}{5}[/tex]

Step-by-step explanation:

You are given:

[tex]\frac{2}{5}m-\frac{4}{5}-\frac{3}{5}m[/tex]

Reorder using commutative property (putting like terms together):

[tex]\frac{2}{5}m-\frac{3}{5}m-\frac{4}{5}[/tex]

Now we are going to bring down the -4/5 (there is nothing to do there).

(2/5)m and -(3/5)m have the same denominator all we have to do is figure out what is 2-3 which is -1

[tex]\frac{-1}{5}m-\frac{4}{5}[/tex]

[tex]\bf \left( \cfrac{~~\begin{matrix} 7 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~\cdot 5\cdot 2}{~~\begin{matrix} 7 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~\cdot 3} \right)^2 \times \left( \cfrac{5^0}{2^{-3}} \right)^3\times 2^{-9}\implies \left( \cfrac{5\cdot 2}{ 3} \right)^2 \times \left( \cfrac{1}{2^{-3}} \right)^3\times 2^{-9}[/tex]

[tex]\bf \left( \cfrac{10}{ 3} \right)^2 \times \left( 2^3 \right)^3\times 2^{-9}\implies \left( \cfrac{10}{ 3} \right)^2 \times 2^9\times 2^{-9}\implies \cfrac{10^2}{3^2}\times 2^{9-9} \\\\\\ \cfrac{100}{9}\times 2^0\implies \cfrac{100}{9}\times 1\implies \cfrac{100}{9}[/tex]

Answer:

audience's perspective of a story - please give brainliest

Answer B. audience's perspective of the story

Answer:

[tex]xy(1+2y\sqrt{x}+\sqrt{y})[/tex]

Step-by-step explanation:

Given expression,

[tex]\sqrt{x^2y^2}+2\sqrt{x^3y^4}+xy\sqrt{y}[/tex]

[tex]=(x^2y^2)^\frac{1}{2} + 2(x^3y^4)^\frac{1}{2} + xy\sqrt{y}[/tex]

[tex]\because (\sqrt{x}=x^\frac{1}{2})[/tex]

[tex]=(x^2)^\frac{1}{2} (y^2)^\frac{1}{2} + 2(x^3)^\frac{1}{2} (y^4)^\frac{1}{2} + xy\sqrt{y}[/tex]

[tex](\because (ab)^n=a^n b^n)[/tex]

[tex]=x^{2\times \frac{1}{2}} y^{2\times \frac{1}{2}} + 2(x^{3\times \frac{1}{2}})(y^{4\times \frac{1}{2}})+xy\sqrt{y}[/tex]

[tex]\because (a^n)^m=a^{mn}[/tex]

[tex]=x^1 y^1 + 2x^{1\frac{1}{2}} y^2 + xy\sqrt{y}[/tex]

[tex]=xy+2x.(x)^\frac{1}{2} y^2 + xy\sqrt{y}[/tex]

[tex]=xy+2xy^2\sqrt{x}+xy\sqrt{y}[/tex]

[tex]=xy(1+2y\sqrt{x}+\sqrt{y})[/tex]

Answer:

B is the right option

Step-by-step explanation:

On edg :))

Answer:

x² + 6x + 9  = (x + 3)(x + 3)

Step-by-step explanation:

It is given a quadratic equation

x² + 6x + 9

To find the factors of given expression

By using middle term splitting

Let f(x) = x² + 6x + 9

 = x² + 3x  + 3x + 9

 = x(x + 3) + 3(x + 3)

 = (x + 3)(x + 3)

Therefore the factors of x² + 6x + 9

(x + 3)(x + 3)

The expression [tex]\(x^2 + 6x + 9\)[/tex] factors completely to [tex]\((x + 3)^2\)[/tex].

To factor the expression [tex]\(x^2 + 6x + 9\)[/tex] completely, we can look for a pair of numbers that multiply to 9 (the constant term) and add up to 6 (the coefficient of the linear term).

