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3 years ago

Show me the steps to solve and the solution y-2>1/2(x-2)

1 answer:

4 0

<span>y-2>1/2(x-2) can be expanded as follows: y - 2 > (1/2)x - 1

Mult. all terms by 2 to remove the fractions:

2y - 4 > x - 2

Add 4 to both sides: 2y > x + 2

Div. both sides by 2: y > (1/2)x + 1 (answer)</span>

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10 minus three times a number is no more than 61

Elena L [17]

Answer:

n ≥ -17

Step-by-step explanation:

Writing a symbolic inequality, we get:

10 - 3n ≤ 61

Solve this for 3n by adding 3n to both sides of this equation:

10 ≤ 61 + 3n

Solve for n by subtracting 61 from both sides and then dividing all of the resulting terms by 3:

-51 ≤ 3n (divide both sides by 3):

-17 ≤ n, or

n ≥ -17

8 0

2 years ago

An equivalent expression for 12x+1 and 2(2x+5) please help

zhannawk [14.2K]

<h2>Answer:</h2>

8x+9

<h2>Step-by-step explanation:</h2>

12x+1 = 2(2x+5)

<h3><em><u>Handle the brackets first</u></em></h3>

12x+1 = 4x+10

<h3><em><u>Collect like terms</u></em></h3>

=8x+9

3 0

3 years ago

Find the argument of the complex number z=1+iv3

elena-14-01-66 [18.8K]

Given:

The complex number is:

$z=1+i\sqrt{3}$

To find:

The argument of the given complex number.

Solution:

If a complex number is $z=x+iy$ , then the argument of the complex number is:

$\theta=\tan^{-1}\dfrac{y}{x}$

We have,

$z=1+i\sqrt{3}$

Here, $x=1$ and $y=\sqrt{3}$ . So, the argument of the given complex number is:

$\theta =\tan^{-1}\dfrac{\sqrt{3}}{1}$

$\theta =\tan^{-1}\sqrt{3}$

$\theta =\tan^{-1}\left(\tan \dfrac{\pi}{3}\right)$

$\theta =\dfrac{\pi}{3}$

Therefore, the argument of the given complex number is $\theta =\dfrac{\pi}{3}$ .

6 0

2 years ago

I need to find the area in the shape

dalvyx [7]

Answer:

32 yd

Step-by-step explanation:

The area of a triangle is calculated by multiplying each of the sides lengths together, 8x8, and dividing it by two which equals 32.

6 0

2 years ago

If g(x) = 3^(2x) then g'(2) is?<br><br>A. 81<br>B. 162ln3<br>C. 81/ln3<br>D. 6/ln3<br>E. 6lin3

Alina [70]

$\bf a^{u(x)}\qquad \qquad \cfrac{df}{dx}=\cfrac{df}{du}\cdot \cfrac{du}{dx}\\\\
-----------------------------\\\\
g(x)=3^{2x}\qquad \qquad u=2x
\\\\\\
\cfrac{dg}{dx}=3^uln(3)\cdot \cfrac{du}{dx}\implies \cfrac{dg}{dx}=3^{2x}ln(3)\cdot 2
\\\\\\
g'(2)=3^{2\cdot 2}ln(3)\cdot 2\implies g'(2)=2\cdot 81ln(3)\implies g'(2)=162ln(3)$

5 0

3 years ago