All categories

3 years ago

Instructions in photo plz help

Mathematics

1 answer:

Anna71 [15]3 years ago

7 0

YS = 2/3 × ( YV ) *.&.* SV = 1/3 × ( YV )

_____________________________

YS = 2/3 × ( YV )

4x + 14 = 2/3 × ( YV )

Multiply both sides by 3/2

( 4x + 14 ) × 3/2 = ( YV ) × 2/3 × 3/2

YV = 6x + 21 ( ☆ )

¥¥¥¥¥¥¥¥¥¥¥¥¥¥¥¥¥¥¥¥¥¥¥¥¥¥¥

SV = 1/3 × ( YV )

1 + 3x = 1/3 × ( YV )

Multiply both sides by 3

( 1 + 3x ) × 3 = ( YV ) × 1/3 × 3

YV = 9x + 3 ( ♡ )

_____________________________

YV = YV

( ☆ ) = ( ♡ )

6x + 21 = 9x + 3

Subtract both sides 3

6x + 21 - 3 = 9x + 3 - 3

6x + 18 = 9x

Subtract both sides 6x

6x - 6x + 18 = 9x - 6x

18 = 3x

Divide both sides by 3

18 ÷ 3 = 3x ÷ 3

x = 6

₩₩₩₩₩₩₩₩₩₩₩₩₩₩₩₩₩₩

YV = 6x + 21 = 9x + 3

YV = 6(6) + 21 = 9(6) + 3

YV = 57

You might be interested in

What is the range of the following list of ordered pairs?<br> (-3, 2), (1, 5), (-2, 3)

dsp73

R= {2,5,3} Domain is the x values and Range is the y values.

6 0

3 years ago

3√x -2/x^2<br><br>please show step by step of differentiation before combining the terms.

EastWind [94]

Answer:

Step-by-step explanation:

Before we differentiate, let us assign a variable to the function. Let y be equal to the function i.e let y = 3√x -2/x²

In differentiation if $y = ax^{n}$ , then $\frac{dy}{dx} = nax^{n-1}$ where n is a constant and dy/dx means we are differentiating the function y with respect to x.

Applying the formula o the question given;

$y= 3\sqrt{x} -2/x^2\\y = 3{x}^\frac{1}{2} - 2x^{-2} \\\\$

On differentiating the resulting function;

$\frac{dy}{dx} = \frac{1}{2}*3x^{\frac{1}{2}-1 } - (-2)x^{-2-1} \\\\\frac{dy}{dx} = \frac{1}{2}*3x^{-\frac{1}{2}} + 2x^{-3}\\ \\\frac{dy}{dx} = \frac{1}{2}*{\frac{3}{x^{\frac{1}{2} } }} + \frac{2}{x^{3} } \\\\\frac{dy}{dx} = {\frac{3}{2x^{\frac{1}{2} } }} + \frac{2}{x^{3} }\\\\\frac{dy}{dx} = {\frac{3}{2\sqrt{x} }} + \frac{2}{x^{3} }$

To combine the terms, we will add up by finding their LCM.

$\frac{dy}{dx} = {\frac{3}{2\sqrt{x} }} + \frac{2}{x^{3} }\\\frac{dy}{dx} = \frac{3x^3+4\sqrt{x} }{2x^{3} \sqrt{x}}$

3 0

4 years ago

Hello can someone help me with this factoring problem? thank you!

Alex

I think this is correct

8 0

2 years ago

(5+6b^3)^2 expand and combine like terms

zepelin [54]

Answer:

(5+6b^3)(5+6b^3)

25+30b^3+30b^3+36b^6

36b^6+60b^3+25

I hope this helps:

6 0

3 years ago

Can somebody help me pleasee????

Marina86 [1]

Answer:

Step-by-step explanation:

3x + 3y = -x + 5y

3x + x = 5y - 3y

4x = 2y

Divide both by 2

2x = y

(1,2) when x=1, y=2

2(1)=2 is a true statement

(2,4) when x=2, y=4

2(2)=4 is a true statement

8 0

4 years ago