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True [87]
2 years ago
9

Two consecutive even numbers are such that five times the smaller number , subtracted from eight times the bigger gives 58. Find

the two Numbers? show working​
Mathematics
1 answer:
olga55 [171]2 years ago
6 0

Answer:

14\text{ and }16

Step-by-step explanation:

Let the smaller number be x. Since they are consecutive even numbers, the larger number can be represented by x+2.

Therefore, we have the following equation:

8(x+2)-5x=58,\\8x+16-5x=58,\\3x=42,\\x=\boxed{14}

Thus, the larger number is x+2=\boxed{16}

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Sergeu [11.5K]

Answer:

A

Step-by-step explanation:

He needs to collect at least 120 so if c is what he must still collect

64 +c must be greater than or equal to 120

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2 years ago
Whats balance fe+30^2 >fe^2O^3
blsea [12.9K]

Answer:

You are given:

4Fe+3O_2 -> 2Fe_2O_3

4:Fe:4

6:O_2:6

You actually have the same number of Fe on both sides, The same is true for O_2 so yes this equation is properly balanced.

For added benefit consider the following equation:

CH_4+O_2-> CO_2+2H_2O

ASK: Is this equation balanced? Quick answer: No

ASK: So how do we know and how do we then balance it?

DO: Count the number of each atom type you have on each side of the equation:

1:C:1

4:H:4

2:O:4

As you can see everything is balanced except for O To balance O we can simply add a coefficient of 2 in front of O_2 on the left side which would result in 4 O atoms:

CH_4+color(red)(2)O_2-> CO_2+2H_2O

1:C:1

4:H:4

4:O:4

Everything is now balanced.

Step-by-step explanation:

5 0
2 years ago
In the diagram, lines p and q are parallel. Identify the angle pair created in the diagram.
mixas84 [53]

Answer:

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Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Square of a standard normal: Warmup 1.0 point possible (graded, results hidden) What is the mean ????[????2] and variance ??????
LenaWriter [7]

Answer:

E[X^2]= \frac{2!}{2^1 1!}= 1

Var(X^2)= 3-(1)^2 =2

Step-by-step explanation:

For this case we can use the moment generating function for the normal model given by:

\phi(t) = E[e^{tX}]

And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

\phi(t) = C \int_{R} e^{tx} e^{-\frac{x^2}{2}} dx = C \int_R e^{-\frac{x^2}{2} +tx} dx = e^{\frac{t^2}{2}} C \int_R e^{-\frac{(x-t)^2}{2}}dx

And we have that the moment generating function can be write like this:

\phi(t) = e^{\frac{t^2}{2}

And we can write this as an infinite series like this:

\phi(t)= 1 +(\frac{t^2}{2})+\frac{1}{2} (\frac{t^2}{2})^2 +....+\frac{1}{k!}(\frac{t^2}{2})^k+ ...

And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:

E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]

E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...

and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:

\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}

And then we have this:

E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...

And then we can find the E[X^2]

E[X^2]= \frac{2!}{2^1 1!}= 1

And we can find the variance like this :

Var(X^2) = E[X^4]-[E(X^2)]^2

And first we find:

E[X^4]= \frac{4!}{2^2 2!}= 3

And then the variance is given by:

Var(X^2)= 3-(1)^2 =2

7 0
3 years ago
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denpristay [2]
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