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balu736 [363]
3 years ago
9

Analyze the diagram below and complete the instructions that follow.

Mathematics
1 answer:
NNADVOKAT [17]3 years ago
7 0

Answer: Its B i just know it

Step-by-step explanation:

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Solve square root of x + square root of x+2 = 2
andrew-mc [135]

The value of x is \frac{1}{4}

Step-by-step explanation:

Given:

\sqrt{x} +  \sqrt{x + 2} = 2

Rearranging the radical on left side,

\sqrt{x + 2} = 2 - \sqrt{x}

Power on both sides,

(\sqrt{x + 2})^{2}  = (2 - \sqrt{x})^{2}

Simplifying the left,

x + 2 = (2 - \sqrt{x} )^{2}

For the RHS equation, use the property of (a-b)² = (a²-2ab+b²),

=  > x + 2 =  {2}^{2} - (2 \times 2 \sqrt{x}) + ( \sqrt{x})^{2}

Now calculating its powers,

=  > x + 2 = 4 - 4\sqrt{x} + x

Now sending -4√x to the LHS (left side), its sign becomes plus (+),

=  > 4 \sqrt{x} = 4 + x - x - 2

Now the +x and -x will be cancelled,

=  > 4 \sqrt{x} = 4 - 2

=  > 4 \sqrt{x} = 2

Bringing 4 to the right side, it becomes the denominator,

=  >  \sqrt{x} =  \frac{2}{4}

=  >  \sqrt{x} =  \frac{1}{4}

Now powering both sides,

=  > ( \sqrt{x})^{2} = ( \frac{1}{2})^{2}

=  > x =  \frac{1}{4}

4 0
2 years ago
What is the taxidistance between (2, 11) and (17, 40)?
Novosadov [1.4K]

15 and 29 or (15,29)

2 + 15 = 17

11 + 29 = 40

Hope it Helped!


6 0
3 years ago
Help please?! This is graphing rational functions
guapka [62]

Answer:

D: (-∞,2) (2,∞)

R:(-∞,0),(0,∞)

Step-by-step explanation:

From the graph we can see that the graph is approaching 0 horizontally and 2 vertically but not crossing so we can assumer these are the asymptotes. So the domain goes all the way from -∞ to 2 but does not cross 2 so we have to stop there and create another bracket for 2 to ∞. It is the same process for the range but in the y direction.

7 0
3 years ago
Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product o
koban [17]

Answer:  17.6 grams

Step-by-step explanation:

As the problem tells us, the velocity of the reaction is proportional to the product of the quantities of A and B that have not reacted, so from this we get the next equation:

                                                       V = k[A][B]

where [A] represents the remaining amount of A, and [B] represents the remaining amount of B. To solve this equation we have to represent it through a differential equation, which is:

                                              dx/dt = k[α - a(t)][β - b(t)]         (1)

where,

k: velocity constant

a(t): quantity of A consumed in instant t

b(t): quantity of B consumed in instant t

α: initial quantity of A

β: initial quantity of B

Now we need to define the equations for a(t) and b(t), and for this we are going to use the law of conservation of mass by Lavoisier, with which we can say that the quantity of C in a certain instant is equal to the sum of the quantities of A and B that have reacted. Therefore, if we need M grams of A and N grams of B to form a quantity of M+N of C, then we can say that in a certain time, the consumed quantities of A and B are given by the following equations:

                                       a(t) = ( M/M+N) · x(t)

                                       b(t) = (N/M+N) · x(t)

where,

x(t): quantity of C in instant t

So for this problem we have that for 1 gram of B, 2 grams of A are used, therefore the previous equations can be represented as:

                                       a(t) = (2/2+1) · x(t) = 2/3 x(t)

                                       b(t) = (1/2+1) · x(t) = 1/3 x(t)

Now we proceed to resolve the differential equation (1) by substituting values:

                                         dx/dt = k[α - a(t)][β - b(t)]  

                                        dx/dt = k[40 - 2x/3][50 - x/3]

                                         dx/dt = k/9 [120 - 2x][150 - x]

We use the separation of variables method:

                                      dx/[120-2x][150-x] = k/3 · dt

We integrate both sides of the equation:

                                     ∫dx/(120-2x)(150-x) = ∫kdt/9

                                     ∫dx/(15-x)(60-x) = kt/9 + c

Now, to integrate the left side of the equation we need to use the partial fraction decomposition:

                                    ∫[1/90(120-2x) - 1/180(150-x)] = kt/9 + c

                                      1/180 ln(150-x/120-2x) = kt/9 + c

                                           (150-x)/(120-2x) = Ce^{20kt}

Now we resolve by taking into account that x(0) = 0, and x(5) = 10,

for x(0) = 0 ,             (150-0)/(120-0) = Ce^{20k(0)} , C = 1.25

for x(5) = 10 ,           (150-10)/(120-(2·10)) = 1.25e^{20k(5)} , k ≈ 113 · 10^{-5}

Now that we have the values of C and k, we have this equation:

                           (150-x)/(120-2x) = 1.25e^{226·10^{-4}t}

and we have to clear by x, obtaining:

               x(t) = 150 · (1 - e^{226·10^{-4}t} / 1 - 2.5e^{226·10^{-4}t})

Therefore the quantity of C that will be formed in 10 minutes is:

           x(10) = 150 · (1 - e^{226·10^{-4}(10)} / 1 - 2.5e^{226·10^{-4}(10)})

                                            x(10) ≈ 17.6 grams

8 0
3 years ago
Subtract 111 from 11010​
skelet666 [1.2K]

Answer : I think it is 10899

4 0
3 years ago
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