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Sloan [31]
3 years ago
8

What is a short method to show the compound inequality x < 2 AND x > 7

Mathematics
1 answer:
jeyben [28]3 years ago
8 0

These inequalities are not compatible: you are looking for a number that is less than 2, but more than 7. This is impossible, because if a number is less than 2, it can't be more than 7.

Conversely, if a number is more than 7, it can't be less than 2.

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Which one would be the number line ???????
8090 [49]

Answer:

B

Step-by-step explanation:

1/2x - 2 > 0

collect like terms

1/2x > 0 + 2

1/2x > 2

Divide both sides by 1/2

x > 2 divided by 1/2

inverse of 1/2 is 2/1

x > 2 times 2/1

x > 4

8 0
2 years ago
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LGS matematik
NeX [460]

Answer:

si quieres la respuesta pon más puntos

3 0
2 years ago
Total surface area of the drawing
Crazy boy [7]
The pentagon's area is the sum of the 5 triangles that comprise it.  This is

5( \frac{1}{2}bh) \times 2=5(9)(6.2)= 279

where the factor of 2 accounts for the top and bottom.  The walls have an area of 5 rectangles

5(bh)=5(9)(6)=270 \\  \\ 279+270=549km^2
4 0
2 years ago
10 points?...1) must be quick to answer,2) must be reasonable answer,3) Show explanation for your answer
Ilia_Sergeevich [38]
Your answer is
{x}^{2}  \div   {2} {z}^{9}
because the numbers raised to negative powers must be flipped over the divisor to become positive. Then, when multiplying 2 and z, you add their exponents.
3 0
3 years ago
3. Find f[h(a+4)], given the functions below. *
mote1985 [20]

Answer:

C 5a^2 +70a +240

Step-by-step explanation:

We have given these following functions:

f(x) = 5x, g(x) = -2x + 1, h(x) = x^2 + 6x + 8

h(a+4)

This function is:

h(a+4) = (a+4)^2 + 6(a + 4) + 8

h(a+4) = a^2 + 8a + 16 + 6a + 24 + 8

h(a+4) = a^2 + 14a + 48

f[h(a+4)]

f(x) = 5x

Thus

f(h(a+4)) = f(a^2+14a+48) = 5(a^2 + 14a + 48) = 5a^2 + 70a + 240

The correct answer is given by option C.

8 0
3 years ago
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