The pair of numbers that satisfy these conditions is 3 and 3  because [tex]\(3 \times 3 = 9\) and \(3 + 3 = 6\).[/tex]

So, we can rewrite the expression as:

[tex]\[ x^2 + 3x + 3x + 9 \][/tex]

Now, we can group the terms:

[tex]\[ (x^2 + 3x) + (3x + 9) \][/tex]

Now, we can factor out the greatest common factor from each group:

[tex]\[ x(x + 3) + 3(x + 3) \][/tex]

Notice that both terms have a common factor of [tex]\(x + 3\)[/tex], so we can factor that out:

[tex]\[ (x + 3)(x + 3) \][/tex]

[tex]\[ (x + 3)^2 \][/tex]

Answer:

The answer is 7

Step-by-step explanation:

The expression is 10-(1/2)^4 * 48

Here PEMDAS rule applies:

where,

P= parenthesis

E= exponent

M= multiplication

D= division

A= addition

S= subtraction

So according to this rule first we will solve parenthesis and exponent.(PE)

10-(1/2)^4 *48

(1/2)^4 means, multiply 1/2 four times:

1/2*1/2*1/2*1/2=1/16

Therefore the expression becomes:

10-1/16*48

Now we have MD which is multiplication and division:

1/16*48 = 3

Now after solving the multiplication and division the expression becomes:

10-3.

After subtracting the terms we have:

10-3=7

Thus the answer is 7....

Answer:

The equation is equal to

[tex]y=250(0.88^{x})[/tex]

Step-by-step explanation:

we know that

In this problem we have a exponential function of the form

[tex]y=a(b^{x})[/tex]

where

x -----> the time in years

y ----> the population of animals

a is the initial value

b is the base

r is the rate of decreasing

b=(1-r) ----> because is a decrease rate

we have

[tex]a=250\ animals[/tex]

[tex]r=12\%=12/100=0.12[/tex]

[tex]b=(1-0.12)=0.88[/tex]

substitute

[tex]y=250(0.88^{x})[/tex]

The equation which  represents a population of 250 animals that decreases at an annual rate of 12%​ is:

               [tex]f(x)=250(0.88)^x[/tex]

It is given that:

A population of 250 animals decreases at an annual rate of 12%​.

This problem could be modeled with the help of a exponential function.

         [tex]f(x)=ab^x[/tex]

where a is the initial amount.

and b is the change in the population and is given by:

[tex]b=1-r[/tex] if the population is decreasing at a rate r.

and [tex]b=1+r[/tex] if the population is increasing at a rate r.

Here we have:

[tex]a=250[/tex]

and x represents the number of year.

[tex]r=12\%=0.12[/tex]

Hence, we have:

[tex]b=1-0.12=0.88[/tex]

Hence, the population function f(x) is given by:

          [tex]f(x)=250(0.88)^x[/tex]

Answer:

11.8294Step-by-step explanation:

Answer:

Step-by-step explanation:

How many deciliters are equivalent to 5 cups?

2.1097 deciliters

11.85 deciliters

118.5 deciliters

210.97 deciliters

ANSWER IS 11.85

[tex]\huge{\boxed{5}}[/tex]

The Pythagorean theorum states that when [tex]a[/tex] and [tex]b[/tex] are sides and [tex]c[/tex] is the hypotenuse, [tex]a^2 + b^2 = c^2[/tex]

So, let's plug in the values. [tex]3^2 + b^2 = (\sqrt{34})^2[/tex]

Simplify. The square of a square root is the number inside the square root. [tex]9 + b^2 = 34[/tex]

Subtract 9 from both sides. [tex]b^2 = 25[/tex]

Get the square root of both sides. [tex]\sqrt{b^2} = \sqrt{25}[/tex]

[tex]b=\boxed{5}[/tex]

Answer:

Step-by-step explanation:

Use the Pythagorean theorem:

[tex]leg^2+leg^2=hypotenuse^2[/tex]

We have

[tex]leg=3,\ hypotenuse=\sqrt{34}[/tex]

Let's mark the other leg as x.

Substitute:

[tex]3^2+x^2=(\sqrt{34})^2[/tex]      use (√a)² = a

[tex]9+x^2=34[/tex]            subtract 9 from both sides

[tex]x^2=25\to x=\sqrt{25}\\\\x=5[/tex]

Let see.

Numbers which are bigger than 0 are defined as positive numbers and have a prefix of + (plus).

Numbers which are smaller than 0 are defined as negative numbers and have a prefix of - (minus).

