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3 years ago

Solve for Y

Mathematics

1 answer:

Lapatulllka [165]3 years ago

7 0

$\huge{\boxed{\sf{y= 12 - 6x}}}$

Step by step explanation:

Step 1:

$2x + \dfrac{y}{3} = 4$

Step 2:

$\implies \dfrac{2x \times 3 + y}{3} = 4$

Let take LCM

Step 3:

$\implies \dfrac{6x + y}{3} = 4$

Step 4:

$\implies6x + y = 4 \times 3$

Step 5:

$\implies6x + y = 12$

Step 6:

$\blue{\implies y= 12 - 6x✓}$

<h3>So option d is correct</h3>

thank you❤️

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Write the equation of a line perpendicular to 2x + 3y = 4 and passing through (-3,-5).

Tasya [4]

Answer:

Step-by-step explanation:

A line perpendicular to the given line has a slope that is the negative inverse of the reference line.

Rewrite the given equation in the format of y=mx+b, where mi is the slope and b is the y-intercept (the value of y when x = 0.

2x + 3y = 4

3y=-2x+4

y = -(2/3)X + (4/3)

The reference slope is -(2/3). The negative inverse is (3/2), which will be the slope of a perpendicular line. We can write the new line as:

y = (3/2)x + b

Any value of b will still result in a line that is perpendicular. But we want a value of b that will shift the line so that it intersects the point (-3,-5). Simply enter this point in the above equation and solve for b.

y = (3/2)x + b

-5 = (3/2)(-3) + b

-5 = -(9/2) + b

-5 = -4.5 + b

b = - 0.5

The equation of the line that is perpendicular to 2x + 3y = 4 and includes point (-3,-5) is

y = (3/2)x - 0.5

6 0

1 year ago

I need help I can't figure out how to do this. I have to find the value of v. 4/5(v−7)=2

Nina [5.8K]

9.5 because 9.5-7 is 2.5. 2.5 times 4/5 is 10/5 which is equivalent to 2. 4/5(2.5-7)=2

7 0

3 years ago

Find the indicated limit, if it exists.

kondor19780726 [428]

Answer:

d) The limit does not exist

General Formulas and Concepts:

Calculus

Limits

Right-Side Limit: $\displaystyle \lim_{x \to c^+} f(x)$
Left-Side Limit: $\displaystyle \lim_{x \to c^-} f(x)$

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

Limit Property [Addition/Subtraction]: $\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)$

Step-by-step explanation:

*Note:

In order for a limit to exist, the right-side and left-side limits must equal each other.

Step 1: Define

Identify

$\displaystyle f(x) = \left\{\begin{array}{ccc}5 - x,\ x < 5\\8,\ x = 5\\x + 3,\ x > 5\end{array}$

Step 2: Find Right-Side Limit

Substitute in function [Limit]: $\displaystyle \lim_{x \to 5^+} 5 - x$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \lim_{x \to 5^+} 5 - x = 5 - 5 = 0$

Step 3: Find Left-Side Limit

Substitute in function [Limit]: $\displaystyle \lim_{x \to 5^-} x + 3$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \lim_{x \to 5^+} x + 3 = 5 + 3 = 8$