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otez555 [7]
2 years ago
11

2/3y + k = j. Solve for y

Mathematics
1 answer:
iVinArrow [24]2 years ago
7 0

The value of y from the expression is y = 3/2(j-k)

<h3>Subject of formula</h3>

Given the expression below

2/3y + k = j.

Subtract k from both sides

2/3y = j - k

Divide both sides by 2/3

(2/3y)/2/3 =  (j-k)/(2/3)

y = 3/2(j-k)

Hence the value of y from the expression is y = 3/2(j-k)

Learn more on subject of formula here; brainly.com/question/657646

#SPJ1

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The constraints of a problem are listed below. What are the vertices of the feasible region?<br>​
Mamont248 [21]

Answer:

Option 4 : (0.\frac{3}{2} ) \ , \ (0,2) \ , \ (6,0) \ , \ (\frac{9}{4} ,0)

Step-by-step explanation:

<u>See the attached figure:</u>

To find the vertices of the feasible region, we need to graph the constraints, then find the area included by them, then calculate the vertices which is the intersection between each two of them.

As shown, the shaded area represents the solution of the constraints

So, the vertices of the feasible region are:

(0.\frac{3}{2} ) \ , \ (0,2) \ , \ (6,0) \ , \ (\frac{9}{4} ,0)

8 0
3 years ago
Read 2 more answers
Given \qquad m \angle LONm∠LONm, angle, L, O, N is a straight angle. \qquad m \angle MON = 8x - 13^\circm∠MON=8x−13 ∘ m, angle,
Tamiku [17]

Answer:

<h3>\boxed{99°}</h3><h3 />

Step-by-step explanation:

m<MON = 8x - 13°

m<LOM = 7x - 17°

To find : m <MON

First, we have to find the value of x :

Create an equation

\mathrm{8x - 13 + 7x - 17 = 180} ( sum of angle in straight line )

Collect like terms

\mathrm{15x - 13 - 17 = 180}

Calculate

\mathrm{15x - 30 = 180}

Move constant to R.H.S and change its sign

\mathrm{15x = 180 + 30}

Calculate the sum

\mathrm{15x = 210}

Divide both sides of the equation by 15

\mathrm{ \frac{15x}{15}  =  \frac{210}{15} }

Calculate

\mathrm{x = 14}

Now, let's find the value of m<MON

\mathrm{8x - 13}

Plug the value of x

\mathrm{ = 8 \times 14 - 13}

Calculate the product

\mathrm{ = 112 - 13}

Calculate the difference

\mathrm{ = 99} °

Hope I helped!

Best regards!

6 0
3 years ago
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