Question

11y^2+18y+7>0 срочнооооо

Answer 1

Ответ:

(-бесконечности,-1),(-7/11,+бесконечности)

Пошаговое объяснение:

11y^2+18y+7>0,

11y^2+11y+7y+7>0,

11y(y+1)+7(y+1)>0,

(y+1)(11y+7)>0,

1. y+1>0, y>-1, (-7/11,+бесконечность)

11y+7>0 y>-7/11

2. y+1<0, y<-1, (-бесконечность,-1)

11y+7<0 y<-7/11

Находим объединение:

(-бесконечности,-1),(-7/11,+бесконечности)

Answer 2

To solve the inequality $11y^2 + 18y + 7 > 0$, we can use factoring or the quadratic formula. Let's proceed with factoring:

Step 1: Write down the inequality $11y^2 + 18y + 7 > 0$

Step 2: Factor the quadratic expression on the left side of the inequality $11y^2 + 18y + 7 = (11y + 7)(y + 1)$

Step 3: Set each factor to zero and solve for $y$: $11y + 7 = 0 \Rightarrow y = -\frac{7}{11}$ $y + 1 = 0 \Rightarrow y = -1$

Step 4: Now, we have three intervals to test:

a) $y < -\frac{7}{11}$ b) $-\frac{7}{11} < y < -1$ c) $y > -1$

Step 5: Pick a test point from each interval and check if it satisfies the original inequality:

a) Test with $y = -1$, for example: $11(-1)^2 + 18(-1) + 7 = -11 + 18 + 7 = 14$ (which is positive, but it should be negative for this interval). So, this interval does not satisfy the inequality.

b) Test with $y = 0$, for example: $11(0)^2 + 18(0) + 7 = 7$ (which is positive, but it should be negative for this interval). So, this interval does not satisfy the inequality.

c) Test with $y = 1$, for example: $11(1)^2 + 18(1) + 7 = 36$ (which is positive, but it should be negative for this interval). So, this interval does not satisfy the inequality.

Step 6: Determine the solution: Since the original inequality does not hold for any of the test points in the three intervals, the solution is the complement of those intervals. The solution is:

$y < -\frac{7}{11}$ or $y > -1$

This means that the original inequality $11y^2 + 18y + 7 > 0$ is satisfied when $y$ is less than $-\frac{7}{11}$ or greater than $-1$.