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Ostrovityanka [42]

4 years ago

15

Solve for x & y: x^2+y^2=20. 3x+y=2

2 answers:

Norma-Jean [14]4 years ago

6 0

You're solving a system here, easy peasy. Solve the second equation for y to get y = 2 - 3x. Now in the first equation, sub in this y value where you see a y and now the equation is in terms of x only: x^2 + (2-3x)^2 = 20. You have to FOIL out the (2-3x)^2 first to get x^2 + 4 - 12x + 9x^2 = 20. Now combine like terms to get 10x^2 - 12x + 4 = 20. Move the 20 over by subtraction to get
10x^2 - 12x - 16 = 0. Factor this now, taking a 2 out first: 2(5x^2 - 6x -8)=0 and you get factors of 2(x-2)(5x+4)=0. x=2 or x = -4/5. If x = 2, then y = -4. If x = -4/5, then y = 22/5

Lerok [7]4 years ago

5 0

Answer:

1

Step-by-step explanation:

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3 years ago

What’s 10.546 divided by 52?

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Answer:

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2 years ago

Find the value of x which satisfies the following equation.<br> log2(x−1)+log2(x+5)=4

weqwewe [10]

$\quad \huge \quad \quad \boxed{ \tt \:Answer }$

$\qquad \tt \rightarrow \: x = 3$

____________________________________

$\large \tt Solution \: :$

$\qquad \tt \rightarrow \: log_{2}(x - 1) log_{2}(x + 5) = 4$

$\qquad \tt \rightarrow \: log_{2} \{(x - 1)(x + 5) \} = 4$

[ log (x) + log (y) = log (xy) ]

$\qquad \tt \rightarrow \: ( x - 1)(x + 5) = {2}^{4}$

$\qquad \tt \rightarrow \: {x}^{2} + 5x - x - 5 = 16$

$\qquad \tt \rightarrow \: {x}^{2} + 4x - 5 - 16 = 0$

$\qquad \tt \rightarrow \: {x}^{2} + 4x -21 = 0$

$\qquad \tt \rightarrow \: {x}^{2} + 7x - 3x - 21 = 0$

$\qquad \tt \rightarrow \: x(x + 7) - 3(x + 7) = 0$

$\qquad \tt \rightarrow \: (x + 7)(x - 3) = 0$

$\qquad \tt \rightarrow \: x = - 7 \: \: or \: \: x = 3$

The only possible value of x is 3, since we can't operate logarithm with a negative integer in it.

$\qquad \tt \rightarrow \: x = 3$

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

4 0

2 years ago

What is the math problem answer?

azamat

To find the mid point add both X coordinates together and divide by 2 and then do the same with the y coordinates.

X coordinate: 4 + -6 = 4-6 = -2 / 2 = -1

Y coordinate: -2 + 3 = 1 / 2 = 1/2

Midpoint: ( -1, 1/2)

3 0

3 years ago

Given that cos (x) = 1/3, find sin (90 - x)

ddd [48]

Answer:

$\sin(90^{\circ} - x)=\frac{1}{3}$

Step-by-step explanation:

Given: $\cos (x)=\frac{1}{3}$

We have to find the value of $\sin(90^{\circ} - x)$

Since Given $\cos (x)=\frac{1}{3}$

Using trigonometric identity,

$\sin(90^{\circ} - \theta)=\cos\theta$

Thus, for $\sin(90^{\circ} - x)$ comparing , we have,

$\theta=x$

We get,

$\sin(90^{\circ} - x)=\cos x=\frac{1}{3}$

Thus, $\sin(90^{\circ} - x)=\frac{1}{3}$

3 0

3 years ago

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