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Slav-nsk [51]
3 years ago
5

Tell whether xy is positive or negative. If x is negative and y is positive, the product is

Mathematics
1 answer:
mamaluj [8]3 years ago
6 0
Negative. A negative number times a positive number is equal to a negative number. X multiplied by Y is a positive.
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Problem Page
LenKa [72]
2:7 , 2 to 7, 2/7 <====
4 0
3 years ago
What inequality is on the number line?​
Elis [28]
Answer:
y ≥ -5

Explanation:
y is greater than -5 because the darkened line is heading towards the positive side.
(The positives are greater than the negatives.)
(The smaller negative numbers are greater than the larger negative numbers.)
7 0
3 years ago
Given f(x) = -6x - 1 and g(x) = x3, choose<br> the expression for fºg)(x).
Gelneren [198K]

Answer:

- 6x³ - 1

Step-by-step explanation:

Substitute x = g(x) into f(x), that is

(f ○ g)(x)

= f(x³ )

= - 6(x³) - 1

= - 6x³ - 1

4 0
3 years ago
M. Score: 0 of 1 pt
beks73 [17]

Answer:

  3×5×53

Step-by-step explanation:

You can use divisibility rules to find the small prime factors.

The number ends in 5, so is divisible by 5.

  795/5 = 159

The sum of digits is 1+5+9 = 15; 1+5 = 6, a number divisible by 3, so 3 is a factor.

  159/3 = 53 . . . . . a prime number,* so we're done.

795 = 3×5×53

_____

* If this were not prime, it would be divisible by a prime less than its square root. √53 ≈ 7.3. We know it is not divisible by 2, 3, or 5. We also know the closest multiples of 7 are 49 and 56, so it is not divisible by 7. Hence 53 is prime.

5 0
3 years ago
Which graph represents the solution set for the system x + 2y 3, x + y 4, and 3x − 2y 4?
Lerok [7]

x + 2y + 3 = 0
Subtract x from both sides.
2y + 3 = -x
Subtract 3 from both sides.
2y = -x - 3
Divide by 2 on both sides
y = -(x+3)/2

x + y + 4 = 0
Subtract x and 4 from both sides
y = -x - 4

3x - 2y + 4 = 0
Subtract 3x and 4 from both sides.
-2y = -3x -4
Divide by -2 from both sides.
y = -(3x + 4) / 2

The answer is the graph that contains these slopes and lines on the graph, which was not provided.

4 0
3 years ago
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