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3 years ago

How to get y by itself in the equation x+2y=10

Mathematics

1 answer:

Nataly_w [17]3 years ago

4 0

Answer:

The value of y by itself in the equation x+2y=10 is

$y=\frac{10-x}{2}$

Step-by-step explanation:

Given equation is x+2y=10 -------- (1)

To find y with given equation as below :

Equation (1) implies that x+2y=10

Subtracting the above equation by x we get

x+2y-x=10-x

x-x+2y=10-x

2y=10-x

Dividing by 2 on both sides we get

$\frac{2y}{2}=\frac{10-x}{2}$

$y=\frac{10-x}{2}$

Therefore the value of y is $\frac{10-x}{2}$

Therefore the value of y by itself in the equation x+2y=10 is

$y=\frac{10-x}{2}$

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Is .04 repeating a irrational number?

ruslelena [56]

.04 repeating is a rational number.

A irrational number has a non-terminating (no end) and non-repeating decimals, meaning numbers after the decimals do not end and do not repeat.

4 0

4 years ago

Simplify: 5^(2)x 5^(4)x5^(-4)x5^(-2)

Katen [24]

Answer:

$320x^{15}$

Step-by-step explanation:

1. take out constants

( 5 × 2 × 4 × - 4 × - 2) $x^{5} x^{5} x^{5}$

2. simplify 5 × 2 to 10

(10 × 4 - 4 × - 2) $x^{5} x^{5} x^{5}$

3. simplify 10 × 4 to 40

(40 × - 4 × - 2) $x^{5} x^{5} x^{5}$

4. simplify 40 × -4 to -160

(-160 × - 2) $x^{5} x^{5} x^{5}$

5. simplify -160 × -2 to 320

$320x^{5} x^{5} x^{5}$

6. use product rule: $x^{a} x^{b} =x^{a+b}$

$320x^{5+5+5}$

7. simplify 5 + 5 to 10

$320x^{10+5}$

8. simplify 10 + 5 to 15 $320x^{15}$

3 0

3 years ago

How to divide (x^2+5x-36)÷(x-4)

mr_godi [17]

Answer:

x + 9

Step-by-step explanation:

Since the divisor is in the form of x - c, use what is called Synthetic Division. Remember, in this formula, -c gives you the OPPOSITE terms of what they really are, so do not forget it. Anyway, here is how it is done:

4| 1 5 -36

↓ 4 36

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

1 9 0 → x + 9

You start by placing the c in the top left corner, then list all the coefficients of your dividend [x² + 5x - 36]. You bring down the original term closest to c, then begin your multiplication. Now depending on what symbol your result is tells you whether the next step is to subtract or add, then you continue this process starting with multiplication all the way up until you reach the end. Now, when the last term is 0, that means you have NO REMAINDER. Finally, your quotient is one degree less than your dividend, so that 1 in your quotient can be an x, and the 9 follows right behind it, giving you the other factor of x + 9.

If you are ever in need of assistance, do not hesitate to let me know by subscribing to my channel [USERNAME: MATHEMATICS WIZARD], and as always, I am joyous to assist anyone at any time.

4 0

3 years ago

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

marshall27 [118]

Answer:

1. 0.0000454

2. 0.01034

3. 0.0821

4. 0.918

Step-by-step explanation:

Let X be the random variable denoting the number of passengers arriving in a minute. Since the mean arrival rate is given to be 10,

$X \sim Poi(\lambda = 10)$

1. Requires us to compute

$P(X = 0) = e^{-10} \frac{10^0}{0!} = 0.0000454$

2. We need to compute $P(X \leq 3) = P(X =0) + P(X =1) + P(X =2) + P(X =3)$

$P(X =1) = e^{-10} \frac{10^1}{1!} = 0.000454$

$P(X =2) = e^{-10} \frac{10^2}{2!} = 0.00227$

$P(X =3) = e^{-10} \frac{10^3}{3!} = 0.00757$

$P(X \leq 3) =0.0000454+ 0.000454 + 0.00227 + 0.00757 = 0.01034$

3. The expected no. of arrivals in a 15 second period is = $10 \times \frac{1}{4} = 2.5$ . So if Y be the random variable denoting number of passengers arriving in 15 seconds,

$Y \sim Poi(2.5)$

$P(Y=0) = e^{-2.5} \frac{2.5^0}{0!} = 0.0821$

4. Here we use the fact that Y can take values $0,1, \dotsc$ . So, the event that "Y is either 0 or $\geq$ 1" is a sure event ( i.e it has probability 1 ).