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bogdanovich [222]
3 years ago
14

Y=3xy-1, Solve for y

Mathematics
1 answer:
Ket [755]3 years ago
3 0

Answer:

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

Step-by-step explanation:

Given

y = 3xy - 1

Collect terms in y on the left side by subtracting 3xy from both sides

y - 3xy = - 1 ← factor out y on the left side

y(1 - 3x) = - 1 ( divide both sides by 1 - 3x

y = \frac{-1}{1-3x} ← multiply numerator/denominator by - 1

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

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In the diagram below, P and Q are points on a circle with centre O.
Gnoma [55]

Answer:

<h2>∠PQT = 72°</h2>

Step-by-step explanation:

According to the diagram shown, ∠OPQ = ∠OQP = 18°. If PQT is a tangent to the circle, it can be inferred that line OQ is perpendicular to line QT. Ths shows that ∠OQT = 90°.

Also from the diagram, ∠OQP + ∠PQT = ∠OQT;

∠PQT  = ∠OQT -  ∠OQP

Given ∠OQP = 18° and ∠OQT = 90°

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4 years ago
Find the missing angle measures.<br> What is m∠2?<br> a 40<br> b 60<br> c 50<br> d 100
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What is the domain and range of the function y = 2x2 - 4x - 10?
11111nata11111 [884]

Answer:

Step-by-step explanation:

The domain of all polynomials is all real numbers.  To find the range, let's solve that quadratic for its vertex.  We will do this by completing the square.  To begin, set the quadratic equal to 0 and then move the -10 over by addition. The first rule is that the leading coefficient has to be a 1; ours is a 2 so we factor it out.  That gives us:

2(x^2-2x)=10

The second rule is to take half the linear term, square it, and add it to both sides.  Our linear term is 2 (from the -2x).  Half of 2 is 1, and 1 squared is 1.  So we add 1 into the parenthesis on the left.  BUT we cannot ignore the 2 sitting out front of the parenthesis.  It is a multiplier.  That means that we didn't just add in a 1, we added in a 2 * 1 = 2.  So we add 2 to the right as well, giving us now:

2(x^2-2x+1)=10+2

The reason we complete the square (other than as a means of factoring) is to get a quadratic into vertex form.  Completing the square gives us a perfect square binomial on the left.

x^2-2x+1=(x-1)^2 and on the right we will just add 10 and 2:

2(x-1)^2=12

Now we move the 12 back over by subtracting and set the quadratic back to equal y:

2(x-1)^2-12=y

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4 0
3 years ago
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