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lilavasa [31]
2 years ago
8

Solve for x. -9x + 2 > 18 OR 13r + 15 < -4

Mathematics
1 answer:
Gemiola [76]2 years ago
6 0

Answer:

x < -1.778 or. r < -1.461

Step-by-step explanation:

-9x + 2 > 18

-9x > 18-2

-9x > 16

x < -(16/9)

x < -1 778

13r + 15 < -4

13r <-4 -15

13r < -19

r < -1.461

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Are the equations m + 3 = -5 and m = -2 equivalent? Explain why or why not?
patriot [66]

Answer:

Step-by-step explanation:

When m+3=-5, collection of like terms says; -5-3

Now,-5+-3 = -8 and -2≠-8 hence,they aren't equivalent

7 0
3 years ago
Pens and Pencils
Lina20 [59]
Pen=x
Pencil=y
y+0.15=x
y+x=0.69
y+y+0.15=0.69
2y=0.69-0.15
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y=0.54/2
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Supplier is right about a priče.

6 0
2 years ago
Roger Brown works for the sanitation department. He earns a salary of $721.00 biweekly. His boss gave him
liq [111]

Answer:

2249 dollars more

Step-by-step explanation:

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3 0
3 years ago
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
A small tree is 80 cm tall.
ruslelena [56]
1.2 meters = 120 centimeters
amount of increase = 120-80 = 40 centimeters
40 ÷ 80 = 0.5 = 50%
There was a 50% increase.
7 0
3 years ago
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