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Elena-2011 [213]
3 years ago
14

Write the equation of a line that is parallel to y=0.6x+3 and that passes through the point

Mathematics
1 answer:
a_sh-v [17]3 years ago
3 0
A line parallel to y = 0.6x + 3 will be of the form:

y = 0.6x + k

where k is a constant.

It passes through (-3,-5)

Thus,

y = 0.6x + k \\ \\ -5 = 0.6 \times (-3) + k \\ \\ -5 = -1.8 + k \\ \\k = -5 + 1.8 \\ \\ k = -3.2

\textbf{The equation is \: \: y = 0.6x - 3.2}
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Solve the system:<br> 5x-7y=-41<br> -3x-5y=-3
-Dominant- [34]

Answer:

x = -4 , y = 3

Step-by-step explanation:

5x - 7y = -41 ... (i)

-3x - 5y = -3 ... (ii)

Multiplying (i) by -3 and (ii) by 5 ;

-15x + 21y = 123 ... (i)

-15x - 25y = -15 ... (ii)

Subtracting (i) by (ii) ;

0 + 46y = 138

46y = 138

y = 138 ÷ 46 = 3

Returning to equation (ii) ;

-3x - 5(3) = -3

-3x = -3 + 15

-3x = 12

x = -4

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3 years ago
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The answer to this is 1.

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Practice order of operations 8+3×2-4
REY [17]

Answer:

10

Step-by-step explanation:

The acronym PEMDAS is helpful when learning order of operations. It stands for Parenthesis, Exponent, Multiplication, Division, Addition, and Subtraction which is the order in which you solve an equation. For this set of numbers there are no parenthesis or exponents so you would start by multiplying 3*2=6 Then add the 8 to the 6 and you get 14 Subtract the 4 and the answer is 10

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2 years ago
Teresa runs each lap in 5 minutes. She will run less than
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Answer:

11 laps i think

Step-by-step explanation:

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5 0
2 years ago
Find all solutions to
BARSIC [14]

Answer:

x= 0 , \frac{1}{14} , \frac{-1}{12}

Step-by-step explanation:

Given, equation is \sqrt[3]{15x-1} + \sqrt[3]{13x+1} = 4\sqrt[3]{x}. →→→ (1)

Now, by cubing the equation on both sides, we get

( \sqrt[3]{15x-1} + \sqrt[3]{13x+1} )³ = (4\sqrt[3]{x})³

⇒ (15x-1) + (13x+1) + 3×\sqrt[3]{15x-1}×\sqrt[3]{13x+1} (\sqrt[3]{15x-1} + \sqrt[3]{13x+1}) = 64 x.

⇒ 28x + 3×\sqrt[3]{15x-1}×\sqrt[3]{13x+1} (4\sqrt[3]{x}) = 64x.        

(since from (1),  \sqrt[3]{15x-1} + \sqrt[3]{13x+1} = 4\sqrt[3]{x})

⇒ 12× \sqrt[3]{15x-1}×\sqrt[3]{13x+1} (\sqrt[3]{x})= 36x.

⇒ 3x = \sqrt[3]{(15x-1)(13+1)(x)}.

Now, once again cubing on both sides, we get

(3x)³ = (\sqrt[3]{(15x-1)(13+1)(x)})³.

⇒ 27x³ = (15x-1)(13x+1)(x).

⇒ 27x³ = 195x³ + 2x² - x

⇒ 168x³ + 2x² - x = 0

⇒ x(168x² + 2x -1) = 0

⇒ by, solving the equation we get ,

x = 0 ; x = \frac{1}{14} ; x = \frac{-1}{12}

therefore, solution is x= 0 , \frac{1}{14} , \frac{-1}{12}

7 0
3 years ago
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