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GaryK [48]
3 years ago
7

Is 7/12 less than 1/2

Mathematics
2 answers:
LekaFEV [45]3 years ago
7 0
No. Make the fractions have a common denominator

1/2=6/12

(Multiply by 6/6, which equals 1)

7/12 is not less than 6/12
fenix001 [56]3 years ago
6 0
It is greater because it is more than 1/2
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A sampling of hemlock trees revealed that 7 out of 190 trees are infected with insects. About how many of the 950 hemlock trees
nignag [31]

Answer:

About 35 trees in a state park are infected with insects

Step-by-step explanation:

we know that

7 out of 190 trees are infected with insects

so

using proportion

Find out about how many of the 950 hemlock trees in a state park are infected with insects

\frac{190}{7}=\frac{950}{x} \\\\x=7(950)/190\\\\x= 35

therefore

About 35 trees in a state park are infected with insects

3 0
3 years ago
wat factor is paired of 5 to give 45 wat factor is paired with 3 to give 24 wat factor is paired with 6 to give 54
Lady bird [3.3K]
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6 0
3 years ago
Which expression is equivalent to (x^1/2y^16)^1/2
Irina-Kira [14]
Xy^16 / 4

I don't really know how to explain it, but this is the answer
5 0
4 years ago
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Inga [223]
56000 or 5.6 * 10^4

Hope this helps!
8 0
4 years ago
If f(x)=x^3-x+2, then (f^-1)'(2)
yawa3891 [41]

Note that f(x) as given is <em>not</em> invertible. By definition of inverse function,

f\left(f^{-1}(x)\right) = x

\implies f^{-1}(x)^3 - f^{-1}(x) + 2 = x

which is a cubic polynomial in f^{-1}(x) with three distinct roots, so we could have three possible inverses, each valid over a subset of the domain of f(x).

Choose one of these inverses by restricting the domain of f(x) accordingly. Since a polynomial is monotonic between its extrema, we can determine where f(x) has its critical/turning points, then split the real line at these points.

f'(x) = 3x² - 1 = 0   ⇒   x = ±1/√3

So, we have three subsets over which f(x) can be considered invertible.

• (-∞, -1/√3)

• (-1/√3, 1/√3)

• (1/√3, ∞)

By the inverse function theorem,

\left(f^{-1}\right)'(b) = \dfrac1{f'(a)}

where f(a) = b.

Solve f(x) = 2 for x :

x³ - x + 2 = 2

x³ - x = 0

x (x² - 1) = 0

x (x - 1) (x + 1) = 0

x = 0   or   x = 1   or   x = -1

Then \left(f^{-1}\right)'(2) can be one of

• 1/f'(-1) = 1/2, if we restrict to (-∞, -1/√3);

• 1/f'(0) = -1, if we restrict to (-1/√3, 1/√3); or

• 1/f'(1) = 1/2, if we restrict to (1/√3, ∞)

6 0
2 years ago
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