Let say number a is equal to the expression,

[tex]a=3xy[/tex]

Since y is negative we can change its prefix to -,

[tex]a=3x\cdot(-y)[/tex]

Any number (in this case 3x) multiplied by negative number will produce a negative number.

Therefore the sign or prefix of number a will be -.

Hope this helps.

r3t40

When you multiply a positive number and a negative number, the result is a negative number. Therefore, the sign of 3xy, when x > 0 and y < 0, is negative.

The question is asking for the sign of the product of two numbers, x and y, when x is positive (x > 0) and y is negative (y < 0). In mathematics, when you multiply a positive number and a negative number, the result is always a negative number.

So, the product of x and y or 3xy in this case, would be negative. This is due to the principle that the product of different signs (in this case, positive and negative) is always negative.

https://brainly.com/question/34274159

#SPJ3

[tex]x^2-y^2=(x-y)(x+y)\\\\x^2-y^2=3\cdot12=36[/tex]

Answer: is an irrational number

Step-by-step explanation:

Like adding three to pi (3.14159265358979323846264....)is still going to be irrational

Answer:

D

Step-by-step explanation:

I took the test

[tex]|\Omega|=\dfrac{6!}{3!3!}=\dfrac{4\cdot5\cdot6}{2\cdot3}=20\\A=\{RBRBRB,BRBRBR\}\\|A|=2\\\\P(A)=\dfrac{2}{20}=\dfrac{1}{10}[/tex]

Answer:

2(n-6)=12

(I went too far in my explanation; I'm not going to erase it because I think it is important to have an example on solving these)

Step-by-step explanation:

Twice the difference of a number and 6 is the same as 12.

Twice means 2 times

Difference means the result of subtracting something.

is the same as means equal to (=).

So we are given 2(n-6)=12.

You can start by dividing 2 on both sides are distributing 2 to terms in the ( ).

I will do it both ways and you can pick your favorite.

2(n-6)=12

Divide both sides by 2.

 n-6  =6

Add 6 on both sides

 n     =12

OR!

2(n-6)=12

Distribute 2 to both terms in the ( )

2n-12=12

Add 12 on both sides

2n    =24

Divide both sides by 2

n       =12

Answer:

25

Step-by-step explanation:

Those parallel lines tell us our triangles are similar. So that means the corresponding sides are proportional.

So we have that x corresponds to x+15 and

40 corresponds to 24+40.

So we have this proportion to solve:

[tex]\frac{x}{x+15}=\frac{40}{24+40}[/tex]

Let's simplify what we can:

[tex]\frac{x}{x+15}=\frac{40}{64}[/tex]

Cross multiply:

[tex](64)(x)=(x+15)(40)[/tex]

Multiply/distribute:

[tex]64x=40x+600[/tex]

Subtract 40x on both sides:

[tex]24x=600[/tex]

Divide both sides by 24:

[tex]x=\frac{600}{24}=25[/tex]

x=25

Answer:

x = 25.

Step-by-step explanation:

24/40 = 15/x

x = (40*15) / 24

x = 600/24

= 25.

Answer:

True.

Step-by-step explanation:

The arcs subtend the same equal angles at the center of the circle, i.e ∠YOZ=∠WOX= 27°. They are also bound by radii. All radii of the same circle are equal thus he two arcs are equal in length. OY= WO= XO= OZ

Therefore the two arcs wx and yz are congruent

Answer:

True

Step-by-step explanation:

just had the same question

Answer:

Step-by-step explanation:

the domain of x represents the values that x can be without the function being undefined.  the function of square rooting is undefined for negative numbers.  so in order to find the domain, you must ensure that the "stuff" in the square root is greater than, or equal, to zero.  hence, (4x+9)>= 0.  the answer is B

For this case we have the following function:

[tex]f (x) = \sqrt {4x + 9} +2[/tex]

By definition, the domain of a function is given by all the values for which the function is defined.

For the given function to be defined, then the root argument must be positive, that is:

[tex]4x + 9 \geq0[/tex]

Answer:

Option B

How much does one song cost?

Answer:

The average rate of change is 1.275

Step-by-step explanation:

The average rate of change of f(x) from x=a to x=b is given by:

[tex]\frac{f(b)-f(a)}{b-a}[/tex]

The money Terry invested is modeled by the function [tex]f(x)=0.01(2)^x[/tex] where x represents number of days.

