All categories

3 years ago

Can someone please help me

Mathematics

1 answer:

aniked [119]3 years ago

5 0

#1 √ 75

Answer : 5√3 or ≈ 8.66025

#2 √ 48

Answer : 4√3 or ≈ 6.9282

#3 √ 128

Answer : 8√2 or ≈ 11.31371

#4 √ 300

Answer : 10√3 or ≈ 17.32051

~~

#5 √⅔

Answer :

$\frac{ \sqrt{6} }{3} = 0.816497$

#6 √16/4

Answer : 2

#7 √ 6/8

Answer :

$\frac{ \sqrt{3} }{2} = 0.866025$

#8 √27/9

Answer : √ 2 or ≈ 1.73205

~~

I hope that helps you out!!

Any more questions, please feel free to ask me and I will gladly help you out!!

~Zoey

You might be interested in

What is the simplified form of the quantity of x plus 7, all over the quantity of x plus 3 + the quantity of x minus 4, all over

Hunter-Best [27]

Multiply all of those and minus and do what is said and of not serch for links ok

3 0

2 years ago

Four people have measured the height of a wall using different methods. Their results are shown in the table. Order their measur

slavikrds [6]

Answer:

5/2, 2.6, 1 + sqrt(3), sqrt(8)

Step-by-step explanation:

Use a calculator:

sqrt(8) = 2.83

5/2 = 2.5

2.6 = 2.6

1 + sqrt(3) = 2.73

Answer:

5/2, 2.6, 1 + sqrt(3), sqrt(8)

3 0

3 years ago

Find all zeros x^3+2x^2+4x+21

dlinn [17]

Answer:

see explanation

Step-by-step explanation:

note when x = - 3

(- 3)³ + 2(- 3)² + 4(- 3) + 21 = - 27 + 18 - 12 + 21 = 0

hence x = - 3 is a zero and (x + 3) is a factor and dividing gives

$\frac{x^3+2x^2+4x+21}{x+3}$ = (x + 3)(x² - x + 7)

For zeros equate to zero

(x + 3)(x² - x + 7) = 0

equate each factor to zero and solve for x

x + 3 = 0 ⇒ x = - 3

x² - x + 7 = 0 ← solve using quadratic formula

x = (1 ± $\sqrt{1-28}$ ) / 2 = (1 ± 3i $\sqrt{3}$ ) / 2

x = $\frac{1}{2}$ ± $\frac{3i\sqrt{3} }{2}$

zeros are x = - 3, x = $\frac{1}{2}$ ± $\frac{3i\sqrt{3} }{2}$

5 0

2 years ago

X + 23 = −4 need help

laila [671]

Answer:

-27

Step-by-step explanation:

x+23−23=−4−23

x=−27

6 0

2 years ago

Read 2 more answers

Find two unit vectors orthogonal to both given vectors. i j k, 4i k

Maurinko [17]

The cross product of two vectors gives a third vector $\mathbf v$ that is orthogonal to the first two.

$\mathbf v=(\vec i+\vec j+\vec k)\times(4\,\vec i+\vec k)=\begin{vmatrix}\vec i&\vec j&\vec k\\1&1&1\\4&0&1\end{vmatrix}=\vec i+3\,\vec j-4\,\vec k$

Normalize this vector by dividing it by its norm:

$\dfrac{\mathbf v}{\|\mathbf v\|}=\dfrac{\vec i+3\,\vec j-4\,\vec k}{\sqrt{1^2+3^2+(-4)^2}}=\dfrac1{\sqrt{26}}(\vec i+3\,\vec j-4\vec k)$

To get another vector orthogonal to the first two, you can just change the sign and use $-\mathbf v$ .

6 0

3 years ago