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ruslelena [56]
3 years ago
12

So Michael had 55 candy and he divide it to 5. How much candy he had now?

Mathematics
2 answers:
Andru [333]3 years ago
7 0

Answer:

11 pieces of candy

Step-by-step explanation:

55 divided by 5=11

valentinak56 [21]3 years ago
6 0

Answer:

11

Step-by-step explanation:

dude just use a calculator lol

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Answer ASAP 15 points
malfutka [58]
A or 5/7, they’re both odd numbers so it cannot be simplified any further
8 0
3 years ago
Wich of the following is the correct graph of solution to the inequality -8 >-5x+2>-38
zimovet [89]
-5x + 2 > -38
-5x > -40
x < 8

-5x + 2 < -8
-5x < -10
x > 2

so the inequality becomes
2 < x < 8

so the graph will be a straight line with open circles on 2 and 8 and shading in between these values.
3 0
3 years ago
Is .72 repeating rational or irrational​
Kamila [148]

Answer:

.72 is a rational number

Hope this helps!

4 0
3 years ago
Read 2 more answers
Find the lateral surface area of the figure.
Brut [27]

Surface area of the cylinder = 816.4 sq. ft.

Solution:

Height of the cylinder = 21 ft

Diameter of the cylinder = 10 ft

Radius of the cylinder = 10 ÷ 2 = 5 ft

Use the value of \pi = 3.14

Surface area of the cylinder = 2 \pi r h+2 \pi r^{2}

Substitute the given values in the surface area formula, we get

Surface area of the cylinder $=2\times3.14\times 5\times 21+2 \times3.14\times 5^2

                                               =659.4+157

                                               =816.4 ft²

Hence the surface area of the cylinder is 816.4 sq. ft.

7 0
3 years ago
Read 2 more answers
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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