What's the solution of the following linear system? y = 4x + 1 y = 4x

1 answer:

7 0

The solution of a system of equations is the point at which the two lines intersect. Since y is already isolated, we can set the equations equal to each other by doing:
4x=4x+1
Subtract 4x from both sides
0=1
Since 0 cannot equal 1 in any case, the lines never intersect. There is no solution to the problem.

You might be interested in

18-22 evens please and show how to do the work

marta [7]

#18
Answer C
4*7 = 28 which is the most closest one to 27.8

#19
perimeter = 2(L+W)
175.33 feet = 2103. 96 in
2 ( 2103.96 +48)
2 * 2151.96
4303.92 inches

#20
L*W = A
2103.96 * 48
100990.08 inch^2

to lazy to do the rest sorry

5 0

2 years ago

subtract the product of 3 and 4 from 36. then divide the quotient of 9 and 3, subtracted from fifteen

Butoxors [25]

The first one is 24, and the second one is 12

4 0

2 years ago

Read 2 more answers

Sean bought a box of graham crackers for $3.99. He also bought 3 bags

leonid [27]

Answer:

$3.08

Explanation:

You subtract 3.99 from 13.23 which equals 9.24. Because there are three bags of marshmallows, divide 9.24 by 3, which equals $3.08.

4 0

3 years ago

Read 2 more answers

What’s the distance from c to d ?

Anon25 [30]

It’s 5 since you’d use Pythagorean theorem

3 0

3 years ago

A box contains 3 coins. One coin has 2 heads and the other two are fair. A coin is chosen at random from the box and flipped. If

Blababa [14]

Answer: Our required probability is $\dfrac{1}{2}$

Step-by-step explanation:

Since we have given that

Number of coins = 3

Number of coin has 2 heads = 1

Number of fair coins = 2

Probability of getting one of the coin among 3 = $\dfrac{1}{3}$

So, Probability of getting head from fair coin = $\dfrac{1}{2}$

Probability of getting head from baised coin = 1

Using "Bayes theorem" we will find the probability that it is the two headed coin is given by

$\dfrac{\dfrac{1}{3}\times 1}{\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times 1}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{3}}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{2}{3}}\\\\=\dfrac{1}{2}$

Hence, our required probability is $\dfrac{1}{2}$

No, the answer is not $\dfrac{1}{3}$

5 0

2 years ago