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lesya [120]
2 years ago
8

4x^2+2xy-10y^2+9=0 what is the value of a,b,c and the discriminant?

Mathematics
1 answer:
zavuch27 [327]2 years ago
6 0

basically you treat y like a number and not a variable

Answer:

a is 4

b is 2y

c is -10y²+9

discriminant is 4(41y²-36)

Step-by-step explanation:

4x²+2xy-10y²+9=0

rewrite in standard form of a quadratic equation like ax² + bx + c = 0

4x²+2yx-10y²+9=0

basically you treat y like a number and not a variable

a is the number with the x²

right away we know a is 4 because of 4x²

b is the one with x so in this formula b is 2y

c is the number without the x which in this case is -10y²+9

discriminant is

b² - 4ac

(2y)²- (4)(4)(-10y²+9)

4y²-(16)(-10y²+9)

4y²-(16)(-10y²+9)

4y²+160y²-144

164y²-144 =

4(41y²-36)

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3 years ago
Two lighthouses are located 75 miles from one another on a north-south line. If a boat is spotted S 40o E from the northern ligh
yuradex [85]

Answer:

The northern lighthouse is approximately 24.4\; \rm mi closer to the boat than the southern lighthouse.

Step-by-step explanation:

Refer to the diagram attached. Denote the northern lighthouse as \rm N, the southern lighthouse as \rm S, and the boat as \rm B. These three points would form a triangle.

It is given that two of the angles of this triangle measure 40^{\circ} (northern lighthouse, \angle {\rm N}) and 21^{\circ} (southern lighthouse \angle {\rm S}), respectively. The three angles of any triangle add up to 180^{\circ}. Therefore, the third angle of this triangle would measure 180^{\circ} - (40^{\circ} + 21^{\circ}) = 119^{\circ} (boat \angle {\rm B}.)

It is also given that the length between the two lighthouses (length of \rm NS) is 75\; \rm mi.

By the law of sine, the length of a side in a given triangle would be proportional to the angle opposite to that side. For example, in the triangle in this question, \angle {\rm B} is opposite to side \rm NS, whereas \angle {\rm S} is opposite to side {\rm NB}. Therefore:

\begin{aligned} \frac{\text{length of NS}}{\sin(\angle {\rm B})} = \frac{\text{length of NB}}{\sin(\angle {\rm S})} \end{aligned}.

Substitute in the known measurements:

\begin{aligned} \frac{75\; \rm mi}{\sin(119^{\circ})} = \frac{\text{length of NB}}{\sin(21^{\circ})} \end{aligned}.

Rearrange and solve for the length of \rm NB:

\begin{aligned} & \text{length of NB} \\ =\; & (75\; \rm mi) \times \frac{\sin(21^{\circ})}{\sin(119^{\circ})} \\ \approx\; & 30.73\; \rm mi\end{aligned}.

(Round to at least one more decimal places than the values in the choices.)

Likewise, with \angle {\rm N} is opposite to side {\rm SB}, the following would also hold:

\begin{aligned} \frac{\text{length of NS}}{\sin(\angle {\rm B})} = \frac{\text{length of SB}}{\sin(\angle {\rm N})} \end{aligned}.

\begin{aligned} \frac{75\; \rm mi}{\sin(119^{\circ})} = \frac{\text{length of SB}}{\sin(40^{\circ})} \end{aligned}.

\begin{aligned} & \text{length of SB} \\ =\; & (75\; \rm mi) \times \frac{\sin(40^{\circ})}{\sin(119^{\circ})} \\ \approx\; & 55.12\; \rm mi\end{aligned}.

In other words, the distance between the northern lighthouse and the boat is approximately 30.73\; \rm mi, whereas the distance between the southern lighthouse and the boat is approximately 55.12\; \rm mi. Hence the conclusion.

4 0
2 years ago
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Solve

d = \sqrt {\left( {-4} \right)^2 + \left( {-3} \right)^2

\sqrt{16 + 9}

\sqrt{25} = 5\\\\ d=5

So, the distance between the points given is 5

4 0
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