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2 years ago

8

WILL GIVE BRAINLEST!!!!!!!!!!!!!!! The three graphs represent the types of solutions that are possible for a system of equations

. Which graph represents a system of equations with no solution?

1 answer:

masha68 [24]2 years ago

4 0

Answer:

its the 2nd one

Step-by-step explanation:

just did it on edge

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For f(x)=2x+1 and g(x)=x^2-7, find (f+g)(x) A. 2x^2-15 B.X^2+2x-6 C.2x^3-6 D.x^2+2x+8

Advocard [28]

Answer:

C

Step-by-step explanation:

(f + g)(x) = f(x) + g(x) , thus

f(x) + g(x)

= 2x + 1 + x² - 7 ← collect like terms

= x² + 2x - 6 → C

5 0

3 years ago

Correlation is a measure of the direction and strength of the linear (straight-line) association between two quantitative variab

MariettaO [177]

Answer:

D

Step-by-step explanation:

The correlation coefficient r=-0.84 denotes that there is inverse relationship between x and y. It means that as the x values increase the y values decrease whereas as the x values decreases the y-values increases. Also, r=-0.84 denotes the strong relationship between x and y because it is close to 1. So, r=-0.84 denotes that there is strong linear relationship between x and y with smaller x values tending to be associated with larger y values.

7 0

2 years ago

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

2 years ago

What is 20/5 in simplest form?

Anna007 [38]

It would be equal to 4

Hope this helps!

5 0

2 years ago

Kate jogged a total distance of 8 1/2 miles during the months of January and February if katie only jogged 1/4 mile everyday whi

swat32

Since in this equation we are dividing 17/2 by 1/4, we can think of this as multiplying 17/2 with 4/1 (or 4).

17 4
---- X ----- = 34
2 1

Our final answer is 34 days.

3 0

2 years ago