The average rate of change from day 2 to day 10 is given by:

[tex]\frac{f(10)-f(2)}{10-2}[/tex]

[tex]f(10)=0.01(2)^{10}=10.24[/tex]

[tex]f(2)=0.01(2)^{2}=0.04[/tex]

The average rate of change becomes:

[tex]\frac{10.24-0.04}{8}[/tex]

[tex]=\frac{10.2}{8}=1.275[/tex]

Answer:

The average rate of change is 1.275

Answer:

Exponential decay

Step-by-step explanation:

b = 0.73

Since the b is less than 1 (b<1), the rate is decreasing.

Answer:

[tex]x\sqrt{7} - 49\sqrt{x}[/tex]

Step-by-step explanation:

We have to simplify the following expression: [tex]\sqrt{7x}(\sqrt{x} - 7\sqrt{7})[/tex]

Using distributive property:

[tex]\sqrt{7x}(\sqrt{x} - 7\sqrt{7}) = \sqrt{7x}\sqrt{x} - 7\sqrt{7}\sqrt{7x}[/tex]

⇒ [tex]x\sqrt{7} - 49\sqrt{x}[/tex]

So the most simplified form of the expression is the following: [tex]x\sqrt{7} - 49\sqrt{x}[/tex]

Certainly! To solve this problem, we will need to simplify the expression √(7x)(√x - 7√7).
First, let us consider the individual square root terms: √(7x) and √x. Recall that the square root of a product can be separated into the product of the square roots of the factors, so √(7x) can be expressed as √7 * √x.
Now we have:
√(7x)(√x - 7√7) = (√7√x)(√x - 7√7)
Next, we distribute √7√x into the terms within the parentheses:
(√7√x)(√x) - (√7√x)(7√7)
Simplify each term:
First term: (√7√x)(√x) = √7 * (√x * √x) = √7 * x
This is because the square root of a number multiplied by itself is just the number.
Second term: (√7√x)(7√7) = 7√7 * (√7√x) = 7 * (√7 * √7) * √x
               = 7 * 7 * √x
               = 49√x
This is because the square root of 7 squared is just 7.
Now we combine the two terms:
√7x = √7 * x - 49√x
And that is the simplified form of the given expression.

Answer:

[tex]\large\boxed{x^\frac{2}{3}}[/tex]

Step-by-step explanation:

[tex]\left(x^\frac{4}{3}x^\frac{2}{3}\right)^\frac{1}{3}\qquad\text{use}\ a^na^m=a^{n+m}\\\\=\left(x^{\frac{4}{3}+\frac{2}{3}}\right)^\frac{1}{3}=\left(x^{\frac{6}{3}\right)^\frac{1}{3}=(x^2)^\frac{1}{3}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=x^{(2)\left(\frac{1}{3}\right)}=x^\frac{2}{3}[/tex]

Answer:

Part A) [tex]15x+25y\geq 700[/tex] and [tex]x> 3y[/tex]

Part B) The point (45,10) and the point (40,5)  satisfy the system

Step-by-step explanation:

Part A) Determine which inequalities represent the constraints for this situation

Let

x -----> the number of small prints sold

y -----> the number of large prints sold

we know that

The system of inequalities that represent this situation is equal to

[tex]15x+25y\geq 700[/tex] ----> inequality A

[tex]x> 3y[/tex] ----> inequality B

Part B) With combinations of small prints and large prints satisfy this system?

we know that

If a ordered pair is a solution of the system, then the ordered pair must satisfy both inequalities

Verify each case

case 1) (45,10)

For x=45, y=10

Inequality A

[tex]15x+25y\geq 700[/tex]

[tex]15(45)+25(10)\geq 700[/tex]

[tex]925\geq 700[/tex] ----> is true

Inequality B

[tex]x> 3y[/tex]

[tex]45> 3(10)[/tex]

[tex]45> 30[/tex] ----> is true

therefore

The point (45,10) satisfy the system

case 2) (35,15)

For x=35, y=15

Inequality A

[tex]15x+25y\geq 700[/tex]

