Is the origin a solution to the system graphed?

Look at the picure then awser please!

Arlecino [84]

Answer:

yes

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .

kogti [31]

Answer:

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

$b = \sqrt{1 + 2\cdot a \cdot c}$ (2)

$c$ is another arbitrary prime number. (3)

Step-by-step explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if $(x+y)^{2} = x^{2} + 2\cdot x\cdot y + y^{2}$ . From statement, we must fulfill the following identity:

$a^{2} + b^{2} + c^{2} - 1 = x^{2} + 2\cdot x\cdot y + y^{2}$

By Associative and Commutative properties, we can reorganize the expression as follows:

$a^{2} + (b^{2}-1) + c^{2} = x^{2} + 2\cdot x \cdot y + y^{2}$ (1)

Then, we have the following system of equations:

$x = a$ (2)

$(b^{2}-1) = 2\cdot x\cdot y$ (3)

$y = c$ (4)

By (2) and (4) in (3), we have the following expression:

$(b^{2} - 1) = 2\cdot a \cdot c$

$b^{2} = 1 + 2\cdot a \cdot c$

$b = \sqrt{1 + 2\cdot a\cdot c}$

From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then, $a, b, c > 1$ . If $a$ , $b$ and $c$ are prime numbers, then $2\cdot a\cdot c$ must be an even composite number, which means that $a$ and $c$ can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.

In addition, $b$ must be a natural number, which means that:

$1 + 2\cdot a\cdot c \ge 4$

$2\cdot a \cdot c \ge 3$

$a\cdot c \ge \frac{3}{2}$

But the lowest possible product made by two prime numbers is $2^{2} = 4$ . Hence, $a\cdot c \ge 4$ .

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

$b = \sqrt{1 + 2\cdot a \cdot c}$ (2)

$c$ is another arbitrary prime number. (3)

Example: $a = 2$ , $c = 2$

$b = \sqrt{1 + 2\cdot (2)\cdot (2)}$

$b = 3$

4 0

3 years ago

Please help need help on this

UkoKoshka [18]

Looking at the data it looks like a line. So the answer will be in y+mx+b format.

Knowing it is a line eliminates answers b and c.
looking at options a is says it starts at about .1 when x is zero. We can see this looks pretty close as the left-most dot is below 1 already.

The slope on a is 1/2--> increasing 1 vertically for every 2 horizontally. The pattern mostly follows that rule.

Therefore a (y=1/2x+.1) would be the correct answer

7 0

3 years ago

Is 5361 divisible by three how do you know just by looking at it

sleet_krkn [62]

Answer:

Yes!

Step-by-step explanation:

Here is a trick. If you add up all of the numbers in the number, and if that is divisible by three, the number is divisible by three. So...

5+3+6+1=15

15 is divisible by 3,

so 5361 is divisible by 3

(if you wanted to know, it is 1787)

8 0

3 years ago

PLEASE ANSWER i WILL PUT MOST BRAINLIEST!

galina1969 [7]

Answer:

The correct option is;

c. Because the p-value of 0.1609 is greater than the significance level of 0.05, we fail to reject the null hypothesis. We conclude the data provide convincing evidence that the mean amount of juice in all the bottles filled that day does not differ from the target value of 275 milliliters.

Step-by-step explanation:

Here we have the values

μ = 275 mL

275.4

276.8

273.9

275

275.8

275.9

276.1

Sum = 1928.9

Mean (Average), = 275.5571429

Standard deviation, s = 0.921696159

We put the null hypothesis as H₀: μ₁ = μ₂

Therefore, the alternative becomes Hₐ: μ₁ ≠ μ₂

The t-test formula is as follows;

$t=\frac{\bar{x}-\mu }{\frac{s}{\sqrt{n}}}$

Plugging in the values, we have,

Test statistic = 1.599292

$t_{\alpha /2}$ at 7 - 1 degrees of freedom and α = 0.05 = ±2.446912

Our p-value from the the test statistic = 0.1608723≈ 0.1609

Therefore since the p-value = 0.1609 > α = 0.05, we fail to reject our null hypothesis, hence the evidence suggests that the mean does not differ from 275 mL.

4 0

3 years ago