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4 years ago

Solve for x X/3-6.2=9

Mathematics

1 answer:

Nadusha1986 [10]4 years ago

8 0

Answer:

x=45.6

Step-by-step explanation:

X/3-6.2=9

Add 6.2 to each side

x/3-6.2+6.2 = 9+6.2

x/3 = 15.2

Multiply both sides by 3

x/3*3 = 15.2 *3

x = 45.6

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3 years ago

Please help me solve the system of equations.

Iteru [2.4K]

Answer: (-2, 0) and (0, -2)

Step-by-step explanation:

This system is:

y + x = -2

y = (x + 1)^2 - 3

To solve this we first need to isolate one of the variables in one fo the equations, in the second equation we have already isolated the variable y, so we can just replace it in the first equation:

(x + 1)^2 - 3 + x = -2

Now we can solve this for x.

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x^2 + 2*x + 0 = 0

The solutions of this equation are given by the Bhaskara's formula, then the solutions are:

$x = \frac{-2 +- \sqrt{2^2 -4*1*0} }{2*1} = \frac{-2 +- 2}{2}$

The two solutions are:

x = (-2 - 2)/2 = -2

In this case, we replace this value of x in the first equation and get:

y - 2 = -2

y = -2 + 2 = 4

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The other solution for x is:

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3 years ago

Root 3 cosec140° - sec140°=4 prove that

Lorico [155]

Answer:

Step-by-step explanation:

We are to show that $\sqrt{3} cosec140^{0} - sec140^{0} = 4\\$

Proof:

From trigonometry identity;

$cosec \theta = \frac{1}{sin\theta} \\sec\theta = \frac{1}{cos\theta}$

$\sqrt{3} cosec140^{0} - sec140^{0} \\= \frac{\sqrt{3} }{sin140} - \frac{1}{cos140} \\= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\$

From trigonometry, 2sinAcosA = Sin2A

$= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\\\= \frac{\sqrt{3}cos140-sin140 }{sin280/2}\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{2sin280}\\\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{sin280}\\since sin420 = \sqrt{3}/2 \ and \ cos420 = 1/2 \\ then\\\frac{4(sin420cos140-cos420sin140) }{sin280}$