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2 years ago

What percent of 125 is 24

Mathematics

2 answers:

bonufazy [111]2 years ago

5 0

125/24=5.21
24 is 5.21% of 125

melisa1 [442]2 years ago

4 0

The answer is 5.21. You just divide 125 24

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nadya68 [22]

Answer:

1. Area=225 m2, 2. Carpet required is 225m2. 3. Cost= Rs 18000

Step-by-step explanation:

1. Area=l*l=15*15=225 m2

2. Carpet has to be 225m2 in order to cover the floor

3. Cost= 225*80=18000

6 0

3 years ago

Which of the following nonlinear inequalities is graphed below

Rashid [163]

Answer:

(x²/4) + (y²/16) ≤ 1

Step-by-step explanation:

The general form of the ellipse in the graph is

x²/4 + y²/16 = 1

we want to find the inequality which represents the shaded area this will determine whether which of the following 2 choices are correct

Choice 1: x²/4 + y²/16 ≥ 1

Choice 2: x²/4 + y²/16 ≤ 1

we note that point (0,0) lies inside the shaded area. Which means if we substitute (0,0), it should create an equality that is valid.

try choice 1: 0²/4 + 0²/16 = 0 ≥ 1 (not valid because obviously zero is smaller than 1)

try choice 1: 0²/4 + 0²/16 = 0 ≤ 1 (this is valid, hence choice 2 is the answer)

4 0

2 years ago

What is the number that not the same in a fish

stira [4]

Answer:

Step-by-step explanation:

6 0

2 years ago

If you help ill give you some of my banana milk...

snow_tiger [21]

Given:

The function is:

$f(x)=4x^2-25x-21$

To find:

The domain of the given function.

Solution:

Domain is the set of input values.

We have,

$f(x)=4x^2-25x-21$

It is a quadratic polynomial.

We know that a quadratic polynomial is defined for all real values of x. So, the given function is defined for all real values of x and the domain of the given function is:

Domain = Set of all real number

Domain = (-∞,∞)

Therefore, the correct option is B.

3 0

2 years ago

Let z=3+i, then find a. Z² b. |Z| c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq

zysi [14]

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

z = √10 (cos(arctan(1/3)) + i sin(arctan(1/3))

Any complex number has 2 square roots. Using the polar form from part (d), we have

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 2)

Then in standard rectangular form, we have

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)$

and

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)$

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

Now, arctan(1/3) is an angle y such that tan(y) = 1/3. In a right triangle satisfying this relation, we would see that cos(y) = 3/√10 and sin(y) = 1/√10. Then

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

So the two square roots of z are

$\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

and

$\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

3 0

3 years ago

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