[tex]15(35)+25(15)\geq 700[/tex]

[tex]900\geq 700[/tex] ----> is true

Inequality B

[tex]x> 3y[/tex]

[tex]35> 3(15)[/tex]

[tex]35> 45[/tex] ----> is not true

therefore

The point (35,15) does not satisfy the system

case 3) (30,10)

For x=30, y=10

Inequality A

[tex]15x+25y\geq 700[/tex]

[tex]15(30)+25(10)\geq 700[/tex]

[tex]700\geq 700[/tex] ----> is true

Inequality B

[tex]x> 3y[/tex]

[tex]30> 3(10)[/tex]

[tex]30> 30[/tex] ----> is not true

therefore

The point (30,10) does not satisfy the system

case 4) (40,5)

For x=40, y=5

Inequality A

[tex]15x+25y\geq 700[/tex]

[tex]15(40)+25(5)\geq 700[/tex]

[tex]725\geq 700[/tex] ----> is true

Inequality B

[tex]x> 3y[/tex]

[tex]40> 3(5)[/tex]

[tex]40> 15[/tex] ----> is true

therefore

The point (40,5)  satisfy the system

The question refers to the relationship between x (number of small prints sold) and y (number of large prints sold), in an uncertain context. Constraints would be limitations or restrictions on the values that x and y can take, often expressed as inequalities. As this problem lacks specific details (like total prints or budget limits), it's impossible to define specific inequalities representing these constraints without them.

In this situation, x and y are variables that represent the quantities of small and large prints sold respectively. As such, they can also be understood as independent and dependent variables. In Mathematics, an independent variable is one that stands alone and isn't changed by the other variables you are trying to measure. In contrast, the dependent variable is what you measure in the experiment and what is affected.

In this context, constraints would be limitations or restrictions on the values that x and y can have, often represented by inequalities.

Without specific detail in the context such as the total prints available, thus giving a limit on total sales, or the price information for small and large prints leading to a budget constraint, it is impossible to define any specific inequalities to represent these constraints. Additional information such as these is required to formulate appropriate inequalities.

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let's use the conjugate of the denominator and multiply top and bottom by it, recall the conjugate of a binomial is simply the same binomial with a different sign in between.

[tex]\bf \cfrac{2\sqrt{x}-3\sqrt{y}}{\sqrt{x}+\sqrt{y}}\cdot \cfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}\implies \cfrac{2\sqrt{x}\sqrt{x}-2\sqrt{x}\sqrt{y}~~-~~3\sqrt{x}\sqrt{y}+3\sqrt{y}\sqrt{y}}{\underset{\textit{difference of squares}}{(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}} \\\\\\ \cfrac{2\sqrt{x^2}-2\sqrt{xy}-3\sqrt{xy}+3\sqrt{y^2}}{(\sqrt{x})^2-(\sqrt{y})^2}\implies \cfrac{2x-5\sqrt{xy}+3y}{x-y}[/tex]

Answer:

[tex]\dfrac{2x-5\sqrt{xy}+3y}{x-y}\\[/tex]

Step-by-step explanation:

In Rationalize the denominator we multiply both numerator and denominator by the conjugate of denominator.

In Conjugate we change the sign of middle operator.

Example: Congugate of (a + b) = a - b

Now Solving the given expression,

[tex]\dfrac{2\sqrt{x} - 3\sqrt{y}}{\sqrt{x} + \sqrt{y}} = \dfrac{2\sqrt{x} - 3\sqrt{y}}{\sqrt{x} + \sqrt{y}}\times \dfrac{\sqrt{x} - \sqrt{y}}{\sqrt{x} - \sqrt{y}}\\\\\Rightarrow \dfrac{(2\sqrt{x} - 3\sqrt{y})(\sqrt{x} - \sqrt{y})}{( \sqrt{x} + \sqrt{y}){(\sqrt{x} - \sqrt{y}})}\ \ \ \ \ \ \ \ \ \ \ [\because (a-b)(a+b)=(a^{2} +b^{2})]\\\Rightarrow \dfrac{2x-2\sqrt{xy}-3\sqrt{xy}+3y}{x-y}\\\\ \Rightarrow \dfrac{2x-5\sqrt{xy}+3y}{x-y}\\[/tex]