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stepladder [879]
3 years ago
13

Solve for y: x(y-3)=2y+1

Mathematics
2 answers:
polet [3.4K]3 years ago
7 0

Answer: y=(3x+1)/(x-2)

Step-by-step explanation: Step by step with principles included in the image.

lana [24]3 years ago
6 0

Answer:

y =  \frac{3x + 1}{x - 2}

Step-by-step explanation:

Add -2y to both sides:

xy−3x+<u>−2y</u>=2y+1+<u>−2y</u>

Simplify:

xy−3x−2y=1

Add 3x to both sides:

xy−3x−2y+<u>3x</u>=1+<u>3x</u>

Simplify:

xy−2y=3x+1

Factor out variable y:

y(x−2)=3x+1

Divide both sides by x-2:

\frac{y(x - 2)}{x - 2}  =  \frac{3x + 1}{x - 2}

Simplify:

y =  \frac{3x + 1}{x - 2}

Hope this helps :)

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Nadia bought 4.3 pounds of ham on Tuesday and 1.7 pounds of ham on Wednesday. She spent a total of $41.94. What was the price of
zhuklara [117]

Answer:

Step-by-step explanation:

41.94/6=6.99

6 0
3 years ago
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Simplify the expression to a polynomial in standard form: (3x+1)(3x^2 −6x−8)
nikklg [1K]

Step-by-step explanation:

3x^3-18x^2-24x+3x^2-6x-8

3x^3-15x^2-30x-8

is the required answer for your question

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3 years ago
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marshall27 [118]
I think it’s C , good luck .
3 0
3 years ago
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5^3+5-6(4)<br><br> HELP FAST I’M IN CLASS
vodomira [7]

To solve: 5^3+5-6(4)

5^3+5-6(4) < first step, 5 to the third power is 125

125 + 5 - 6(4) < next step, 6(4) or 6 x 4 is 24

125 + 5 - 24 < third step, add/subtract based off what comes first.

130 - 24 < final step. 130 - 24 = 106

106 is your final answer.

5 0
3 years ago
For p = $3000, r = 3.5%,and t = 5 years, Find the balance in an account When interest is compounded (a) quarterly, (b) monthly,a
MAVERICK [17]

Answer:

a) $3,571.02

b) $3,572.9

c) $3,573.74

Step-by-step explanation:

Data provided in the question:

p = $3000,

r = 3.5%,

t = 5 years

a) quarterly

number of periods in a year, n = 4

Interest rate per period = 3.5% ÷ 4 = 0.875%

Now,

A = p\times \left( 1 + \frac{r}{n} \right)^{\Large{n \cdot t}}

A = total amount

n = number of times compounded per year

on substituting the respective values, we get

A = 3000 × \left( 1 + \frac{ 0.035 }{ 4 } \right)^{\Large{ 4 \cdot 5 }}

A = 3000 × [/tex]\cdot { 1.00875 } ^ { 20 }[/tex]

A = 3000 × 1.19034

A = $3,571.02

b) monthly

number of periods in a year, n = 12

Now,

A = p\times \left( 1 + \frac{r}{n} \right)^{\Large{n \cdot t}}

on substituting the respective values, we get

A = 3000 × \left( 1 + \frac{ 0.035 }{ 12 } \right)^{\Large{ 12 \cdot 5 }}

A = 3000 × [/tex]\cdot { 1.002917} ^ { 60 }[/tex]

A = 3000 × 1.190967

A = $3,572.9

c) continuously

A = pe^{r\times t}

on substituting the respective values, we get

A = 3,000 × e^{0.035\times 5}

or

A = 3,000 × e^{0.175}

or

A = 3,000 × 1.1912

or

A = $3,573.74

5 0
3 years ago
Read 2 more answers
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