All categories

3 years ago

What is equal to 4/5

Mathematics

2 answers:

nikitadnepr [17]3 years ago

7 0

8/10 is equal to 4/5.

Vika [28.1K]3 years ago

6 0

95% 95% 95% 95% 95% 95%

You might be interested in

Find the radius of a circle with a circumference of 23.55 feet. Use 3.14 for

Oksi-84 [34.3K]

Answer:

3.75

Step-by-step explanation:

8 0

2 years ago

Hello Please Find The Answer To This Question (easy)

irina [24]

The equation of the line that is perpendicular to y = -5/2 x + 1 and passing through (10, 3) will be y = 2/5 x + 1.

<h3>What is the linear system?</h3>

A linear system is one in which the parameter in the equation has a degree of one. It might have one, two, or even more variables.

The equation of the perpendicular line will be given as

y = mx + c

The given equation is y = -5/2 x + 1.

The slope of the line is -5/2.

Then the product of the slope of the perpendicular lines is a negative one.

-5/2 × m = - 1

m = 2/5

Then the equation will be

y = 2/5 x + c

Then the equation is passing through (10, 3). Then we have

3 = 2/5 × 10 + c

3 = 2 + c

c = 1

Then the equation is y = 2/5 x + 1.

More about the linear system link is given below.

brainly.com/question/20379472

#SPJ1

7 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

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