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Advocard [28]
3 years ago
9

Helppppppp! Find the length of PQ if PQR is an isosceles triangle with PQ ≈ QR.​

Mathematics
1 answer:
KATRIN_1 [288]3 years ago
5 0

Answer:

43

Step-by-step explanation:

3x + 10 = 5x - 12

       10 = 2x - 12

      22 = 2x

        x = 11

check:

33 + 10 = 55 - 12

43 = 43

correct

x = 11 so bring this into 3x + 10

we get 43

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ASAP!!! Please help!!!
Sunny_sXe [5.5K]

Answer:

-\frac{28}{99}

Step-by-step explanation:

To divide a fraction by a fraction, flip the second fraction and change the divide sign into a multiplication sign.

\frac{4}{9}\div -\left(\frac{11}{7}\right)=\frac{4}{9} *-\frac{7}{11}

Multiply:

-\frac{4*7}{9*11}

-\frac{28}{99}

6 0
3 years ago
Amanda worked 10 hours Monday through Friday. She worked an additional 5 hours on Saturday. She makes $14.25 an hour. What are A
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Answer:

10x5days+5=55

55÷14.25

=50.35

Step-by-step explanation:

Amanda earn $50.35. Amanda's wage is $50.35

5 0
1 year ago
9x-3y=27<br> -6x+9y=24<br> idek what this is but i need help
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Answer:

Step-by-step explanation:

if x=6 y=9 that is answer. I just know, but think. BOI

8 0
3 years ago
To rent a carpet cleaner at the hardware store, there is a set fee and an hourly rate. The rental cost, c, can be determined usi
hichkok12 [17]

3h

The set cost is 25, because c=25 when h=0. So, this leaves 3h as the hourly rate.

7 0
3 years ago
Read 2 more answers
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
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