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Elina [12.6K]
3 years ago
13

4% of 50 is what number​

Mathematics
2 answers:
Anon25 [30]3 years ago
5 0

Answer:

2

Step-by-step explanation:

Percentage Calculator: What is 4 percent of 50? = 2.

hope this helps pls give brainliest

valentinak56 [21]3 years ago
4 0
50 = 100% so then divide 50 by 100 = 0.5 then multiply it by the percent you want to find so 0.5 x 4 = 2 so 4% of 50 is 2!
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What is the LCM of 16a 3 b and 24ab 2?
vitfil [10]

Do the operation:

3, 16, 24 | 2

3, 8, 12 | 2

3, 4, 6 | 2

3, 2, 3 | 2

3, 1, 3 | 3

1, 1

Answer:

\boxed{\bf~The~answer~is~2^{4}~*~3~or~48.}

Hope it helped,

Happy homework/ study/ exam!

6 0
3 years ago
An earthquake measuring 6.4 on the Richter scale struck Japan in July 2007, causing extensive damage. Earlier that year, a minor
a_sh-v [17]

Answer:

10

Step-by-step explanation

The earthquake  measures 6.4 on the Richter scale which struck Japan in Jullu 2007 and caused and extensive damage. Earlier that year, a minor earthquake measuring 3.1 in the Richter scale has stroked in parts of Pennsylvania.

Fomular:

The magnitude of an earthquake is M log(I/S)

where I donates the intensity of the earthquake and S be the intensity of the standard earthquake.

Calculation:

Consider that M1 be the magnitude Japanese earthquake and M2 be the magnitude of the Pennsylvania earthquake and L1 be the intensity of the Japanese earthquake and L2 the intensity of the Pennsylvania  earthquake.

Here the magnitude of the Japanese earthquake is M1 = 6.14 and the magnitude of the Pennsylvania is M2 = 3.1

By the use of magnitude of the earthquake fomular M = log I1/S, the intensity of the Japanese earthquake is calculated as follows .

M1 = log I1/S

I1/s = 10

3 0
3 years ago
What is the volume (in cubic units) of a cone with a radius of 6
kykrilka [37]
Evaluate the volume with the given measurements.

Answer: 376.8 m^3
3 0
2 years ago
4x^2-21x-18=0<br> Find the discriminate
SpyIntel [72]

a = 4, b = -21, c = -18

to keep from getting "mixed up", evaluate the discriminant first ...

b<sup>2</sup> - 4ac = (-21)<sup>2</sup> - 4(4)(-18) = 729

sqrt(729) = 27

x = (21 +/- 27)/8

x = -3/4, x = 6

since the discriminant is a perfect square, the original quadratic will factor ...

4x<sup>2</sup> - 21x - 18 = 0

(4x + 3)(x - 6) = 0

x = -3/4, x = 6

6 0
2 years ago
Read 2 more answers
Suppose Upper F Superscript prime Baseline left-parenthesis x right-parenthesis equals 3 x Superscript 2 Baseline plus 7 and Upp
Sedaia [141]

It looks like you're given

<em>F'(x)</em> = 3<em>x</em>² + 7

and

<em>F</em> (0) = 5

and you're asked to find <em>F(b)</em> for the values of <em>b</em> in the list {0, 0.1, 0.2, 0.5, 2.0}.

The first is done for you, <em>F</em> (0) = 5.

For the remaining <em>b</em>, you can solve for <em>F(x)</em> exactly by using the fundamental theorem of calculus:

F(x)=F(0)+\displaystyle\int_0^x F'(t)\,\mathrm dt

F(x)=5+\displaystyle\int_0^x(3t^2+7)\,\mathrm dt

F(x)=5+(t^3+7t)\bigg|_0^x

F(x)=5+x^3+7x

Then <em>F</em> (0.1) = 5.701, <em>F</em> (0.2) = 6.408, <em>F</em> (0.5) = 8.625, and <em>F</em> (2.0) = 27.

On the other hand, if you're expected to <em>approximate</em> <em>F</em> at the given <em>b</em>, you can use the linear approximation to <em>F(x)</em> around <em>x</em> = 0, which is

<em>F(x)</em> ≈ <em>L(x)</em> = <em>F</em> (0) + <em>F'</em> (0) (<em>x</em> - 0) = 5 + 7<em>x</em>

Then <em>F</em> (0) = 5, <em>F</em> (0.1) ≈ 5.7, <em>F</em> (0.2) ≈ 6.4, <em>F</em> (0.5) ≈ 8.5, and <em>F</em> (2.0) ≈ 19. Notice how the error gets larger the further away <em>b </em>gets from 0.

A <em>better</em> numerical method would be Euler's method. Given <em>F'(x)</em>, we iteratively use the linear approximation at successive points to get closer approximations to the actual values of <em>F(x)</em>.

Let <em>y(x)</em> = <em>F(x)</em>. Starting with <em>x</em>₀ = 0 and <em>y</em>₀ = <em>F(x</em>₀<em>)</em> = 5, we have

<em>x</em>₁ = <em>x</em>₀ + 0.1 = 0.1

<em>y</em>₁ = <em>y</em>₀ + <em>F'(x</em>₀<em>)</em> (<em>x</em>₁ - <em>x</em>₀) = 5 + 7 (0.1 - 0)   →   <em>F</em> (0.1) ≈ 5.7

<em>x</em>₂ = <em>x</em>₁ + 0.1 = 0.2

<em>y</em>₂ = <em>y</em>₁ + <em>F'(x</em>₁<em>)</em> (<em>x</em>₂ - <em>x</em>₁) = 5.7 + 7.03 (0.2 - 0.1)   →   <em>F</em> (0.2) ≈ 6.403

<em>x</em>₃ = <em>x</em>₂ + 0.3 = 0.5

<em>y</em>₃ = <em>y</em>₂ + <em>F'(x</em>₂<em>)</em> (<em>x</em>₃ - <em>x</em>₂) = 6.403 + 7.12 (0.5 - 0.2)   →   <em>F</em> (0.5) ≈ 8.539

<em>x</em>₄ = <em>x</em>₃ + 1.5 = 2.0

<em>y</em>₄ = <em>y</em>₃ + <em>F'(x</em>₃<em>)</em> (<em>x</em>₄ - <em>x</em>₃) = 8.539 + 7.75 (2.0 - 0.5)   →   <em>F</em> (2.0) ≈ 20.164

4 0
3 years ago
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