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Anika [276]
4 years ago
10

Literal equations 2x-3y=8 solve for y

Mathematics
2 answers:
ludmilkaskok [199]4 years ago
7 0
2x - 3y = 8
- 3y = 8 - 2x
y = ( 8 - 2x ) / -3

 or

2x - 3y = 8 
2x = 8 + 3y
2x - 8 = 3y
(2x - 8) / 3 = y


vichka [17]4 years ago
4 0
2x-3y = 8

Add 3y on both sides

2x = 8 + 3y

Subtract 8 on both sides

3y = 2x - 8

Divide 3 on both sides

y= \frac{2}{3}x- \frac{8}{3}
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The price of a shirt was marked up from $24 to $30What is the percent of increase?
deff fn [24]

Answer:

25% increase

Step-by-step explanation:

First take the new price and subtract the original price

30-24 = 6

Divide by the original price

6/24 = 1/4 = .25

Change to percent form

25%

The price went up so it is an increase

25% increase

8 0
3 years ago
Read 2 more answers
HELP WITH ALL <br> Determine the composite functions of the following set of functions
elixir [45]

QUESTION 1

i) Given f(x)=5x-2 and g(x)=x^2.

(f\circ g)=f(g(x))

(f\circ g)=f(x^2)

We plug in x^2 into f(x)=5x-2

(f\circ g)=5x^2-2

ii) Given f(x)=5x-2 and g(x)=x^2.

(g\circ f)=g(f(x))

(g\circ f)=g(5x-2)

We plug in 5x-2 into g(x)=x^2

(g\circ f)=(5x-2)^2

(g\circ f)=25x^2-10x-10x+4

(g\circ f)=25x^2-20x+4

QUESTION 2

i) Given g(x)=x^2-4 and h(x)=\sqrt{2x+1}.

(h\circ g)=h(g(x))

(h\circ g)=h(x^2-4)

We plug in x^2-4 into h(x)=\sqrt{2x+1}

(h\circ g)=\sqrt{2(x^2-4)+1}

(h\circ g)=\sqrt{2x^2-8+1}

(h\circ g)=\sqrt{2x^2-7}

ii) Given g(x)=x^2-4 and h(x)=\sqrt{2x+1}.

(g\circ h)=g(h(x))

(g\circ h)=g(\sqrt{2x+1})

We plug in \sqrt{2x+1 into g(x)=x^2-4

(g\circ h)=(\sqrt{2x+1})^2-4

(g\circ h)=2x+1-4

(g\circ h)=2x-3

QUESTION 3

i) Given g(x)=x+5 and h(x)=\sqrt{x-4}.

(g\circ h)=g(h(x))

(g\circ h)=g(\sqrt{x-4})

We plug in \sqrt{x-4} into g(x)=x+5

(g\circ h)=\sqrt{(x-4)+5}

ii) Given g(x)=x+5 and h(x)=\sqrt{x-4}.

(g\circ g)=g(g(x))

(g\circ g)=g(x+5)

We plug in x+5 into g(x)=x+5

(g\circ g)=(x+5)+5}

(g\circ g)=x+10}

QUESTION 4

The given functions are;

f(x)=-2x+1 and g(x)=-x^2  

f\circ f=f(f(x))

f\circ f=f(-2x+1)

f\circ f=-2(-2x+1)+1)

f\circ f=4x-2+1)

f\circ f=4x-1)

ii)  

g\circ g=g(g(x))

g\circ g=g(-x^2)

g\circ g=-(x^2)^2

g\circ g=-x^4

QUESTION 5

f(x)=5x-2;g(x)=x^2;h(x)=-3x

h\circ g\circ f=h\circ (g(f(x))

h\circ g\circ f=h\circ (g(5x-2))

h\circ g\circ f=h\circ ((5x-2)^2)

h\circ g\circ f=h\circ (25x^2-20x+4)

h\circ g\circ f=h(25x^2-20x+4)

h\circ g\circ f=-3(25x^2-20x+4)

h\circ g\circ f=-75x^2+60x-12)

ii.

f\circ f\circ g=f\circ (f(g(x))

f\circ f\circ g=f\circ (f(x^2)

f\circ f\circ g=f\circ (5x^2-2)

f\circ f\circ g=f(5x^2-2)

f\circ f\circ g=5(5x^2-2)-2

f\circ f\circ g=25x^2-10-2

f\circ f\circ g=25x^2-12

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3 years ago
What is the relationship between angle 1 and angle 5​
Sauron [17]

Answer:

○ b Corresponding Angles

Step-by-step explanation:

These two angles are a mirror, so they <em>correspond</em>.

I am joyous to assist you anytime.

4 0
4 years ago
220x + 260y &lt; 4,500 solve
Allisa [31]

Answer:

255-13y

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11

Step-by-step explanation:

4 0
3 years ago
A driver is 15 feet below the surface of the water 10 seconds after he entered the water and 90 feet below the surface after 40
spin [16.1K]

Answer:

\frac{-75ft}{30sec} or simplified would be \frac{-2.5ft}{sec}

Step-by-step explanation:

In order to calculate the rate of change, we must subtract his end position and end time by the starting position and starting time. We are using the -15 feet since it is below the surface area as the starting position and the 10 seconds as the starting time. We make each of these into a fraction and subtract like so...

\frac{-90ft - (-15ft)}{40sec - 10sec} = rate of change

\frac{-75ft}{30sec} = rate of change

\frac{-2.5ft}{sec} = rate of change

Therefore, we can see that the rate of change of the diver is \frac{-75ft}{30sec} or simplified would be \frac{-2.5ft}{sec}

6 0
3 years ago
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