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Anestetic [448]
2 years ago
13

40 is 60% of what number?

Mathematics
2 answers:
Bad White [126]2 years ago
6 0

Answer:

We have, 60% × x = 40

or,

60

100

× x = 40

Multiplying both sides by 100 and dividing both sides by 60,

we have x = 40 ×

100

60

x = 66.67

If you are using a calculator, simply enter 40×100÷60, which will give you the answer.

11111nata11111 [884]2 years ago
4 0
40(60/100) 4(6/10) =24/10 = 2.4
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Which of the following lists all of the positive factors of 18?
babymother [125]

The square root of 18 is 4.2426, rounded down to the closest whole number is 4. Testing the integer values 1 through 4 for division into 18 with a 0 remainder we get these factor pairs: (1 and 18), (2 and 9), (3 and 6). The factors of 18 are 1, 2, 3, 6, 9, 18.

Step-by-step explanation:

4 0
3 years ago
Write an algebraic expression for 84 more than the product of 167 and b
Ket [755]

Answer:

  167b + 84

Step-by-step explanation:

Multiplication is indicated between a number and a variable by writing them next to each other. A "product" is the result of multiplication, so "the product of 167 and b" is written 167b.

If you want a value that is 84 more than that, you get it by adding 84.

  167b + 84

4 0
2 years ago
Are the triangles shown congruent?
boyakko [2]

Answer:

the second one i think

Step-by-step explanation:

because both have 2 sides with 1 angle

5 0
2 years ago
Find e^cos(2+3i) as a complex number expressed in Cartesian form.
ozzi

Answer:

The complex number e^{\cos(2+31)} = \exp(\cos(2+3i)) has Cartesian form

\exp\left(\cosh 3\cos 2\right)\cos(\sinh 3\sin 2)-i\exp\left(\cosh 3\cos 2\right)\sin(\sinh 3\sin 2).

Step-by-step explanation:

First, we need to recall the definition of \cos z when z is a complex number:

\cos z = \cos(x+iy) = \frac{e^{iz}+e^{-iz}}{2}.

Then,

\cos(2+3i) = \frac{e^{i(2+31)} + e^{-i(2+31)}}{2} = \frac{e^{2i-3}+e^{-2i+3}}{2}. (I)

Now, recall the definition of the complex exponential:

e^{z}=e^{x+iy} = e^x(\cos y +i\sin y).

So,

e^{2i-3} = e^{-3}(\cos 2+i\sin 2)

e^{-2i+3} = e^{3}(\cos 2-i\sin 2) (we use that \sin(-y)=-\sin y).

Thus,

e^{2i-3}+e^{-2i+3} = e^{-3}\cos 2+ie^{-3}\sin 2 + e^{3}\cos 2-ie^{3}\sin 2)

Now we group conveniently in the above expression:

e^{2i-3}+e^{-2i+3} = (e^{-3}+e^{3})\cos 2 + i(e^{-3}-e^{3})\sin 2.

Now, substituting this equality in (I) we get

\cos(2+3i) = \frac{e^{-3}+e^{3}}{2}\cos 2 -i\frac{e^{3}-e^{-3}}{2}\sin 2 = \cosh 3\cos 2-i\sinh 3\sin 2.

Thus,

\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2-i\sinh 3\sin 2\right)

\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2\right)\left[ \cos(\sinh 3\sin 2)-i\sin(\sinh 3\sin 2)\right].

5 0
2 years ago
What does it mean when it ask hat is the constant of proportionality in pages per hour?
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What question is that from? I'm talking about ur grade
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3 years ago
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