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Step2247 [10]
3 years ago
5

What is the gcf of 18a and 24ab

Mathematics
1 answer:
Leona [35]3 years ago
7 0
The gcf of 18a and 24ab is 6a
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a sprinter running in the olympics starts at the 0 meter mark and ends at the 400 meter mark. it took him 65 seconds to run this
solniwko [45]

Answer:

I am not too sure but I think the answer would be 6

Step-by-step explanation:

3 0
2 years ago
Look at the two patterns below: Pattern A: Starts at 5 and follows the rule "add 6" Pattern B: Starts at 5 and follows the rule
OleMash [197]

Answer:

The first five terms in Pattern A are 5, 11, 17, 23, and 29

Step-by-step explanation:

Pattern A:

5 + 6a

Pattern B:

5 + 3a

Take a = 1 and compare ;

Pattern A:

5 + 6(1) = 11

Pattern B:

5 + 3(1) = 8

A ≠ 2B

B ≠ 1/3A

First 5 terms in pattern A :

Pattern A:

5

5 + 6(1) = 11

5 + 6(2)= 17

5 + 6(3) = 23

5 + 6(4) = 29

5 + 6(5) = 35

Pattern B cannot have a term 0 as it starts from 5

8 0
2 years ago
(8 × 7 − 3) × 7 Match each expression to another expression with the same value
kodGreya [7K]

Hey there!

(8 * 7 - 3) * 7

= 8(7)(7) - 3(7) ⬅️ POSSIBLY BE AN EQUIVALENT EXPRESSION

= 56(7) - 3(7)

= 392 - 3(7)

= 392 - 21

= 371 ⬅️ OVERALL ANSWER/EQUATION ANSWER

Most likely your answer should be:

• 8(7)(7) - 3(7)

• 56(7) - 3(7)

• 392 - 3(7)

• 392 - 21

EITHER OF THESE SHOULD WORK ⬆️

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

8 0
2 years ago
Read 2 more answers
Two negative integers are 5 units apart on the number line, and their product is 126. What is the sum of the two integers?
Tatiana [17]
If b>a, b-a=5 so b=5+a

ab=126, using b from above

a(5+a)=126

5a+a^2=126

a^2+5a-126=0
 
(a+14)(a-9)=0, since a<0 the numbers are

-14 and -9


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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