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stepan [7]
3 years ago
12

What are the solutions to the inequality (x-3)(x+5) ≤ 0

Mathematics
2 answers:
AleksAgata [21]3 years ago
8 0

Answer:

d

Step-by-step explanation:

DaniilM [7]3 years ago
7 0
1. Solve for x
x= 3 or x=-5

2. From the values of x above, we have these 3 intervals to test
x<=-5
-5<=x<=3
x<=3

3. Pick a test point for each interval
For the interval x<=-5:
Lets pick x=-6. Then, (-6-3)(-6+5)<=0. After simplifying, we get 9<=0, which is false. So we drop this interval.

For the interval -5<=x<=3:
Lets pick x=0. Then (0-3)(0+5)<=0. After simplifying, we get -15<=0, which is true. So we keep this interval.

For the interval x>=3:
Lets pick x=4. Then, (4-3)/4+5)<=0. After simplifying, we get 9<=0, which is false. So we drop this interval.

4. Therefore, 
-5<=x<=3.
The answer is therefore number 3 :D
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Answer:

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Step-by-step explanation:

well 3/3 is full so it's just like 12/12, so 12 pieces of 1/12 piece are in 3/3

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Josh invested $450 at 0.5% simple interest for 2 years. How much is his total account balance at the end of 2 years?
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The total account balance would be $454.50

Step-by-step explanation:

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2 years ago
Write the equation of the line that passes through the point (2,-9)<br> and has a slope of -5.
balu736 [363]

Step-by-step explanation:

Find equation

y - y1 = m(x - x1)

y + 9 = -5(x - 2)

y = -5x + 10 - 9

y = -5x + 1 → Equaton

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2 years ago
Simplify (7ty)(3t)<br> (answer fast please)
attashe74 [19]
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4 0
3 years ago
Please help me on this problem
Anna71 [15]

Answer:

The pairs of integer having two real solution forax^{2} -6x+c = 0 are

  1. a = -4, c = 5
  2. a = 1, c = 6
  3. a = 2, c = 3
  4. a = 3, c = 3

Step-by-step explanation:

Given

ax^{2} -6x+c = 0

Now we will solve the equation by putting all the 6 pairs so we get the  following

-3x^{2} -6x-5 = 0 for a = -3 , c=-5

-4x^{2} -6x+5 = 0 for a = -4 , c=5

1x^{2} -6x+6 = 0 for a = 1 , c=6

2x^{2} -6x+3 = 0 for a = 2 , c=3

3x^{2} -6x+3 = 0 for a = 3 , c=3

5x^{2} -6x+4 = 0 for a = 5 , c=4

The above  all are Quadratic equations inn general form ax^{2} +bx+c=0

where we have a,b and c constant values

So for a real Solution we must have

Disciminant , b^{2} -4\timesa\timesc \geq 0

for a = -3 , c=-5 we have

Discriminant =-24 which is less than 0 ∴ not a real solution.

for a = -4 , c=5 we have

Discriminant = 116 which is greater than 0 ∴ a real solution.

for a = 1 , c=6 we have

Discriminant =12 which is greater than 0 ∴ a real solution.

for a = 2 , c=3 we have

Discriminant =12 which is greater than 0 ∴ a real solution.

for a = 3 , c=3 we have

Discriminant =0 which is equal to 0 ∴ a real solution.

for a = 5 , c=4 we have

Discriminant =-44 which is less than 0 ∴ not a real solution.

7 0
3 years ago
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